An isosceles triangle,on Trigonometry, I.S.I Entrance 2016, Solution to Subjective problem no. 6

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. If \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\), then prove that the triangle is isosceles.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4" link_option_url="https://cheenta.com/isicmientrance/" link_option_url_new_window="on"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 6 from 2016 
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]Trigonometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]7 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"] Plane Trigonometry by S.L. Loney

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. Given \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\)    

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

\(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})\) \(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\)  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

\(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\) \(\Rightarrow  b\sin A\cos B=a\cos A \sin B\) \(\Rightarrow  (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \)

 

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

\(\Rightarrow  (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}\)  [since \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)] \(\Rightarrow \frac{\cos B}{\cos A}=1\) \(\cos B=\cos A\) \(A=B\).

\(\Rightarrow \Delta ABC\) is isosceles  (Proved).

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="9" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Four Points on a Circle, ISI Entrance 2017, Subjective Problem no 2

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25" link_option_url="https://drive.google.com/file/d/11l85putXeAnmPGa2zOudC7D8sqi7m4zW/view?usp=drivesdk"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Consider a circle of radius 6 as given in the diagram below. Let \(B,C,D\) and \(E\) be points on the circle such that \(BD\) and \(CE\), when extended, intersect at \(A\). If \(AD\) and \(AE\) have length 5 and 4 respectively, and \(DBC\) is a right angle, then show that the length  of \(BC\) is \(\frac{12+\sqrt{15}}{5}\). Diagram : click here. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" link_option_url="https://cheenta.com/isicmientrance/" link_option_url_new_window="on"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.3.4" hover_enabled="0"]
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Plane Geometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]6 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]

'An excursion in Mathematics' published by Bhaskaracharya Pratishthana [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.7" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Given, \(AD=5, AE=4\) and \(\angle DBC=90^\circ \). As \(D,B,C\) are points on the circle having radius 6. Therefore \(DC\) is the diameter of the circle \(\Rightarrow DC=6×2=12\).

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

Now  DC is diameter of the circle \(\Rightarrow \angle DEC=90^\circ \). Therefore \(\angle DEA\) is also right angle.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

The length of \(DE=\sqrt{5^2-4^2}=3\).  And the length of \(EC=\sqrt{12^2-3^2}=3\sqrt{15}\). Therefore \(AC=AE+EC=4+3\sqrt{15}\).

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

From \(∆AED\) and \(∆ABC\) we have, \(\frac{AD}{DE}=\frac{AC}{BC}  \Rightarrow  BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}\). Therefore the length of \(BC\) is \(\frac{12+9\sqrt{15}}{5}\).   (Ans.)

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="59px||48px|||" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Watch the video ( Coming Soon ... )

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="9" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Product of Digits, ISI Entrance 2017

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="4.3.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]  Let \(g : \mathbb{N} \to \mathbb{N} \) with \( g(n) \) being the product of digits of \(n\).        (a) Prove that \( g(n)\le n\) for all \( n \in \mathbb{N} \) .        (b) Find all \(n \in \mathbb{N} \) , for which \( n^2-12n+36=g(n) \). [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" link_option_url="https://cheenta.com/isicmientrance/" link_option_url_new_window="on"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4" hover_enabled="0"]
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 5 from 2017
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Inequality

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]9 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]

Challenge and Thrill of Pre-College Mathematics by V.Krishnamuthy , C.R.Pranesachar, ect. [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" custom_padding="|||12px||"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Let \(n\) be a \(k\) digit(s) number number , then \(n\) can be written as                             \(n=a_0+10a_1+10^2a_2+\cdots+10^{k-1}a_{k-1}\) Where ,\(a_0,a_1,...,a_{k-1}\) are digits of \(n\).

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

\(a_0,a_1,...,a_{k-1} \in [1,9] \) as the range of the function \(g\) is \(\mathbb{N}\) \(\Rightarrow a_0,a_1,...,a_{k-1}\neq 0\) . Now \(g(n)=a_0a_1a_2\cdots a_{k-1}\le \underbrace{10\cdot10\cdot10\cdots 10}_{(k-1) times}  \cdot a_{k-1}\)   [Since \(a_0,a_1,...,a_{k-1} \le 10 \)]  Equality holds when \(k=1\) .

