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Difference between revisions of "2022 AMC 10A Problems/Problem 13"

(Created page with "==Problem== Let <math>\triangle ABC</math> be a scalene triangle. Point <math>P</math> lies on <math>\overline{BC}</math> so that <math>\overline{AP}</math> bisects <math>\an...")
 
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Let <math>AB=AY=2x.</math> By the Angle Bisector Theorem, we have <math>AC=3x,</math> or <math>YC=x.</math>  
 
Let <math>AB=AY=2x.</math> By the Angle Bisector Theorem, we have <math>AC=3x,</math> or <math>YC=x.</math>  
  
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By parallel lines, we get <math>\angle YAD=\angle YCB</math> and <math>\angle YDA=\angle YBC.</math> Note that <math>\triangle ADY \sim \triangle CBY</math> by the Angle-Angle Similarity, with the ratio of similitude <math>\overline{AY}{CY}=2.</math> It follows that <math>AD=2CB=2(BP+PC)=\boxed{\textbf{(C) } 10}.</math>
+
By parallel lines, we get <math>\angle YAD=\angle YCB</math> and <math>\angle YDA=\angle YBC.</math> Note that <math>\triangle ADY \sim \triangle CBY</math> by the Angle-Angle Similarity, with the ratio of similitude <math>\frac{AY}{CY}=2.</math> It follows that <math>AD=2CB=2(BP+PC)=\boxed{\textbf{(C) } 10}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 22:16, 11 November 2022

Problem

Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$

$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution

DIAGRAM IN PROGRESS.

WILL BE DONE TOMORROW, WAIT FOR ME THANKS.

Suppose that $\overline{BD}$ intersect $\overline{AP}$ and $\overline{AC}$ at $X$ and $Y,$ respectively. By Angle-Side-Angle, we conclude that $\triangle ABX\cong\triangle AYX.$

Let $AB=AY=2x.$ By the Angle Bisector Theorem, we have $AC=3x,$ or $YC=x.$

By parallel lines, we get $\angle YAD=\angle YCB$ and $\angle YDA=\angle YBC.$ Note that $\triangle ADY \sim \triangle CBY$ by the Angle-Angle Similarity, with the ratio of similitude $\frac{AY}{CY}=2.$ It follows that $AD=2CB=2(BP+PC)=\boxed{\textbf{(C) } 10}.$

~MRENTHUSIASM

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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