Number Theory, Ireland MO 2018, Problem 9

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]The sequence of positive integers $a_1, a_2, a_3, ...$ satisfies $a_{n+1} = a^2_{n} + 2018$ for $n \ge 1$ .
Prove that there exists at most one $n$ for which $a_n$ is the cube of an integer.

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Ireland MO 2018, Problem 9 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" hover_enabled="0" open="off"]Number Theory [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" hover_enabled="0" open="off"]8/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" hover_enabled="0" open="off"]Excursion in Mathematics by Bhaskaryacharya Prathisthan [/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"]

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" hover_enabled="0"][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="4.0" hover_enabled="0"], wIt is so important to know and use the modulo technqiue at the right time. We will use the modulo technique, i.e. we will see the problem through the lens of modulo some number. What is that number? If you visit this website, you will understand that to handle cubes modulo something is 9. So, we will deal the whole equation modulo 9.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0" hover_enabled="0"]

Definition: kth power residue of a number n is the complete residue system modulo n. For eg: Quadratic Residue (2nd power) of 4 is {0,1}.

Cubic(3rd) Power Residue of 9 is {0,1,-1}.
6th Power Residue of 9 is {0,1}
Quadratic(2nd Power) Residue of 9 is {0,1,4,7}

We will use these ideas here. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0" hover_enabled="0"]Let $a_k$ be the smallest integer which is a cube; let $a_k=a^3$ . Note that, $a_{k+1}=a^6+2018$ . Now, the modulo picture comes in. Starting from this cube. We will observe the sequence modulo 9. Case 1: $ a_k = 0$ mod 9 Then, the sequence modulo 9 will be $0 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9 are {0,1,-1}. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0" hover_enabled="0"]Case 2: $ a_k = 1,-1$ mod 9 Then, the sequence modulo 9 will be $\pm 1 \mapsto 3 \mapsto 2 \mapsto 6 \mapsto 2 \mapsto \dots$ Hence, there are no further cubes possible as the cubic residues of 9 are {0,1,-1}. 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" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" 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"]

Polynomial Functional Equation - Random Olympiad Problem

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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" _i="1" _address="0.0.0.1"]Find all the real Polynomials P(x) such that it satisfies the functional equation: $latex P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x $.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][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" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.29.2" _i="0" _address="0.1.0.0.0"]

Unknown [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.29.2" _i="1" _address="0.1.0.0.1" open="off"]Functional Equation, Polynomials[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.29.2" _i="2" _address="0.1.0.0.2" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.29.2" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics Challenges and Thrills in Pre College Mathematics[/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" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.29.2" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.29.2" _i="1" _address="0.1.0.2.1"]Well, it is really good that the information polynomial is given! You should use that. What is the first thing that you check in a Polynomial Identity? Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same $latex n^2 $. But did you observe something fishy? [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.29.2" _i="2" _address="0.1.0.2.2"]Now rewrite the equation as $latex P(2P(x)) - 2P(P(x)) = P(x)^{2}$. Do the Degree trick now... You see it right? Yes, on the left it is $latex n^2 $ and on the RHS it is $latex 2n $. So, there are two cases now... Figure them out!

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.29.2" _i="3" _address="0.1.0.2.3"]

Case 1: $latex 2n = n^2 $... i.e. P(x) is either a quadratic or a constant function. Case 2: $latex P(2P(x)) - 2P(P(x)) $ has coefficient zero till $latex x^2n$. We will study case 1 now. Case 1: $latex 2n = n^2$... i.e. P(x) is either a quadratic or a constant function. $latex P(2P(x)) - 2P(P(x)) = P(x)^{2}$ = $latex P(2y) - 2P(y) = y^{2}$ where $latex y = P(x) $. Now, expand using $latex P(x) = ax^2 + bx +c$, it gives $latex 2ay^2 -c = y^2 $... Now find out all such polynomials satisfying this property. For e.g. $latex \frac{x^2}{2}$ is a solution. If P(x) is constant, prove that $latex P(x) = 0 / \frac{-1}{2} $. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.29.2" _i="4" _address="0.1.0.2.4"]Case 2: $latex P(2y) - 2P(y) = y^{2}$. Assume a general form of P(x) = $latex $and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS. Now, we have already solved it for quadratic or less degree. [/et_pb_tab][et_pb_tab title="Techniques Revisited" _builder_version="3.29.2" _i="5" _address="0.1.0.2.5"]

Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.

