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

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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 1 (Generalization)==
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==Solution 1 (The extra line)==
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Let the intersection of <math>AC</math> and <math>BD</math> be <math>M</math>, and the intersection of <math>AP</math> and <math>BD</math> be <math>N</math>. Draw a line from <math>M</math> to <math>BC</math>, and label the point <math>O</math>. We have <math>\triangle BPN \sim \triangle BOM</math>, with a ratio of <math>2</math>, so <math>BO = 4</math> and <math>OC = 1</math>. We also have <math>\triangle COM \sim \triangle CAP</math> with ratio <math>3</math>.
  
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Suppose that <math>\overline{BD}</math> intersect <math>\overline{AP}</math> and <math>\overline{AC}</math> at <math>X</math> and <math>Y,</math> respectively. By Angle-Side-Angle, we conclude that <math>\triangle ABX\cong\triangle AYX.</math>
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Suppose the area of <math>\triangle COM</math> is <math>x</math>. Then, <math>[\triangle CAP] = 9x</math>. Because <math>\triangle CAP</math> and <math>\triangle BAP</math> share the same height and have a base ratio of <math>3:2</math>, <math>[\triangle BAP] = 6x</math>. Because <math>\triangle BOM</math> and <math>\triangle COM</math> share the same height and have a base ratio of <math>4:1</math>, <math>[\triangle BOM] = 4x</math>, <math>[\triangle BPN] = x</math>, and <math>[OMNP] = 4x - x = 3x</math>. Thus, <math>[\triangle MAN] = [\triangle BAN] = 5x</math>. Finally, because <math>BPN \sim ADN</math> and the ratio <math>AN : PN</math> is <math>5</math> (because <math>[\triangle BAN] : [\triangle BPN] = 5</math> and they share a side), <math>AD = 2 \cdot 5 = \boxed{\textbf{(C) } 10}.
  
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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>
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~mathboy100
  
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By alternate interior angles, 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>
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==Solution 2 (Generalization)==
 +
 
 +
Suppose that </math>\overline{BD}<math> intersect </math>\overline{AP}<math> and </math>\overline{AC}<math> at </math>X<math> and </math>Y,<math> respectively. By Angle-Side-Angle, we conclude that </math>\triangle ABX\cong\triangle AYX.<math>
 +
 
 +
Let </math>AB=AY=2x.<math> By the Angle Bisector Theorem, we have </math>AC=3x,<math> or </math>YC=x.<math>
 +
 
 +
By alternate interior angles, 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}.$
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 2 (Assumption) ==
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== Solution 3 (Assumption) ==
 
<asy>
 
<asy>
 
size(300);
 
size(300);

Revision as of 01:16, 26 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 $AD?$

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

Diagram

[asy] /* Made by MRENTHUSIASM */ size(300); real r = 4*sqrt(114)/13; pair A, B, C, D, P, X, Y; A = origin; B = (2,r); C = (3/2*sqrt(2^2+r^2),0); D = A + 2*(C-B); P = B + 2*dir(C-B); X = intersectionpoint(B--D,A--P); Y = intersectionpoint(B--D,A--C); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*N,linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot(X^^Y,linewidth(4)); markscalefactor=0.03; draw(rightanglemark(B,X,A),red); draw(anglemark(P,A,B,20), red); draw(anglemark(C,A,P,20), red); add(pathticks(anglemark(P,A,B,20), n = 1, r = 0.1, s = 7, red)); add(pathticks(anglemark(C,A,P,20), n = 1, r = 0.1, s = 7, red)); draw(A--B--C--cycle^^A--P^^B--D^^A--D); draw(B--C,MidArrow(0.3cm,Fill(red))); draw(A--D,MidArrow(0.3cm,Fill(red))); label("$2$",midpoint(B--P),rotate(90)*dir(midpoint(P--B)--P),red); label("$3$",midpoint(P--C),rotate(90)*dir(midpoint(C--P)--C),red); [/asy]

~MRENTHUSIASM

Solution 1 (The extra line)

Let the intersection of $AC$ and $BD$ be $M$, and the intersection of $AP$ and $BD$ be $N$. Draw a line from $M$ to $BC$, and label the point $O$. We have $\triangle BPN \sim \triangle BOM$, with a ratio of $2$, so $BO = 4$ and $OC = 1$. We also have $\triangle COM \sim \triangle CAP$ with ratio $3$.

Suppose the area of $\triangle COM$ is $x$. Then, $[\triangle CAP] = 9x$. Because $\triangle CAP$ and $\triangle BAP$ share the same height and have a base ratio of $3:2$, $[\triangle BAP] = 6x$. Because $\triangle BOM$ and $\triangle COM$ share the same height and have a base ratio of $4:1$, $[\triangle BOM] = 4x$, $[\triangle BPN] = x$, and $[OMNP] = 4x - x = 3x$. Thus, $[\triangle MAN] = [\triangle BAN] = 5x$. Finally, because $BPN \sim ADN$ and the ratio $AN : PN$ is $5$ (because $[\triangle BAN] : [\triangle BPN] = 5$ and they share a side), $AD = 2 \cdot 5 = \boxed{\textbf{(C) } 10}.

~mathboy100

==Solution 2 (Generalization)==

Suppose that$ (Error compiling LaTeX. Unknown error_msg)\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.$By alternate interior angles, we get$\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 3 (Assumption)

[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, \[\bigtriangleup DAB \sim \bigtriangleup ABP\] and therefore $\frac{DA}{AB} = \frac{AB}{BP}$, so $AD=\boxed{\textbf{(C) }10}$, and we're done!

~Bxiao31415

Video Solution 1

https://youtu.be/_0_EGdkhOFg

- Whiz

Video Solution by OmegaLearn

https://youtu.be/77JIN0iVizA

~ pi_is_3.14

Video Solution (Quick and Simple)

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See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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