Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1 (Angle Bisector Theorem and Similar Triangles))
 
(25 intermediate revisions by 12 users not shown)
Line 36: Line 36:
 
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>
−
 
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
Line 49: Line 48:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
Minor grammar edit also added.
+
==Solution 2 (No Angle Bisectors!)==
 +
 
 +
[[File:Amc10a2022q13.png | 800px | center]]
 +
 
 +
We can draw lines from \( C \) to a new point \( E \), and extend the line \( BC \) to a point \( F \) such that the line \( FD \) has the same slope of the line \( CE \) and line \( AB \).
  
−
==Solution 2 (Auxiliary Lines)==
+
Notice how we have two parallelograms, but more specifically, these parallelograms are just made up of \( 2 \times \triangle ABC \). We see that the parallelograms \( ABCE \) and \( ECFD \) are congruent, and therefore \( BC \cong AE \cong CF \cong ED \). \( BC = 5 \), so \( AE = 5 \), and because \( AE \cong ED \), we have \( ED = 5 \).
 +
 
 +
Finally, we have \( AE + ED = AD \), so \( AD = 5 + 5 =\) <math>\boxed{\textbf{(C) } 10}.</math>
 +
 
 +
And just like that! No angle bisectors needed! 😁
 +
 
 +
~Pinotation
 +
 
 +
~Image adopted from MRENTHUSIASM, thank you so much!
 +
 
 +
==Solution 3 (Auxiliary Lines)==
 
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 of intersection <math>O</math>.  
 
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 of intersection <math>O</math>.  
  
Line 62: Line 75:
 
~mathboy100
 
~mathboy100
  
−
== Solution 3 (Slopes) ==
+
== Solution 4 (Slopes) ==
 
Let point <math>B</math> be the origin, with <math>C</math> being on the positive <math>x</math>-axis and <math>A</math> being in the first quadrant.
 
Let point <math>B</math> be the origin, with <math>C</math> being on the positive <math>x</math>-axis and <math>A</math> being in the first quadrant.
  
Line 92: Line 105:
 
~mathboy100
 
~mathboy100
  
−
== Solution 4 (Assumption) ==
+
== Solution 5 (Assumption) ==
 
<asy>
 
<asy>
 
size(300);
 
size(300);
Line 122: Line 135:
 
~[[User:Bxiao31415|Bxiao31415]]
 
~[[User:Bxiao31415|Bxiao31415]]
  
−
==Video Solution (THINKING CREATIVELY!!!)==
+
==Video Solution 1 ==
 
https://youtu.be/m1-7E8T_i_E
 
https://youtu.be/m1-7E8T_i_E
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
 +
== Video Solution 3 ==
  
−
== Video Solution 1 ==
+
https://youtu.be/77JIN0iVizA
−
 
 
−
https://youtu.be/_0_EGdkhOFg
 
−
 
 
−
- Whiz
 
  
−
== Video Solution ==
+
== Video Solution 4 ==
  
−
https://youtu.be/77JIN0iVizA
+
https://youtu.be/G8NRcVxSdz0
  
−
==Video Solution by SpreadTheMathLove==
+
==Video Solution 5 by SpreadTheMathLove==
  
 
https://www.youtube.com/watch?v=nhlpSATltRU
 
https://www.youtube.com/watch?v=nhlpSATltRU
  
 
~Ismail.maths93
 
~Ismail.maths93
 +
 +
== Video Solution 6 by Lucas637 ==
 +
https://www.youtube.com/watch?v=R1CtcZ2pWVk
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:38, 11 September 2025

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 (No Angle Bisectors!)

Amc10a2022q13.png

We can draw lines from \( C \) to a new point \( E \), and extend the line \( BC \) to a point \( F \) such that the line \( FD \) has the same slope of the line \( CE \) and line \( AB \).

Notice how we have two parallelograms, but more specifically, these parallelograms are just made up of \( 2 \times \triangle ABC \). We see that the parallelograms \( ABCE \) and \( ECFD \) are congruent, and therefore \( BC \cong AE \cong CF \cong ED \). \( BC = 5 \), so \( AE = 5 \), and because \( AE \cong ED \), we have \( ED = 5 \).

Finally, we have \( AE + ED = AD \), so \( AD = 5 + 5 =\) $\boxed{\textbf{(C) } 10}.$

And just like that! No angle bisectors needed! 😁

~Pinotation

~Image adopted from MRENTHUSIASM, thank you so much!

Solution 3 (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 4 (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 5 (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 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_13&oldid=260357"