# Difference between revisions of "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. ! Missing $inserted.)\triangle CKP$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$(Error compiling LaTeX. ! Missing$ inserted.)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. ! Missing $inserted.)AB = 30 \sqrt{331}.$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:

<cmath>(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \cdot LC \cdot \frac{7}{10}</cmath> <cmath>(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}</cmath>

The first equation gives$(Error compiling LaTeX. ! Missing$ inserted.)LC = 252$or$378$and the second gives$LC = 252 or 168$. The only value that satisfies both equations is$ (Error compiling LaTeX. ! Missing $inserted.)LC = 252$, and since$LP = LC - PC$, we have $$LC = 252 - 180 = \boxed{072}.$$

~IceMatrix

~Shreyas S