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

  \(\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}\) \(\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}+\cdots+10^2a_2+10a_1+a_0\)    [Since \(a_0,a_1,...,a_{k-1} >0\)]  \(\Rightarrow g(n)\le n\)        (Proved) .  

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

\(n^2-12n+36=g(n)\) \(\Rightarrow n^2-12n+36 \le n\)     [Since \(g(n) \le n\) ] \(\Rightarrow n^2-13n+36 \le 0\) \(\Rightarrow (n-9)(n-4) \le 0\) \(\Rightarrow 4\le n\le 9\) \(\Rightarrow n={4,5,6,7,8,9}\) (Ans.)  . 

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problems

[/et_pb_text][et_pb_post_slider include_categories="9" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Three Primes, ISI Subjective Entrance 2017

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let \(p_1,p_2,p_3\)  be primes with \(p_2\neq p_3\), such that \(4+p_1p_2\) and \(4+p_1p_3\) are perfect squares. Find all possible values of \(p_1,p_2,p_3\).   

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.7" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" min_height="204px" custom_margin="||119px|||" custom_padding="|||18px||"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Let \(4+p_1p_2=m^2\) and \(4+p_1p_3=n^2\), where \( m,n \in \mathbb{N}\). \(\Rightarrow p_1p_2=(m-2)(m+2)\) and \(p_1p_3=(n+2)(n-2)\). Since \(p_1,p_2,p_3\) are primes with \(p_2\neq p_3\) \(\Rightarrow m\neq n\).

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

 Case -I: \(p_2<p_3   \Rightarrow m<n\). Clearly, \(p_1=m+2=n-2,    \Rightarrow n=m+4\)              \(p_2=m-2\)    and                \(p_3=n+2=m+6\). Therefore, \((m+2),(m-2),(m+6)\) are all prime numbers.               We see that \(m=5\) satisfy the above condition. And then,\( p_1=7,p_2=3,p_3=11\) .    

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.7"]

Case-II: \(p_2>p_3    \Rightarrow   m>n\). \(\Rightarrow   p_1=m-2=n+2     \Rightarrow     m=n+4\)                        \( p_2=m+2=n+6\)   and                          \(p_3=n-2\). Now \((n+2),(n+6), (n-2)\) all are primes. Again , \(n=5\) satisfy this condition. Hence \(p_1=7,p_2=11,p_3=3\).  

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Thus all possible values of \(p_1,p_2,p_3\) are ( 7,3,11)  and (7,11,3). Now need to conclude that there does not exist any more triple of prime numbers satisfying the given condition. Consider these numbers:            \(p_1=m+2,p_2=m-2,p_3=m+6\) , now the gaps between \(p_1,p_2,p_3\) are given by:                 \(p_1-p_2=4,   p_3-p_2=8  \)  and  \(p_3-p_1=4\). We see that for \(m>9\) these three gaps cannot be 4,8 and 4 simultaneously . That is at least one of these three gaps is greater than 4 for \(m>9\) . And between 1 to 9 only \(m=5\) satisfy the given condition. Hence there does not exist any more triples.   

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4" hover_enabled="0"]

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2017. Subjective Problem no. 6.

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]

8.5 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]

Elementary Number Theory by David M. Burton [/et_pb_accordion_item][/et_pb_accordion][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

System of n equations of Real Analysis , I.S.I Entrance 2008, Solution to Subjective Problem No. 9.

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]

 For \(n\ge3 \), determine all real solutions of the system of \(n\) equations :

                                               \(x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\)

                                                                 ...............................

                      \(x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n-1}+x_n=\frac{1}{x_i}\)

                                                                 ................................

                                               \(x_2+x_3+\cdots+x_{n-1}+x_n=\frac{1}{x_1}\).

