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

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draw(A--B--C--cycle^^A--P^^B--D^^A--D);
 
draw(A--B--C--cycle^^A--P^^B--D^^A--D);
 
draw(B--C,MidArrow(0.3cm,Fill(red)));
 
draw(B--C,MidArrow(0.3cm,Fill(red)));
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drawe(90)*dir(midpoint(P--B)--P),red);
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draw(A--D,MidArrow(0.3cm,Fill(red)));
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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);
 
label("$3$",midpoint(P--C),rotate(90)*dir(midpoint(C--P)--C),red);
 
</asy>
 
</asy>

Revision as of 01:28, 4 October 2024

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 (Angle Bisector Theorem and Similar Triangles)

Suppose that $\overline{BD}$ intersects $\overline{AP}$ 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 2 (Auxiliary Lines)

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 of intersection $O$.

By adding this extra line, we now have many pairs of similar triangles. 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 $3$. Additionally, $\triangle BPN \sim \triangle ADN$ (with an unknown ratio). It is also true that $\triangle BAN \cong \triangle MAN$.

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 thus $[OMNP] = 4x - x = 3x$. Thus, $[\triangle MAN] = [\triangle BAN] = 5x$.

Finally, we have $\frac{[\triangle BAN]}{[\triangle BPN]} = \frac{5x}{x} = 5$, and because these triangles share the same height $\frac{AN}{PN} = 5$. Notice that these side lengths are corresponding side lengths of the similar triangles $BPN$ and $ADN$. This means that $AD = 5\cdot BP = \boxed{\textbf{(C) } 10}$.

~mathboy100

Solution 3 (Slopes)

Let point $B$ be the origin, with $C$ being on the positive $x$-axis and $A$ being in the first quadrant.

By the Angle Bisector Theorem, $AB:AC = 2:3$. Thus, assume that $AB = 4$, and $AC = 6$.

Let the perpendicular from $A$ to $BC$ be $AM$.

Using Heron's formula, \[[ABC] = \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} = \frac{15}{4}\sqrt{7}.\]

Hence, \[AM = \frac{\frac{15}{4}\sqrt{7}}{\frac{5}{2}} = \frac{3}{2}\sqrt{7}.\]

Next, we have \[BM^2 + AM^2 = AB^2\] \[\therefore BM = \sqrt{16 - \frac{63}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \textrm{ and } PM = \frac{3}{2}.\]

The slope of line $AP$ is thus \[\frac{-\frac{3}{2}\sqrt{7}}{\frac{3}{2}} = -\sqrt{7}.\]

Therefore, since the slopes of perpendicular lines have a product of $-1$, the slope of line $BD$ is $\frac{1}{\sqrt{7}}$. This means that we can solve for the coordinates of $D$:

\[y = \frac{3}{2}\sqrt{7}\] \[y = \frac{1}{\sqrt{7}}x\] \[\frac{1}{\sqrt{7}}x = \frac{3}{2}\sqrt{7}\] \[x = \frac{7 \cdot 3}{2} = \frac{21}{2}\] \[D = \left(\frac{21}{2}, \frac{3}{2}\sqrt{7}\right).\]

We also know that the coordinates of $A$ are $\left(\frac{1}{2}, \frac{3}{2}\sqrt{7}\right)$, because $BM = \frac{1}{2}$ and $AM = \frac{3}{2}\sqrt{7}$.

Since the $y$-coordinates of $A$ and $D$ are the same, and their $x$-coordinates differ by $10$, the distance between them is $10$. Our answer is $\boxed{\textbf{(C) }10}.$

~mathboy100

Solution 4 (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/m1-7E8T_i_E

~Education, the Study of Everything

Video Solution 2

https://youtu.be/_0_EGdkhOFg

- Whiz

Video Solution 3

https://youtu.be/77JIN0iVizA

Video Solution 4

https://youtu.be/G8NRcVxSdz0

Video Solution 5 by SpreadTheMathLove

https://www.youtube.com/watch?v=nhlpSATltRU

~Ismail.maths93

Video Solution 6 by Lucas637

https://www.youtube.com/watch?v=R1CtcZ2pWVk

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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