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Difference between revisions of "2022 AMC 10A Problems/Problem 13"
(→‎Video Solution 4 by SpreadTheMathLove)
 
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<math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12</math>
 
<math>\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12</math>
  
−
==Solution==
+
==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>
  
−
<b>DIAGRAM IN PROGRESS.
+
~MRENTHUSIASM
  
−
WILL BE DONE TOMORROW, WAIT FOR ME THANKS.</b>
+
==Solution 1 (Angle Bisector Theorem and Similar Triangles)==
  
−
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>
+
Suppose that <math>\overline{BD}</math> intersects <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>  
 
Let <math>AB=AY=2x.</math> By the Angle Bisector Theorem, we have <math>AC=3x,</math> or <math>YC=x.</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>
+
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>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 3 (Cheaty solution if you are almost out of time) ==
+
==Solution 2 (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>.
 +
 
 +
By adding this extra line, we now have many pairs of similar triangles. 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>. Additionally, <math>\triangle BPN \sim \triangle ADN</math> (with an unknown ratio). It is also true that <math>\triangle BAN \cong \triangle MAN</math>.
 +
 
 +
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 thus <math>[OMNP] = 4x - x = 3x</math>. Thus, <math>[\triangle MAN] = [\triangle BAN] = 5x</math>.
 +
 
 +
Finally, we have <math>\frac{[\triangle BAN]}{[\triangle BPN]} = \frac{5x}{x} = 5</math>, and because these triangles share the same height <math>\frac{AN}{PN} = 5</math>. Notice that these side lengths are corresponding side lengths of the similar triangles <math>BPN</math> and <math>ADN</math>. This means that <math>AD = 5\cdot BP = \boxed{\textbf{(C) } 10}</math>.
 +
 
 +
~mathboy100
 +
 
 +
== Solution 3 (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.
 +
 
 +
By the Angle Bisector Theorem, <math>AB:AC = 2:3</math>. Thus, assume that <math>AB = 4</math>, and <math>AC = 6</math>.
 +
 
 +
Let the perpendicular from <math>A</math> to <math>BC</math> be <math>AM</math>.
 +
 
 +
Using Heron's formula, <cmath>[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}.</cmath>
 +
 
 +
Hence, <cmath>AM = \frac{\frac{15}{4}\sqrt{7}}{\frac{5}{2}} = \frac{3}{2}\sqrt{7}.</cmath>
 +
 
 +
Next, we have <cmath>BM^2 + AM^2 = AB^2</cmath>
 +
<cmath>\therefore BM = \sqrt{16 - \frac{63}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \textrm{ and } PM = \frac{3}{2}.</cmath>
 +
 
 +
The slope of line <math>AP</math> is thus <cmath>\frac{-\frac{3}{2}\sqrt{7}}{\frac{3}{2}} = -\sqrt{7}.</cmath>
 +
 
 +
Therefore, since the slopes of perpendicular lines have a product of <math>-1</math>, the slope of line <math>BD</math> is <math>\frac{1}{\sqrt{7}}</math>. This means that we can solve for the coordinates of <math>D</math>:
 +
 
 +
<cmath>y = \frac{3}{2}\sqrt{7}</cmath>
 +
<cmath>y = \frac{1}{\sqrt{7}}x</cmath>
 +
<cmath>\frac{1}{\sqrt{7}}x = \frac{3}{2}\sqrt{7}</cmath>
 +
<cmath>x = \frac{7 \cdot 3}{2} = \frac{21}{2}</cmath>
 +
<cmath>D = \left(\frac{21}{2}, \frac{3}{2}\sqrt{7}\right).</cmath>
 +
 
 +
We also know that the coordinates of <math>A</math> are <math>\left(\frac{1}{2}, \frac{3}{2}\sqrt{7}\right)</math>, because <math>BM = \frac{1}{2}</math> and <math>AM = \frac{3}{2}\sqrt{7}</math>.
 +
 
 +
Since the <math>y</math>-coordinates of <math>A</math> and <math>D</math> are the same, and their <math>x</math>-coordinates differ by <math>10</math>, the distance between them is <math>10</math>. Our answer is <math>\boxed{\textbf{(C) }10}.</math>
 +
 
 +
~mathboy100
 +
 
 +
== Solution 4 (Assumption) ==
 
<asy>
 
<asy>
 
size(300);
 
size(300);
Line 45: Line 116:
 
draw(rightanglemark(D,XX,P));
 
draw(rightanglemark(D,XX,P));
 
</asy>
 
</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 \sim \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 \sim \bigtriangleup ABP</cmath> and therefore <math>\frac{DA}{AB} = \frac{AB}{BP}</math>, so <math>AD=\boxed{\textbf{(C) }10}</math>, and we're done!
  
 
~[[User:Bxiao31415|Bxiao31415]]
 
~[[User:Bxiao31415|Bxiao31415]]
  
−
==Video Solution 1 (Quick and Simple)==
+
==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 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
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:21, 23 January 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

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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