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" body_font="Raleway||||||||" toggle_font="||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"]

I.S.I. (Indian Statistical Institute) B.Math.(Hons.) Entrance Examination 2008. Subjective Problem no. 9.

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"]

System of \(n\) Equations

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" open="off" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"]

9 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" body_text_shadow_horizontal_length="0em" body_text_shadow_vertical_length="0em" body_text_shadow_blur_strength="0em"]

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_font="||||||||" tab_text_color="#ffffff" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

 Let \( s = x_1+x_2+\cdots+x_i+\cdots +x_{n-1}+x_n\).

Then the system of \(n\) equations are equivalent to \(x_i^2-sx_i+1=0\) for \(i=1,2,....,n\).

It follows that \(x_1,x_2,.....,x_{n-1},x_n\) are solutions to the quadratic equation: \(u^2-su+1=0\).                                                   \(\cdots\cdots\cdots\cdots (i)\)

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

Now we have two possible cases.

Case - I :  Two roots of the equation (i) are equal, then , all \(x_i\) are equal. 

i.e. , \(x_1=x_2=\cdots=x_{n-1}=x_n=u\).

\(\Rightarrow s=nu\).

Putting this value of \(s\) in equation (i) we get, \((n-1)u^2=1\).

\(\Rightarrow u=\frac{1}{\pm \sqrt{n-1}}\).

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

Case - II : Two roots of equation (i) are not equal. Then let these two roots are \(u_1 \) and \(u_2\). Also let among {\(x_1,x_2,......,x_{n-1},x_n\)}   \( k\)  of them equal to \(u_1\) and \((n-k) \) of them equal to \(u_2\), where \(0<k<n\).

      In this case, we have , (a) \(u_1+u_2=s\).    [Sum of roots of equation (i)]

                                              (b) \(u_1\cdot u_2=1\).    [ Product of roots of equation (i)].

Now,.   \(s=x_1+x_2+\cdots+x_{n-1}+x_n=k\cdot u_1+(n-k)\cdot u_2\)

        \(\Rightarrow s=(u_1+u_2)+(k-1)\cdot u_1+(n-k-1)\cdot u_2\) 

        \(\Rightarrow (k-1)\cdot u_1+(n-k-1)\cdot u_2=0\)   [using (a)]

        \(\Rightarrow  (k-1)\cdot u_1^2=(k+1-n)\cdot u_1\cdot u_2=k+1-n\le 0\) [using (b) and since ,\( k<n\Rightarrow k+1\le n\)].

      \(\Rightarrow u_1^2\le 0\)

But \(u_1\neq 0\) as the coefficient of \(u^0\) is 1(\(\neq 0\)) in equation (i).

        \(\Rightarrow u_1^2<0\)

\(\Rightarrow u_1\) is not real. So we have no real solution to the system of \(n\) equations in this case.

 

 

 

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Thus the total no. of real solutions to the system of \(n\) equations is two and these two solutions are:

                        \(x_1=x_2=\cdots=x_{n-1}=x_n=\frac{1}{\sqrt{n-1}}\)      and

                        \(x_1=x_2=\cdots=x_{n-1}=x_n=-\frac{1}{\sqrt{n-1}}\).(Ans.)

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px"]

Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Trigonometric Substitution, ISI Entrance 2019 - Problem 6

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] For all natural numbers\(n\), let          \(A_n=\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}\)           (\( n\) many radicals) (a) Show that for \(n\ge 2,  A_n=2\sin \frac{π}{2^{n+1}}\). (b) Hence, or otherwise, evaluate the limit                              \(\displaystyle\lim _{n\to \infty} 2^n A_n\) .

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4" hover_enabled="0"]
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 6.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Trigonometric Substitution 

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]7.5 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca. [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

For \(n\ge 2\), \(\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}\)       (\( n\) many radicals) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+0}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2+2\cos \frac{π}{2}}}}\).   

