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

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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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Let <math>\angle DBT = \alpha</math> and <math>\angle BEM = \beta</math>.   
 
Extend AD intersects BC at point T, then TC = TA, TE is perpendicular to AC
 
Extend AD intersects BC at point T, then TC = TA, TE is perpendicular to AC
  
 
Thus, AB is the tangent of the circle BEM
 
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
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Then the question is equivalent as  the <math>\angle ABT</math> is the auxillary angle of <math>\angle BEM</math>.
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as <math>/betta &= 180-B</math>
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''ontinued''
  
 
==See Also==
 
==See Also==
 
{{USAMO newbox|year=2024|num-b=4|num-a=6}}
 
{{USAMO newbox|year=2024|num-b=4|num-a=6}}
 +
{{USAJMO newbox|year=2024|num-b=5|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:39, 22 March 2025

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

Let $\angle DBT = \alpha$ and $\angle BEM = \beta$. 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 $\angle BEM$.

ontinued

See Also

2024 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
2024 USAJMO (Problems • Resources)
Preceded by
Problem 5 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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