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Difference between revisions of "2009 AIME I Problems/Problem 5"

(→‎Solution 3 (Law of Cosines Bash))
(→‎Solution 3 (Law of Cosines Bash))
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<cmath>(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \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 <math>LC = 252</math> or <math>378</math> and the second gives <math>LC = 252 or 168</math>.
+
The first equation gives <math>LC = 252</math> or <math>378</math> and the second gives <math>LC = 252</math> or <math>168</math>.
  
 
The only value that satisfies both equations is <math>LC = 252</math>, and since <math>LP = LC - PC</math>, we have <cmath>LC = 252 - 180 = \boxed{072}.</cmath>
 
The only value that satisfies both equations is <math>LC = 252</math>, and since <math>LP = LC - PC</math>, we have <cmath>LC = 252 - 180 = \boxed{072}.</cmath>

Revision as of 00:11, 6 March 2021

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 \[LC = 252 - 180 = \boxed{072}.\]

Video Solution

https://youtu.be/2Xzjh6ae0MU

~IceMatrix

Video Solution

https://youtu.be/kALrIDMR0dg

~Shreyas S

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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