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

=\(\sqrt{2+\sqrt{2+\cdots +\sqrt{2(1+\cos \frac{π}{2})}}}\) =\(\sqrt{2+\sqrt{2+\cdots +\sqrt{4\cos^2 \frac{π}{2^2}}}}\) =\(\sqrt{2+\sqrt{2+\cdots +2\cos \frac{π}{2^2}}}\). (\(n-1\) many radicals)      ........    ...........     ..........      ...   .......  ...      .......     .........     ............    .....     ......... =\(2\cos \frac{π}{2^n}\)        \([n\ge 2]\).  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

Now \(A_n=\sqrt{2-2\cos \frac{π}{2^n}}\) \(\Rightarrow  A_n= \sqrt{2(1-\cos \frac{π}{2^n})}\) \(\Rightarrow  A_n= \sqrt{4\sin^2 \frac{π}{2^{n+1}}}\)    \(\Rightarrow  A_n= 2\sin \frac{π}{2^{n+1}}\). 

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Now \(\displaystyle\lim_{n\to \infty} 2^nA_n=\displaystyle\lim_{n\to \infty} 2^{n+1}\cdot \sin \frac{π}{2^{n+1}}\). Since , \(n \to \infty \) \(\Rightarrow 2^{n+1}\to \infty \) \(\Rightarrow \frac{π}{2^{n+1}}\to 0\) Let \(\frac{π}{2^{n+1}}=z \Rightarrow z \to 0\), when \(n \to \infty\). Therefore, \(\displaystyle\lim_{n\to \infty} 2^n A_n \)=\(\displaystyle\lim_{z \to 0}\frac{\sin z}{z}\cdot π =1\times π=π\). (Ans.)               

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Two Similar Triangles, ISI Entrance Subjective 2018

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Suppose that \(PQ\) and \(RS\) are two chords of a circle intersecting at a point \(O\) , It is given that \(PO=3\) cm and \( SO=4\) cm . Moreover, the area of the triangle \(POR\) is \(7 cm^2 \) . Find the are of the triangle \(QOS\) .  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.22.4"]
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 2.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.3.4" hover_enabled="0" open="on"]

Plane Geometry [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]

5.5 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.22.4" open="off"]


'Challenge and Thrill of Pre-College Mathematics' by V,Krishnamurthy, C.R.Pranesachar, ect.

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" custom_padding="|||34px||"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

\(PQ\) and \( RS\) are two chords of the circle \(C\) , intersecting at the point \(O\). See figure: click here.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

Given \(PO=3\) cm \(SO=4\) cm \([\triangle POR]= 7 cm^2\).  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

From the triangles \(POS\) and \(QOS\) we have,                 \(\angle POR=\angle SOQ\) [Opposite angles]                 \(\angle SRP=\angle SQO \) [Angle on the same semi-circle \(STP\)]                 \(\angle QSO= \angle OPR\) [Angle on the same semi-circle \(ST'P\)]  Therefore the \(\triangle POR\) and \(\triangle SOQ\) are similar triangles .  

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

\(\frac{[\triangle POR]}{OP^2}=\frac{[\triangle SOQ]}{SO^2}.\) \(\Rightarrow [\triangle SOQ]=\frac{SO^2}{PO^2}\cdot [\triangle POR]\)=\(\frac{4^2}{3^2}\cdot 7=12\frac{4}{9}\).(Ans.)

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Pythagorean Triple, ISI Subjective Entrance 2018, Problem 7

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Let \(a,b,c \in \mathbb{N}\) be such that
\(a^2+b^2=c^2\) and \(c-b=1\).
Prove that
(i) \(a\) is odd,
(ii) \(b\) is divisible by 4,
(iii) \(a^b+b^a\) is divisible by \(c\).  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.3.4" hover_enabled="0"]
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 7.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]8 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]

'Elementary Number Theory' by David M. Burton
'Challenge and Thrill of Pre-College Mathematics' by V,Krishnamurthy, C.R.Pranesachar, ect. [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

\(a^2+b^2=c^2=(b+1)^2=b^2+2b+1\) \(\Rightarrow a^2=2b+1\) Which implies \(a^2\) is odd integer. \(\Rightarrow a\) is also an odd integer =\(2k+1\) (say).