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Find all polynomials $latex P(2P(x)) - 8P(P(x)) = P(x)^{2}$.
Find all polynomials $ P(cP(x)) - d P(P(x)) = P(x)^{2} $ depending on the values of c and d.

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Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" 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" _i="7" _address="0.1.0.7"][/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" _i="8" _address="0.1.0.8"]

SMO(senior)-2014 Problem 2 Number Theory

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]Find, with justification, all positive real numbers $a,b,c$ satisfying the system of equations: $a\sqrt{b}=a+c,b\sqrt{c}=b+a,c\sqrt{a}=c+b.$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.27.4" hover_enabled="0" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]SMO (senior)-2014 stage 2 problem 2

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="on"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27.4" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics

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Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27.4" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Given all three relations are cyclic and symmetric . So without loss of generality it can be assumed that $ a \geq b \geq c >0 $ . Then proceed . [ Note $ (0, 0, 0) $ can't be a solution since $ a , b , c $ are positive reals .] [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27.4" hover_enabled="0" _i="2" _address="0.1.0.2.2"]So $ a \sqrt b = a + c \Rightarrow a(\sqrt b - 1) = c \ [and \ we \ have \ a \geq c] \Rightarrow ( \sqrt b - 1 ) \leq 1 \Rightarrow b \leq 4 $[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27.4" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Similarly $ b \sqrt c = b + a \Rightarrow b(\sqrt c - 1) = a \ [and \ we \ have \ a \geq b ] \Rightarrow \sqrt c - 1 \geq 1 \Rightarrow c \geq 4 $[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27.4" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

Till now we have $ b \leq 4 \ and \ c \geq 4 $ , but we assumed that $ b \geq c $ . So it is clear that $ b = c =4 $ $ \Rightarrow a = 4 $ also. So the only triplet $ (a , b , c)$ is $ (4,4 ,4) $ . [/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" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" 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" _i="5" _address="0.1.0.5"][/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" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="6" _address="0.1.0.6"]

SMO (senior) -2014/problem-4 Number Theory

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]For each positive integer $n$ let $x_n=p_1+\cdots+p_n$ where $p_1,\ldots,p_n$ are the first $n$ primes. Prove that for each positive integer $n$ , there is an integer $k_n$ such that $x_n<k_n^2<x_{n+1}$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.27" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]SMO (senior)-2014 stage 2 problem 4

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Excursion in Mathematics

Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27.4" hover_enabled="0" _i="1" _address="0.1.0.2.1"]We have $ P_1 = 2 , P_=3 , P_3=5 , P_4 =7 , P_5 = 11 \ and \ so \ on .... $. Now to understand the expression $x_n<k_n^2<x_{n+1}$ , observe . $ For \ n=1 \ , \ 2 < 2^2 < 2+3 $ $ For \ n=2 \ , \ 2+3 < 3^2 < 2+3+5 $ $ For \ n=3 \ , \ 2+3+5 < 4^2 < 2+3 +5+7$ $ For \ n=4 \ , \ 2+3 +5+7 < 5^2 <2+3 +5+7 +11 $ Now proceed to prove $ \forall n \geq 5 $ .[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27.4" hover_enabled="0" _i="2" _address="0.1.0.2.2"]Observe $ \forall n \geq 5 $ we have $ P_n > (2n-1) $. [where $ n \in Z^+ $] Then try to use $ x_n = P_1 + P_2 + ...+P_5+..... +P_n > 1 +3 + ....+ 9 +... (2n-1) = n^2 \\ \Rightarrow x_n > n^2 , \forall n \geq 5[where \ n \in Z^+] $ .[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27.4" hover_enabled="0" _i="3" _address="0.1.0.2.3"]Think if $ x_n=P_1 + P_2 + .... + P_5 +...P_n = b^2 for \ some \ n , b \in Z^+ $ , then we are done . If not so , then think $ m $ be the largest non negative integer such that $ (n+m)^2 < x_n $ . Now note that the next perfect square is $ (n+m+1)^2 $ . Observe that if we can prove that $ (n+m+1)^2 - (n+m)^2 = (2n+ 2m +1) \geq P_{n+1} $ , then we are done . Now try to verify this claim .[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27.4" hover_enabled="0" _i="4" _address="0.1.0.2.4"]

