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

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== Problem ==
 
== Problem ==
  
 
Triangle <math>ABC</math> has <math>AC = 450</math> and <math>BC = 300</math>. Points <math>K</math> and <math>L</math> are located on <math>\overline{AC}</math> and <math>\overline{AB}</math> respectively so that <math>AK = CK</math>, and <math>\overline{CL}</math> is the angle bisector of angle <math>C</math>. Let <math>P</math> be the point of intersection of <math>\overline{BK}</math> and <math>\overline{CL}</math>, and let <math>M</math> be the point on line <math>BK</math> for which <math>K</math> is the midpoint of <math>\overline{PM}</math>. If <math>AM = 180</math>, find <math>LP</math>.
 
Triangle <math>ABC</math> has <math>AC = 450</math> and <math>BC = 300</math>. Points <math>K</math> and <math>L</math> are located on <math>\overline{AC}</math> and <math>\overline{AB}</math> respectively so that <math>AK = CK</math>, and <math>\overline{CL}</math> is the angle bisector of angle <math>C</math>. Let <math>P</math> be the point of intersection of <math>\overline{BK}</math> and <math>\overline{CL}</math>, and let <math>M</math> be the point on line <math>BK</math> for which <math>K</math> is the midpoint of <math>\overline{PM}</math>. If <math>AM = 180</math>, find <math>LP</math>.
  
== Solution ==
+
==Diagram==
Sorry, I fail to get the diagram up here, someone help me.
+
<center><asy>
 
+
import markers;
Since <math>K</math> is the midpoint of <math>\overline{PM}, \overline{AC}</math>.
+
defaultpen(fontsize(8));
 
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size(300);
Thus, <math>AK=CK,PK=MK</math> and the opposite angles are congruent.
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pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P;
 
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C = intersectionpoints(Circle(A,450), Circle(B,300))[0];
Therefore, <math>\bigtriangleup{AMK}</math> is congruent to <math>\bigtriangleupCPK</math> because of SAS
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K =  midpoint(A--C);
 +
L = (3*B+2*A)/5;
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P = extension(B,K,C,L);
 +
M = 2*K-P;
 +
draw(A--B--C--cycle);
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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></center>
  
angle <math>KMA</math> is congruent to <math>KPA</math> because of CPCTC
+
== Solution 1==
 +
<center><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));
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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></center>
  
That shows <math>\overline{AM}</math> is parallel to <math>\overline{CP}</math> (also <math>CL</math>)
+
Since <math>K</math> is the midpoint of <math>\overline{PM}</math> and <math>\overline{AC}</math>, quadrilateral <math>AMCP</math> is a parallelogram, which implies <math>AM||LP</math> and  <math>\bigtriangleup{AMB}</math> is similar to <math>\bigtriangleup{LPB}</math>
  
That makes <math>\bigtriangleup{AMB}</math> congruent to <math>\bigtriangleup{LPB}</math>
+
Thus,
 
 
Thus, <math>\frac {AM}{LP}=\frac {AB}{LB}</math>
 
  
 
<cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath>
 
<cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath>
  
Now let apply angle bisector thm.
+
Now let's apply the angle bisector theorem.
  
 
<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>
 
<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>
Line 32: Line 64:
  
 
<cmath>LP=\boxed {072}</cmath>
 
<cmath>LP=\boxed {072}</cmath>
 +
 +
==Solution 2==
 +
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem:
 +
<cmath>\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL</cmath>
 +
So, we can weight <math>A</math> as <math>2</math> and <math>B</math> as <math>3</math> and <math>L</math> as <math>5</math>. Since <math>K</math> is the midpoint of <math>A</math> and <math>C</math>, the weight of <math>A</math> is equal to the weight of <math>C</math>, which equals <math>2</math>.
 +
Also, since the weight of <math>L</math> is <math>5</math> and <math>C</math> is <math>2</math>, we can weight <math>P</math> as <math>7</math>.
 +
 +
By the definition of mass points, <cmath>\frac{LP}{CP}=\frac{2}{5}\implies LP=\frac{2}{5}CP</cmath>
 +
By vertical angles, angle <math>MKA =</math> angle <math>PKC</math>.
 +
Also, it is given that <math>AK=CK</math> and <math>PK=MK</math>.
 +
 +
By the SAS congruence, <math>\triangle MKA</math> = <math>\triangle PKC</math>. So, <math>MA</math> = <math>CP</math> = <math>180</math>.
 +
Since <math>LP=\frac{2}{5}CP</math>, <math>LP = \frac{2}{5}(180) = \boxed{072}</math>
 +
 +
==Video Solution==
 +
https://youtu.be/2Xzjh6ae0MU
 +
 +
~IceMatrix
 +
 +
==Video Solution==
 +
https://youtu.be/kALrIDMR0dg
 +
 +
~Shreyas S
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=4|num-a=6}}
 
{{AIME box|year=2009|n=I|num-b=4|num-a=6}}
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 18:21, 18 June 2020

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 $LP=\frac{2}{5}CP$, $LP = \frac{2}{5}(180) = \boxed{072}$

Video Solution

https://youtu.be/2Xzjh6ae0MU

~IceMatrix

Video Solution

https://youtu.be/kALrIDMR0dg

~Shreyas S

See also

2009 AIME I (ProblemsAnswer KeyResources)
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

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