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

\((2k+1)^2+b^2=c^2\) \(\Rightarrow 4k^2+4k+1+b^2=b^2+2b+1\) \(\Rightarrow b=2k(k+1)\). Now \(k(k+1) \) is always even =\(2l\) (say). Therefore, \(b=4l\), i.e. \(b\) is divisible by 4.  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

\(a^b+b^a=a^{4l}+b^{2k+1}=(a^2)^{2l}+(c-1)^{2k+1}\) =\((2b+1)^{2l}+(c-1)^{2k+1}\) =\((2c-1)^{2l}+(c-1)^{2k+1}.\)

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Now \(2l\) is even , therefore , \((2c-1)^{2l} \) is of the form : \( 2cp+1\) where \(p \in \mathbb{N}\). And \(2k+1\) is odd , therefore \((c-1)^{2k+1} \) is of the form : \(cq-1\), where \(q \in \mathbb{N}\). Therefore \(a^b+b^a=(2cp+1)+(cq-1)=c\cdot (2p+q)\). \(\Rightarrow c|a^b+b^a\).

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

Powers of 2 – I.S.I. Entrance 2019 Subjective Problem 1

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Prove that the positive integers \(n\) that cannot be written as a sum of \(r\) consecutive positive integers, with \(r>1\) ,are of the form \(n=2^l\) for some \(l\ge 0\).  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.3.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.3.4"]

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 1.

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]

8.5 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" open="off"]

Elementary Number Theory by David M. Burton[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.3.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" custom_padding="|||25px||"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Claim: Any positive integer \(n\) can be written as \(n=2^k\cdot m\) , where \(k\ge0\) and \(m\) is an odd positive integer. To prove this claim use the fact :  \(n=2^k\cdot p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}\), where \(k\) and all \(k_i\) are non-negetive integers and all \(p_i\) are odd primes.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

The sum of (any) \(r\) consecutive positive integers is given by, \((q+1)+(q+2)+(q+3)+\cdots+(q+r)\) =   \(q\cdot r+(1+2+3+\cdots+r)\) =   \(q\cdot r+\frac{r(r+1)}{2}\).  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

Equating this sum to \(n\) we get,           \(2^k\cdot m = q\cdot r+\frac{r(r+1)}{2}\)   Or,     \(2^{k+1}\cdot m = 2q\cdot r+r(r+1)\)   Or,     \(2^l\cdot m = r(2q+r+1)\) , where \(l=k+1\ge1\).

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

In particular if we take \(n=2^l\) then \(m\) is equal to 1. Since both \(r\) and \((2q+r+1)\) are greater than 1, so they can't be equal to \(m\) in this case. Again one of \(r, (2q+r+1)\) is odd integer which implies the product \(r(2q+r+1)\) can't be equal to \(2^l\).  \(\Rightarrow 2^l \neq r(2q+r+1) \) \(\Rightarrow 2^l\) can't be expressed as the sum of \(r\) consecutive positive integers with \(r>1\) and \(l\ge 1\). Now, \(n=2^0=1 \) also can't be written in the same manner when \(l=0\). Therefore the positive integers \(n\) that cannot be written as a sum of \(r\) consecutive positive integers, with \(r>1\) , are of the form \(n=2^l\) for some \(l\ge 0\). To show that any number other than of the form \(2^l\) is sum of consecutive integers:
Observe that for \(n=2^l.m\), where \(m\) is an odd number greater than 1, there are two cases:
1. \( m < 2^l \) 
Select \(r\) and \(q\) such that \(r = m\) and \(2q+r+1 = 2^l \). 2. \( m > 2^l \)
Select \(r\) and \(q\) such that \( 2q+r+1 = m\) and \( r = 2^l \). (QED).

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Connected Program at Cheenta

[/et_pb_text][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/isicmientrance/" button_text="Learn More" button_alignment="center" _builder_version="3.22.4" custom_button="on" button_text_color="#ffffff" button_bg_color="#e02b20" button_border_color="#e02b20" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]