Suppose our claim is not true i.e. $ P_n < 2n + 2m +1$ . So, $ P_n < 2n + 2m +1 \\ \Rightarrow 2n+ 2m \geq P_n , \forall n \in Z^+ \\ \Rightarrow (2n +2m-2)+(2n+ 2m -4)+.....2m \geq P_n + P_{n-1}+......+P_1 \\ \Rightarrow n^2 + 2mn -n \geq P_n + P_{n-1}+......+P_1 \\ \Rightarrow n^2 + 2mn -n \geq x_n \\ \Rightarrow n^2 + 2mn +m^2 > n^2 + 2mn -n\geq x_n \\ \Rightarrow (n+m)^2 > x_n $ . Contradiction! since we have assumed $ x_n = P_1 + P_2+......+P_{n-1} > (n+m)^2 $ . Thus ,$ (n+m+1)^2 \in (x_n , x_{n+1}) $ . [/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" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" 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" _i="5" _address="0.1.0.5"][/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" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="6" _address="0.1.0.6"]

Circle in Circle - PRMO 2017 | Problem 27

Let $ \Omega_1 $ be a circle with center O and let AB be a diameter of $ \Omega_1 $. Let P be a point on the segment OB different from O. Suppose another circle $ \Omega_2 $ with center P lies in the interior of $ \Omega_1 $. Tangents are drawn from A and B to the circle $ \Omega_2 $ intersecting $ \Omega_1 $ again at $A_1$ and $B_1$ respectively such that $A_1 $ and $B_1$ are on the opposite sides of AB. Given that $A_1B = 5, AB_1 = 15 $ and $ OP = 10$, find the radius of $ \Omega_1 $.

Start with hints

Do you really need a hint? Try it first!

Hint 1

Draw a diagram carefully.

Hint 2

Suppose the point of tangencies are at C and D. Join PC and PD.

Can you find two pairs of similar triangles?

Hint 3

$ \Delta APC \sim \Delta AA_1B $

Why?

Notice that AC is perpendicular to $ AA_1 $ as the radius is perpendicular to the tangent.

Also $ \angle A $ is common to both triangles. Hence the two triangles are similar (equiangular implies similar).

Similarly $ \Delta BPD \sim \Delta BAB_1 $.

Use the ratio of sides to find OA.

Hint 4

Suppose OA = R (radius of the big circle).

OC = r (radius of the small circle).

We already know OP = 10, $ A_1 B = 5, AB_1 = 15$

Since $ \Delta AA_1B $ and $ ACP $ are similar we have $ \frac{AP}{AB} = \frac{PC}{A_1B}$. This implies $ \frac{R+10}{2R} = \frac{r}{5}$ (1)

Similarlly since $ \Delta BPD $ and $ BAB_1 $ are similar we have $ \frac{BP}{BA} = \frac{PD}{AB_1}$. This implies $ \frac{R-10}{2R} = \frac{r}{15}$ (2)

Multiply the reciprocal of (2) with (1) to get R = 20.

Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year.

Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Real Surds - Problem 2 Pre RMO 2017

Problem

Suppose (a, b) are positive real numbers such that (a \sqrt{a}+b \sqrt{b}=183 . a \sqrt{b}+b \sqrt{a}=182). Find (\frac{9}{5}(a+b)).

Hint 1

This problem will use the following elementary algebraic identity:

$(x+y)^3=x^3+y^3+3 x^2 y+3 x y^2$

Can you identify what is x and what is y?

