Inequality with triangles and squares

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Inequality with triangles and squares

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Brazilian Math Olympiad

Brazilian Math Olympiad 2018

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Source: Brazilian Mathematical Olympiad 2018 - Q1

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#1 h Nov 16, 2018, 3:07 PM • 6 Y

Y by UK2019Project, mathisreal, Maths_Guy, Ruy, Adventure10, Mango247

We say that a polygon $P$ is inscribed in another polygon $Q$ when all vertices of $P$ belong to perimeter of $Q$ . We also say in this case that $Q$ is circumscribed to $P$ . Given a triangle $T$ , let $l$ be the maximum value of the side of a square inscribed in $T$ and $L$ be the minimum value of the side of a square circumscribed to $T$ . Prove that for every triangle $T$ the inequality $L/l \ge 2$ holds and find all the triangles $T$ for which the equality occurs.

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#2 h Nov 18, 2018, 8:14 AM • 2 Y

Y by Maths_Guy, Adventure10

It is easy to get $l = \max \left(\frac{ah_a}{a+h_a},\frac{bh_b}{b+h_b},\frac{ch_c}{c+h_c}\right)$ and $L = \min (\max(a,h_a),\max(b,h_b),\max(c,h_c))$ .

Case 1. $a =\min (\max(a,h_a),\max(b,h_b),\max(c,h_c))$ .

Case 1.1. $a+h_a = \min (a+h_a,b+h_b,c+h_c)$ . Now $\frac{L}{l} = \frac{a(a+h_a)}{ah_a} = \frac{a+h_a}{h_a} = \frac{a}{h_a} + 1\geq 1 + 1 =2.$ Case 1.2. $b+h_b = \min (a+h_a,b+h_b,c+h_c)$ . We have $a\geq h_a$ . Suppose $b\geq h_b$ then, $a\leq b$ giving $h_a \geq h_b$ . Therefore we have $b\geq a \geq h_a \geq h_b$ . Also $b+h_b\leq a+h_a$ and as $ah_a = bh_b$ , therefore $a=b$ and $h_a = h_b$ which is same as case 1.1. Suppose $h_b\geq b$ , then $a\leq h_b$ , also $a\geq h_a$ . Therefore $h_b\geq a \geq h_a$ , also as $bh_b = ah_a$ , $h_a \geq b$ . Again $h_b\geq a\geq h_a\geq b$ but $b+h_b \leq a + h_a$ , so $a=h_b$ and $b = h_a$ again boiling down to case 1.1.

Case 2. $h_a =\min (\max(a,h_a),\max(b,h_b),\max(c,h_c))$ .

Case 2.1. $a+h_a = \min (a+h_a,b+h_b,c+h_c)$ . Now $\frac{L}{l} = \frac{h_a(a+h_a)}{ah_a} = \frac{a+h_a}{a} = \frac{h_a}{a} + 1\geq 1 + 1 =2.$
Case 2.2. $b+h_b = \min (a+h_a,b+h_b,c+h_c)$ . We have $h_a \geq a$ . Suppose $b\geq h_b$ , then $h_a\leq b$ . Then as $ah_a=bh_b$ , $b\geq h_a \geq a\geq h_b$ . But $b+h_b\geq a + h_a$ which gives $b=h_a$ which gives case 2.1.. If $b\leq h_b$ , then $h_a \leq h_b$ and $a\geq b$ . Therefore $h_b\geq h_a \geq a\geq b$ which gives $a=b$ again giving case 2.1.

Equality holds for $a=h_a$ or $b=h_b$ or $c=h_c \qquad \square$

I don't know if this works

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#4 h Nov 18, 2018, 12:13 PM • 2 Y

Y by Adventure10, Mango247

$a$ is x side of the triangle and let $h$ be the height relating to it then
the maximum side of inscribed square is max $\frac{xh}{x+h}$ but the minimum side of the circumscribed square is $\underset{\text{cyclically}}{\Large\text {min}} \ \ \frac{b^2+c^2-a^2}{2\sqrt{b^2+c^2-2bc\sin A}}$ ...

This post has been edited 1 time. Last edited by PROF65, Nov 19, 2018, 9:40 AM
Reason: rectify

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#5 h Nov 19, 2018, 3:04 AM • 3 Y

Y by Maths_Guy, aopsuser305, Adventure10

Sketch: Let $\mathcal{S}$ and $\mathcal{S}'$ be any squares circumscribed about and inscribed in a triangle $\mathcal{T}$ . The idea is to prove the easy inequalities
$2[\mathcal{S}'] \le [\mathcal{T}] \quad \text{and} \quad 2[\mathcal{T}] \le [\mathcal{S}]$ from which the desired inequality follows.

Now suppose equality holds for $\mathcal{S}$ and $\mathcal{S}'$ . Letting $a$ , $b$ , $c$ be the sides of $\mathcal{T}$ and $h_a$ , $h_b$ , $h_c$ be the corresponding altitudes, the equality case for $2[\mathcal{T}] \le [\mathcal{S}]$ is $a = h_a$ , $b = h_b$ , or $c = h_c$ , and if any of these hold it is easy to see that equality holds in $2[\mathcal{S}'] \le [\mathcal{T}]$ . Thus the answer is all triangles satisfying $a = h_a$ , $b = h_b$ , or $c = h_c$ .

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