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Difference between revisions of "2020 USAMO Problems/Problem 1"

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The area of <math>\triangle OO_1O_2</math> is minimized if <math>CX \perp AB</math> because
 
The area of <math>\triangle OO_1O_2</math> is minimized if <math>CX \perp AB</math> because
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<math></math>\frac {[OO_1O_2]} {[ABC]} = (\frac {O_1 O_2} {AB})^2 \ge (\frac {EF} {AB})^2 = \frac {1}{4}$.  
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<cmath>\frac {[OO_1O_2]} {[ABC]} = \left(\frac {O_1 O_2} {AB}\right)^2 \ge \left(\frac {EF} {AB}\right)^2 = \frac {1}{4}.</cmath>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=m157cfw0vdE
 
https://www.youtube.com/watch?v=m157cfw0vdE
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{{MAA Notice}}

Latest revision as of 22:53, 18 October 2022

Problem 1

Let $ABC$ be a fixed acute triangle inscribed in a circle $\omega$ with center $O$. A variable point $X$ is chosen on minor arc $AB$ of $\omega$, and segments $CX$ and $AB$ meet at $D$. Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$, respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.


Solution

2020 USAMO 1.png

Let $E$ be midpoint $AD.$ Let $F$ be midpoint $BD \implies$ \[EF = ED + FD = \frac {AD}{2} + \frac {BD}{2} = \frac {AB}{2}.\] $E$ and $F$ are the bases of perpendiculars dropped from $O_1$ and $O_2,$ respectively.

Therefore $O_1O_2 \ge EF = \frac {AB}{2}.$

\[CX \perp O_1O_2, AX \perp O_1O \implies \angle O O_1O_2 = \angle AXC\] $\angle AXC = \angle ABC (AXBC$ is cyclic) $\implies \angle O O_1O_2 = \angle ABC.$

Similarly $\angle BAC = \angle O O_2 O_1 \implies \triangle ABC \sim \triangle O_2 O_1O.$

The area of $\triangle OO_1O_2$ is minimized if $CX \perp AB$ because \[\frac {[OO_1O_2]} {[ABC]} = \left(\frac {O_1 O_2} {AB}\right)^2 \ge \left(\frac {EF} {AB}\right)^2 = \frac {1}{4}.\] vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://www.youtube.com/watch?v=m157cfw0vdE

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