Hint 2

background_video_pause_outside_viewport="on" tab_text_shadow_style="none" body_text_shadow_style="none"] Set $x=\sqrt{a}, y=\sqrt{b}$. Then the given information translates to

$$
x^3+y^3=183, x^2 y+x y^2=182
$$

This implies $(x+y)^3=(\sqrt{a}+\sqrt{b})^3=x^3+y^3+3\left(x^2 y+x y^2\right)=183+3 \times 182=729$ Finally taking cube root on both sides, we have $\sqrt{a}+\sqrt{b}=9$

Hint 3

Note that $\sqrt{a} b+a \sqrt{b}=182 \Rightarrow \sqrt{a} \sqrt{b}(\sqrt{a}+\sqrt{b})=182 \Rightarrow \sqrt{a b} \times 9=182$ So at this point we know $(\sqrt{a}+\sqrt{b})=9, \sqrt{a b}=\frac{182}{9}$. It should be easy to find the value of $\left.\frac{9}{( } 5\right)(a+b)$ from these relations.

Final Answer

$a+b=(\sqrt{a}+\sqrt{b})^2-2 \sqrt{a b}=9^2-2 \times \frac{182}{9}=\frac{365}{9}$

Hence $\frac{9}{5}(a+b)=\frac{9}{5} \times \frac{365}{9}=73$

Pre RMO 2017 Problem 13 Solution - Kite in a Circle

Cyclic Quadrilaterals are often important objects in a Geometry problem. Recognizing them can lead to a path to the solution. Pre RMO 2017 Problem 13 Solution is a part of our Pre-RMO problem solving series.

Also visit: Math Olympiad Program of Cheenta

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PreRMO and I.S.I. Entrance Open Seminar

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Advanced Mathematics Seminar 2 hours

[/et_pb_text][et_pb_text _builder_version="3.0.82" text_text_color="#d4ccff" text_line_height="1.9em" locked="off"] An Open seminar for Pre-RMO and I.S.I. Entrance 2019 aspirants. We will work on topics from Number Theory, Geometry and Algebra. Registration is free. There are only 25 seats available. Date: 29th June, Friday, 6 PM [/et_pb_text][/et_pb_column][et_pb_column type="1_2"][et_pb_image src="https://cheenta.com/wp-content/uploads/2018/04/IMG_0385.jpg" _builder_version="3.3.1" animation_style="zoom" animation_direction="left" animation_intensity_zoom="20%" /][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section bb_built="1" admin_label="Course Chapters" specialty="on" padding_top_2="0px" _builder_version="3.0.82" custom_margin="|||" custom_padding="0px||120px|" prev_background_color="#000000"][et_pb_column type="2_3" specialty_columns="2"][et_pb_row_inner custom_padding="0px|||" use_custom_gutter="on" gutter_width="4" _builder_version="3.0.82" custom_margin="-80px|||"][et_pb_column_inner type="4_4" saved_specialty_column_type="2_3"][et_pb_blurb admin_label="Chapter" title="Online" url="#" image="https://cheenta.com/wp-content/uploads/2018/06/coding-icon_4.jpg" icon_placement="left" image_max_width="64px" content_max_width="1100px" _builder_version="3.0.82" header_font="|on|||" header_text_color="#2e2545" header_line_height="1.5em" body_text_color="#8585bd" body_font_size="16px" body_line_height="1.9em" background_color="#ffffff" box_shadow_style="preset2" box_shadow_horizontal="0px" box_shadow_vertical="0px" box_shadow_blur="60px" box_shadow_color="rgba(71,74,182,0.12)" custom_margin="|||" custom_padding="30px|40px|30px|40px" animation_style="zoom" animation_direction="bottom" animation_intensity_zoom="20%" animation_starting_opacity="100%" locked="off"] Students outside Calcutta may attend the live online seminar. You will need a laptop/computer with active internet connection, latest browser (chrome or firefox) and audio input/output. [/et_pb_blurb][et_pb_blurb admin_label="Chapter" title="In Calcutta Center" url="#" image="https://cheenta.com/wp-content/uploads/2018/06/coding-icon_12.jpg" icon_placement="left" image_max_width="64px" content_max_width="1100px" _builder_version="3.0.82" header_font="|on|||" header_text_color="#2e2545" header_line_height="1.5em" body_text_color="#8585bd" body_font_size="16px" body_line_height="1.9em" background_color="#ffffff" box_shadow_style="preset2" box_shadow_horizontal="0px" box_shadow_vertical="0px" box_shadow_blur="60px" box_shadow_color="rgba(71,74,182,0.12)" custom_margin="|||" custom_padding="30px|40px|30px|40px" animation_style="zoom" animation_direction="bottom" animation_intensity_zoom="20%" animation_starting_opacity="100%" locked="off"] Students in and around Calcutta, may come directly to our offline center near Tollygunj. Our center is near Bansdroni Metro Station. Address: 26 W/1 A, Khanpur Rd, Sahid Nagar Colony, Netaji Nagar, Kolkata, West Bengal 700047 [/et_pb_blurb][/et_pb_column_inner][/et_pb_row_inner][et_pb_row_inner custom_padding="20px|5px|20px|20px|false|false" _builder_version="3.3.1" box_shadow_style="preset1"][et_pb_column_inner type="4_4" saved_specialty_column_type="2_3"][et_pb_contact_form email="helpdesk@cheenta.com" title="Register Here" success_message="Our team will process your application soon. 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Pre RMO 2017

