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

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== Problem ==
 
== Problem ==
Point <math>D</math> is selected inside acute triangle <math>ABC</math> so that <math>\angle DAC=\angle ACB</math> and <math>\angle BDC=90^\circ+\angle BAC</math>. Point <math>E</math> is chosen on ray <math>BD</math> so that <math>AE=EC</math>. Let <math>M</math> be the midpoint of <math>BC</math>. Show that line <math>AB</math> is tangent to the circumcircle of triangle <math>BEM</math>.
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Point <math>D</math> is selected inside acute triangle <math>ABC</math> so that <math>\angle DAC=\angle ACB</math> and <math>\angle BDC=90^\circ+\angle BAC</math>. Point <math>E</math> is chosen on ray <math>BD</math> so that <math>AE=EC</math>. Let <math>M</math> be the midpoint of <math>BC</math>. Show that line <math>AB</math> is tangent to the circumcircle of triangle <math>BEM</math>..
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 04:27, 1 November 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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