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

(→‎See Also)
(→‎Solution 3 (Cheaty solution if you are almost out of time))
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== Solution 3 (Cheaty solution if you are almost out of time) ==
 
== Solution 3 (Cheaty solution if you are almost out of time) ==
 +
<asy>
 +
size(300);
 +
pair A, B, C, P, XX, D;
 +
B = (0,0);
 +
P = (2,0);
 +
C = (5,0);
 +
A=(0,4.47214);
 +
D = A + (10,0);
 +
draw(A--B--C--cycle, linewidth(1));
 +
dot("$A$", A, N);
 +
dot("$B$", B, SW);
 +
dot("$C$", C, E);
 +
dot("$P$", P, S);
 +
dot("$D$", D, E);
 +
markscalefactor = 0.1;
 +
draw(anglemark(B,A,P));
 +
markscalefactor = 0.12;
 +
draw(anglemark(P,A,C));
 +
draw(P--A--D--B, linewidth(1));
 +
XX = intersectionpoints(A--P,B--D)[0];
 +
dot("$X$", XX, dir(150));
 +
markscalefactor = 0.03;
 +
draw(rightanglemark(A,B,C));
 +
draw(rightanglemark(D,XX,P));
 +
</asy>
 
Since there is only one possible value of <math>AD</math>, we assume <math>\angle{B}=90^{\circ}</math>. By the angle bisector theorem, <math>\frac{AB}{AC}=\frac{2}{3}</math>, so <math>AB=2\sqrt{5}</math> and <math>AC=3\sqrt{5}</math>. Now observe that <math>\angle{BAD}=90^{\circ}</math>. Let the intersection of <math>BD</math> and <math>AP</math> be <math>X</math>. Then <math>\angle{ABD}=90^{\circ}-\angle{BAX}=\angle{APB}</math>. Consequently, <cmath>\bigtriangleup DAB ~ \bigtriangleup ABP</cmath> and therefore <math>\frac{DA}{AB} = \frac{AB}{BP}</math>, so <math>AD=\fbox{(C)10}</math>, and we're done!
 
Since there is only one possible value of <math>AD</math>, we assume <math>\angle{B}=90^{\circ}</math>. By the angle bisector theorem, <math>\frac{AB}{AC}=\frac{2}{3}</math>, so <math>AB=2\sqrt{5}</math> and <math>AC=3\sqrt{5}</math>. Now observe that <math>\angle{BAD}=90^{\circ}</math>. Let the intersection of <math>BD</math> and <math>AP</math> be <math>X</math>. Then <math>\angle{ABD}=90^{\circ}-\angle{BAX}=\angle{APB}</math>. Consequently, <cmath>\bigtriangleup DAB ~ \bigtriangleup ABP</cmath> and therefore <math>\frac{DA}{AB} = \frac{AB}{BP}</math>, so <math>AD=\fbox{(C)10}</math>, and we're done!
  

Revision as of 05:05, 12 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

Solution 2 By Omega Learn Using Similar Triangles and Angle Bisector Theorem

https://youtu.be/77JIN0iVizA

~ pi_is_3.14

Solution 3 (Cheaty solution if you are almost out of time)

[asy] size(300); pair A, B, C, P, XX, D; B = (0,0); P = (2,0); C = (5,0); A=(0,4.47214); D = A + (10,0); draw(A--B--C--cycle, linewidth(1)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, E); dot("$P$", P, S); dot("$D$", D, E); markscalefactor = 0.1; draw(anglemark(B,A,P)); markscalefactor = 0.12; draw(anglemark(P,A,C)); draw(P--A--D--B, linewidth(1)); XX = intersectionpoints(A--P,B--D)[0]; dot("$X$", XX, dir(150)); markscalefactor = 0.03; draw(rightanglemark(A,B,C)); draw(rightanglemark(D,XX,P)); [/asy] Since there is only one possible value of $AD$, we assume $\angle{B}=90^{\circ}$. By the angle bisector theorem, $\frac{AB}{AC}=\frac{2}{3}$, so $AB=2\sqrt{5}$ and $AC=3\sqrt{5}$. Now observe that $\angle{BAD}=90^{\circ}$. Let the intersection of $BD$ and $AP$ be $X$. Then $\angle{ABD}=90^{\circ}-\angle{BAX}=\angle{APB}$. Consequently, \[\bigtriangleup DAB ~ \bigtriangleup ABP\] and therefore $\frac{DA}{AB} = \frac{AB}{BP}$, so $AD=\fbox{(C)10}$, and we're done!

~Bxiao31415

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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