How many positive integers less than $1000$ have the property that the sum of the digits of each such number is divisible by $7$ and the number itself is divisible by $3$ ?
Suppose $a,b$ are positive real numbers such that $a\sqrt{a}+b\sqrt{b}=183$. $a\sqrt{b}+b\sqrt{a}=182$. Find $\frac{9}{5}(a+b)$.
A contractor has two teams of workers: team A and team B. Team A can complete a job in $12$ days and team B can do the same in $36$ days.Team A starts working on the job and team B joins team A after four days.The team A withdraws after two more days. For how many more days should team B work to complete the job?
Let $a,b$ be integers such that all the roots of the equation $(x^2+ax+20)(x^2+17x+b)=0$ are negative integers.What is the smallest possible value of $a+b$?
Let $u,v,w$ be real numbers in geometric progression such that $u>v>w$. Suppose $u^{40}=v^n=w^{60}$.Find the value of $n$.
Let the sum $\sum_{n=1}^{9}\frac{1}{n(n+1)(n+2)}$ written in its lowest terms be $\frac{p}{q}$.Find the value of $q-p$.
Find the number of positive integers $n$, such that $\sqrt{n}+\sqrt{n+1}<11$.
A pen costs ₹ $11$ and a notebook costs ₹ $13$.find the number of ways in which a person can spend exactly ₹ $1000$ to buy pens and notebooks.
There are five cities $A,B,C,D,E$ on a certain island.Each city is connected to every other city by road.In how many ways can a person starting from city $A$ come back to $A$ after visiting some cities without visiting a city more than once and without taking the same road more than once.? (The order in which he visits the cities also matters; e.g., the routes ${A}\to{B}\to{C}\to{A}$ and ${A}\to{C}\to{B}\to{A}$ are different. )
There are eight rooms on the first floor of a hotel,with four rooms on each side of the corridor,symmetrically situated (that is each room is exactly oposite to one other room).Four guests have to be accommodated in four of the eight rooms (that is, one in each) such that no two guests are in adjacent rooms or in oposite rooms.In how many ways can the guests be accommodated?
Let $f(x)=sin\frac{x}{3}+cos\frac{3x}{10}$ for all real $x$. Find the least natural number $n$ such that $f(n\pi+x)=f(x)$ for all real $x$.
In a class, the total number of boys and girls are in the ratio $4:3$. On one day it was found that $8$ boys and $14$ girls were absent from the class, and that the number of boys was the square of the number of girls. What is the total number of students in the class?
In a rectangle $ABCD$. $E$ is the midpoint of $AB$: $F$ is a point on $AC$ such that $BF$ is perpendicular to $AC$: and $FE$ perpendicular to $BD$. Suppose $BC=8\sqrt{3}$. Find $AB$.
Suppose $x$ is a positive real number such that $\{x\}.[x]$ and $x$ are in a geometric progression. Find the least positive integer $n$ such that $x^n>100$. (Here $[x]$ denotes the integer part of $x$ and $\{x\}=x-[x]$.)
Integers $1,2,3,...,n$, where $n>2$, are written on a board. Two numbers $m,k$ such that $1<{m}<{n}$, $1<{k}<{n}$ are removed and the averege of the remaining numbers is found to be $17$. What is the maximum sum of the two removed numbers?
Five distinct $2-digit$ numbers are in a geometric progression. Find the middle term.
