Difference between revisions of "2024 USAMO Problems/Problem 5"

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(Solution 1)
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== Solution 1 ==
 
== Solution 1 ==
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define angle DBT as <math>/alpha</math>, the angle BEM as <math>/betta</math>. 
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Extend AD intersects BC at point T, then TC = TA, TE is perpendicular to AC
  
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Thus, AB is the tangent of the circle BEM
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Then the question is equivalent as  the angle ABT is the auxillary angle of the angle BEM
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as <math>/betta &= 180-B</math>
  
 
==See Also==
 
==See Also==
 
{{USAMO newbox|year=2024|num-b=4|num-a=6}}
 
{{USAMO newbox|year=2024|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:14, 5 May 2024

The following problem is from both the 2024 USAMO/5 and 2024 USAJMO/6, so both problems redirect to this page.

Problem

Point $D$ is selected inside acute triangle $ABC$ so that $\angle DAC=\angle ACB$ and $\angle BDC=90^\circ+\angle BAC$. Point $E$ is chosen on ray $BD$ so that $AE=EC$. Let $M$ be the midpoint of $BC$. Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.

Solution 1

define angle DBT as $/alpha$, the angle BEM as $/betta$. Extend AD intersects BC at point T, then TC = TA, TE is perpendicular to AC

Thus, AB is the tangent of the circle BEM

Then the question is equivalent as the angle ABT is the auxillary angle of the angle BEM as $/betta &= 180-B$ (Error compiling LaTeX. Unknown error_msg)

See Also

2024 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png