2009 AIME I Problems/Problem 5

Problem

Triangle $ABC$ has $AC = 450$ and $BC = 300$ . Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$ , and $\overline{CL}$ is the angle bisector of angle $C$ . Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$ , and let $M$ be the point on line $BK$ for which $K$ is the midpoint of $\overline{PM}$ . If $AM = 180$ , find $LP$ .

Diagram

$[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1)); label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1)); label("$P$",P,(1,1)); label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2)); label("$300$",(B+C)/2,(1,1)); [/asy]$

Solution 1

$[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$A$",A,(-1,-1));label("$B$",B,(1,-1));label("$C$",C,(1,1)); label("$K$",K,(0,2));label("$L$",L,(0,-2));label("$M$",M,(-1,1)); label("$P$",P,(1,1)); label("$180$",(A+M)/2,(-1,0));label("$180$",(P+C)/2,(-1,0));label("$225$",(A+K)/2,(0,2));label("$225$",(K+C)/2,(0,2)); label("$300$",(B+C)/2,(1,1)); [/asy]$

Since $K$ is the midpoint of $\overline{PM}$ and $\overline{AC}$ , quadrilateral $AMCP$ is a parallelogram, which implies $AM||LP$ and $\bigtriangleup{AMB}$ is similar to $\bigtriangleup{LPB}$

Thus,

$\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1$

Now let's apply the angle bisector theorem.

$\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}$

$\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}$

$\frac {180}{LP}=\frac {5}{2}$

$LP=\boxed {072}$

Solution 2

Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: $\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL$ So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$ . Since $K$ is the midpoint of $A$ and $C$ , the weight of $A$ is equal to the weight of $C$ , which equals $2$ . Also, since the weight of $L$ is $5$ and $C$ is $2$ , we can weight $P$ as $7$ .

By the definition of mass points, $\frac{LP}{CP}=\frac{2}{5}\implies LP=\frac{2}{5}CP$ By vertical angles, angle $MKA =$ angle $PKC$ . Also, it is given that $AK=CK$ and $PK=MK$ .

By the SAS congruence, $\triangle MKA$ = $\triangle PKC$ . So, $MA$ = $CP$ = $180$ . Since $LP=\frac{2}{5}CP$ , $LP = \frac{2}{5}(180) = \boxed{072}$

Solution 3 (Law of Cosines Bash)

Using the diagram from solution $1$ , we can also utilize the fact that $AMCP$ forms a parallelogram. Because of that, we know that $AM = CP = 180$ .

Applying the angle bisector theorem to $\triangle CKB$ , we get that $\frac{KP}{PB} =$ \frac{225}{300} = \frac{3}{4}. $So, we can let$ MK = KP = 3x $and$ BP = 4x$.

Now, apply law of cosines on$ (Error compiling LaTeX. Unknown error_msg)\triangle CKP $and$ \triangle CPB. $If we let$ \angle KCP = \angle PCB = \alpha$, then the law of cosines gives the following system of equations:

<cmath>9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha</cmath> <cmath> 16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos a\lpha.</cmath>

Bashing those out, we get that$ (Error compiling LaTeX. Unknown error_msg)x = 15 \sqrt{13} $and$ \cos \alpha = \frac{7}{10}. $Since$ \cos \alpha = \frac{7}{10} $, we can use the double angle formula to calculate that$ \cos 2 \cdot \alpha = -\frac{1}{50}. $Now, apply Law of Cosines on$ \triangle ABC $to find$ AB$.

We get: <cmath>AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).</cmath>

Bashing gives$ (Error compiling LaTeX. Unknown error_msg)AB = 30 \sqrt{331}. $From the angle bisector theorem on$ \triangle ABC $, we know that$ \frac{AL}{BL} = \frac{450}{300} = \frac[3}{2}. $So,$ AL = 18 \sqrt{331} $and$ BL = 12 \sqrt{331}. $Now, we apply Law of Cosines on$ \triangle ALC $and$ \triangle BLC $in order to solve for the length of$ LC$.

We get the following system:

The first equation gives$ (Error compiling LaTeX. Unknown error_msg)LC = 252 $or$ 378 $and the second gives$ LC = 252 or 168$.

The only value that satisfies both equations is$ (Error compiling LaTeX. Unknown error_msg)LC = 252 $, and since$ LP = LC - PC$, we have $LC = 252 - 180 = \boxed{072}.$

Video Solution

https://youtu.be/2Xzjh6ae0MU

~IceMatrix

Video Solution

https://youtu.be/kALrIDMR0dg

~Shreyas S

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2009 AIME I Problems/Problem 5

Contents

Problem

Diagram

Solution 1

Solution 2

Solution 3 (Law of Cosines Bash)

Video Solution

Video Solution

See also