Suppose the altitudes of a triangle are $10$, $12$ and $15$. What is its semi-perimeter?
If the real numbers $x,y,z$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, what is the value of $x^2+y^2+z^2$?
Suppose $1,2,3$ are the roots of the equation $x^4+ax^2+bx=c$. Find the value of $c$.
What is the number of triples $(a,b,c)$ of positive integers such that (i) $a<b<c<10$ and (ii) $a,b,c,10$ form the sides of a quadrilateral?
Find the number of ordered triples $(a,b,c)$ of positive integers such that $abc=108$.
Suppose in the plane $10$ pairwise nonparallel lines intersect one another. What is the maximum possible number of polygons (with finite areas) that can be formed?
Suppose an integer $x$, a natural number $n$ and a prime number $p$ satisfy the equation $7x^2-44x+12=p^n$. Find the largest value of $p$.
Let $P$ be an interior point of a triangle $ABC$ whose side lengths are $26,65,78$. The line through $P$ parallel to $BC$ meets $AB$ in $k$ and $AC$ in $L$. The line through $P$ parallel to $CA$ meets $BC$ in $M$ and $BA$ in $N$. The line through $P$ parallel to $AB$ meets $CA$ in $S$ and $CB$ in $T$. If $KL,MN,ST$ are of equal lengths, find this common length.
Let $ABCD$ be a rectangle and let $E$ and $F$ be points on $CD$ and $BC$ respectively such that area $(ADE)=16$ area $(CEF)=9$ and area $(ABF)=25$. What is the area of triangle $AEF$?
Let $AB$ and $CD$ be two parallel chords in a circle with radius $5$ such that the centre $O$ lies between these chords. Suppose $AB=6$, $CD=8$. Suppose further that the area of the part of the circle lying between the chords $AB$ and $CD$ is $(m\pi+n)/k$, where $m,n,k$ are positive integers with $gcd(m,n,k)=1$. What is the value of $m+n+k$?
Let ${\Omega}_{1}$ be a circle with centre $O$ and let $AB$ be a diameter of ${\Omega}_{1}$. Let $P$ be a point on the segment $OB$ different from $O$. Suppose another circle ${\Omega}_{2}$ with centre $P$ lies in the interior of ${\Omega}_{1}$. Tangents are drawn from $A$ and $B$ to the circle ${\Omega}_{2}$ intersecting ${\Omega}_{1}$ again at $A_1$ and $B_1$ respectively such that $A_1$ and $B_1$ are on the opposite sides of $AB$. Given that ${A_1}B=5$, $A{B_1}=15$ and $OP=10$, find the radius of ${\Omega}_{1}$.
Let $p,q$ be prime numbers such that $n^{3pq}-n$ is a multiple of $3pq$ for all positive integers $n$. Find the least possible value of $p+q$.
For each positive integer $n$, consider the highest common factor $h_n$ of the two numbers $n!+1$ and $(n+1)!$. For $n<100$, find the largest value of $h_n$.
Consider the areas of the four triangles obtained by drawing the diagonals $AC$ and $BD$ of a trapezium $ABCD$. The product of this areas, taken two at time, are computed. If among the six products so obtained two products are $1296$ and $576$, determine the square root of the maximum possible area of the trapezium to the nearest integer.

WB PRE-RMO 2016 PAPER AND ANSWERS

prmo2016

CLICK ON THE ABOVE LINK to get the WB PRE-RMO 2016 PAPER AND ANSWERS.

Number Theory, Ireland MO 2018, Problem 9

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Real Surds - Problem 2 Pre RMO 2017

Pre RMO 2017 Problem 13 Solution - Kite in a Circle

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WB PRE-RMO 2016 PAPER AND ANSWERS