Triangle Inequality Problem - AMC 12B, 2014 - Problem 13

Try this beautiful problem from American Mathematics Competition - 12B ,2014, Problem Number - 13 based on Triangle inequality

Problem - Triangle Inequality

Real numbers a and b are chosen with 1 < a < b such that no triangle with positive area has side lengths 1,a and b or \(\frac {1}{b},\frac {1}{a}\) and 1. What is the smallest possible value of b?

  • $\frac{3+\sqrt{3}}{2}$
  • $\frac{5}{2}$
  • $\frac{3+\sqrt{5}}{2}$
  • $\frac{3+\sqrt{6}}{2}$

Key Concepts

Triangle Inequality

Inequality

Geometry

Check the Answer

Answer: \(\frac {3+\sqrt 5}{2}\)

Try with Hints

Here is the first hint where you can start this sum:

It is given $1>\frac{1}{a}>\frac{1}{b}$. Use Triangle Inequality here :

$$
\begin{aligned}
& a+1>b \
& a>b-1 \
& \frac{1}{a}+\frac{1}{b}>1
\end{aligned}
$$

If we want to find the lowest possible value of $b$, we create we try to create two degenerate triangles where the sum of the smallest two sides equals the largest side. Thus we get : $a=b-1$

Now try to do the rest of the sum……………………

Here is the rest of steps to check your problem :

$$
2 b-1=b^2-b
$$

Now Solving for $\mathbf{b}$ using the quadratic equation, we get

$$
\begin{aligned}
& b^2-3 b+1=0 \
& b=\frac{3+\sqrt{5}}{2} \text { (Answer) }
\end{aligned}
$$

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Triangle Inequality - Mathematical Circles - Problem No. 5

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Try this beautiful problem from Mathematical Circles book based on Triangle inequality.

Problem :

Find a point inside a convex quadrilateral such that the sum of the distances from the point to the vertices is minimal .

Key Concepts

Triangle Inequality

Inequality

Geometry

Check the Answer

Mathematical Circles - Chapter 6 - Triangle Inequality - Problem 6

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints

Do you really need a hint ? You can start thinking about the Triangle Inequality...........

If you have already get the idea about the main concept for this sum then you can start the problem by taking a quadrilateral ABCD with diagonals that intersect at point 'o'.

The distance from 'o' to all vertices is equal.

Triangle Inequaliy image from Mathematical Circles

Here is the diagram where OA + OB + OC + OD = AC + BD - which is sum of the diagonals. We can consider this as one case ...

As a last hint you have to take another point to o' to compare with first case

Now o' be another point inside the quadrilateral. If we use triangle inequality here we have ,

AO' + OC' > AC from Triangle AO'C

BO' + O'D > BD from Triangle BO'D

Hence AO' + O'C + BO' + O'D > AB + BD

Therefore its clear from this that the point that minimizes the sum of the distances is the point of intersection of diagonals.

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[/et_pb_text][/et_pb_column] [/et_pb_row] [/et_pb_section]

Can we prove that the length of any side of a triangle is not more than half of its perimeter?

Can we Prove that ........

The length of any side of a triangle is not more than half of its perimeter

Key Concepts

Triangle Inequality

Perimeter

Geometry

Check the Answer

Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles - Chapter 6 - Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints

We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

Proof based on triangle

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum ................................

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , \(\frac {perimeter}{2} > a \)

\(\frac {perimeter}{2} \) = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

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Inequality, Israel MO 2018, Problem 3

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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"]Determine the minimal and maximal values the expression $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ can take, where $a,b,c$ are real numbers.[/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"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Israel MO 2018, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Algebra, Inequality[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Excursion in Mathematics by Bhaskarcharya 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"][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"]Given the expressions, what inequality comes to your mind first? The triangle inequality right? |x| + |y| \( \geq \) |x+y|. Can you use this inequality to get a maximum bound?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Maximum Bound: Observe that the maximum bound is got by the triangle inequality as explained. $|a+b| \le |a|+|b|$
$|b+c| \le |b|+|c|$  
$|a+c| \le |a|+|c|$ We get, $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|} \le\frac{|a|+|b|+|b|+|c|+|a|+|c|}{|a|+|b|+|c|}=2$ Never forget to mention the equality case: a = b = c is the equality case.  [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]What about the minimum inequality? The idea is that you can observe that keeping the denominator constant, can you reduce the numerator. Let's take the case of the |a| = |b| = |c| = 1. Now, the expression is maximized when a = b = c = 1 or -1. So, obviously one must be positive or two must be negative or vice-versa. In either case, we get \( \frac{2}{3} \). Okay, then maybe we need to deal with the signs and stuff to get a hold on the minimum. Let's fix the signs of a,b,c then, we can break the bonds of the modulus. Let's proceed to the next hint.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Let $a$, $b$, and $c$ be arbitrary real numbers, not all of them equal $0$. By flipping signs, we can assume that at least two of $a$, $b$, and $c$ are non-negative. Actually, without loss of generality, we can assume that $a, b\geq 0$. \( 3|a + b| + 3|b + c| + 3|c + a| \geq 3a + 3b + 3|b+c| + 3|a+c| \geq 2(a+b) + (a + |a + c|) + (b + |b + c|) \) \( = 2(|a| + |b|) + (|-a| + |a + c|) + (|-b| + |b + c|) \geq 2(|a| + |b|) + |c| + |c| = 2|a| + 2|b| + 2|c| \) We have proved that the minimum possible value of $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ is $\frac{2}{3}$. The minimum is $\frac{2}{3}$, which is attained for $a = b = 1$, $c = -1$.  

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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"]

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