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

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Point <math>D</math> is selected inside acute triangle <math>A B C</math> so that <math>\angle D A C=</math> <math>\angle A C B</math> and <math>\angle B D C=90^{\circ}+\angle B A C</math>. Point <math>E</math> is chosen on ray <math>B D</math> so that <math>A E=E C</math>. Let <math>M</math> be the midpoint of <math>B C</math>.
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{{duplicate|[[2024 USAMO Problems/Problem 5|2024 USAMO/5]] and [[2024 USAJMO Problems/Problem 6|2024 USAJMO/6]]}}
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Show that line <math>A B</math> is tangent to the circumcircle of triangle <math>B E M</math>.
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__TOC__
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== Problem ==
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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>.
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== Solution 1 ==
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Let <math>\angle DBT = \alpha</math> and <math>\angle BEM = \beta</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 <math>\angle ABT</math> is the auxillary angle of <math>\angle BEM</math>.
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continue
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==See Also==
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{{USAMO newbox|year=2024|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 04:28, 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 (Problems • Resources)
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

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