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# 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 $\triangle CKP$ and $\triangle CPB.$

If we let $\angle KCP = \angle PCB = \alpha$, then the law of cosines gives the following system of equations:

$$9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha$$ $$16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos \alpha.$$

Bashing those out, we get that $x = 15 \sqrt{13}$ and $\cos \alpha = \frac{7}{10}.$

Since $\cos \alpha = \frac{7}{10}$, we can use the double angle formula to calculate that $\cos \left(2 \alpha \right) = -\frac{1}{50}.$

Now, apply Law of Cosines on $\triangle ABC$ to find $AB$.

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

Bashing gives $AB = 30 \sqrt{331}.$

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:

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

The first equation gives $LC = 252$ or $378$ and the second gives $LC = 252$ or $168$.

The only value that satisfies both equations is $LC = 252$, and since $LP = LC - PC$, we have $$LC = 252 - 180 = \boxed{072}.$$

~IceMatrix

~Shreyas S

## Solution 4(Area Ratios)

Note that we are given that $\overline{MK} = \overline{KP}$, that $\overline{AK} = \overline{CK}.$ Note then that $\angle MKA = \angle CKB$ by vertical angles. From this, we have $\triangle MKA \cong PKC.$ This means that $\overline{CP}$ is 180. Applying angle bisector theorem on $\triangle ACB$ gives $\frac{\overline{AL}}{\overline{LB}} = \frac{450}{300} = \frac{3}{2}.$ Applying it on $\triangle KCB$ yields $$\frac{\overline{KP}}{\overline{PB}} = \frac{225}{300} = \frac{3}{4}$$ Now we can proceed with area ratios. Suppose the area of $\triangle ACB = A.$ This means that $$[\triangle AKL] = \left(\frac{225}{225+225}\right)\left(\frac{3}{5}\right)A = \frac{3}{10}A$$ Continuing on $\triangle LPB$ we have $$[\triangle LPB] = \left(\frac{2}{2+3}\right)\left(\frac{1}{2}\right)\left(\frac{4}{4+3}\right) = \frac{4}{35}A$$ Since $\overline{AK}=\overline{KC}$ $[\triangle KPL] = [\triangle AKB]-[\triangle AKL] - [\triangle LPB] = \frac{1}{2}A - \frac{3}{10}A - \frac{4}{35}A = \frac{3}{35}A.$ Area ratios on $\triangle KCP$ yield $[\triangle KCP] = \left(\frac{1}{2}\right)\left(\frac{3}{3+4}\right) = \frac{3}{14}.$ Now, suppose $\overline{LP} = x.$ We have that the ratio of areas of $\triangle LKP$ and $\triangle PKC$ is $\frac{x}{180}$ and is also $\frac{\frac{3}{35}}{\frac{3}{14}}$ and equating these gives $$x = \boxed{72}$$