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

(Solution 1)
(Solution 1)
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== Solution 1 ==
 
== Solution 1 ==
define angle DBT as <math>/alpha</math>, the angle BEM as <math>/betta</math>.   
+
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
+
Then the question is equivalent as  the <math>\angle ABT</math> is the auxillary angle of <math>\angle BEM</math>.
  
 
continue
 
continue

Revision as of 21:53, 16 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

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 $\angle BEM$.

continue

See Also

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

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