Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (Added Mass Points to the title of solution 2)
 
(32 intermediate revisions by 18 users not shown)
Line 1: Line 1:
−
 
 
== 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==
 +
<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));
 +
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>
 +
 
 +
== 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));
 +
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>
 +
 
 +
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>
 +
 
 +
Thus,
 +
 
 +
<cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath>
 +
 
 +
Now let's apply the angle bisector theorem.
 +
 
 +
<cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath>
 +
 
 +
<cmath>\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}</cmath>
 +
 
 +
<cmath>\frac {180}{LP}=\frac {5}{2}</cmath>
 +
 
 +
<cmath>LP=\boxed {072}</cmath>
 +
 
 +
==Solution 2 (Mass Points)==
 +
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>
 +
 
 +
==Solution 3 (Law of Cosines Bash)==
 +
Using the diagram from solution <math>1</math>, we can also utilize the fact that <math>AMCP</math> forms a parallelogram. Because of that, we know that <math>AM = CP = 180</math>.
 +
 
 +
Applying the angle bisector theorem to <math>\triangle CKB</math>, we get that <math>\frac{KP}{PB} = \frac{225}{300} = \frac{3}{4}.</math> So, we can let <math>MK = KP = 3x</math> and <math>BP = 4x</math>.
 +
 
 +
Now, apply law of cosines on <math>\triangle CKP</math> and <math>\triangle CPB.</math>
 +
 
 +
If we let <math>\angle KCP = \angle PCB = \alpha</math>, 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 \alpha.</cmath>
 +
 
 +
Bashing those out, we get that <math>x = 15 \sqrt{13}</math> and <math>\cos \alpha = \frac{7}{10}.</math>
 +
 
 +
Since <math>\cos \alpha = \frac{7}{10}</math>, we can use the double angle formula to calculate that <math>\cos \left(2 \alpha \right) = -\frac{1}{50}.</math>
 +
 
 +
Now, apply Law of Cosines on <math>\triangle ABC</math> to find <math>AB</math>.
 +
 
 +
We get: <cmath>AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).</cmath>
 +
 
 +
Bashing gives <math>AB = 30 \sqrt{331}.</math>
 +
 
 +
From the angle bisector theorem on <math>\triangle ABC</math>, we know that <math>\frac{AL}{BL} = \frac{450}{300} = \frac{3}{2}.</math> So, <math>AL = 18 \sqrt{331}</math> and <math>BL = 12 \sqrt{331}.</math>
 +
 
 +
Now, we apply Law of Cosines on <math>\triangle ALC</math> and <math>\triangle BLC</math> in order to solve for the length of <math>LC</math>.
 +
 
 +
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 <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>
 +
 
 +
==Video Solution==
 +
https://youtu.be/2Xzjh6ae0MU
 +
 
 +
~IceMatrix
 +
 
 +
==Video Solution==
 +
https://youtu.be/kALrIDMR0dg
 +
 
 +
~Shreyas S
 +
 
 +
== Solution 4(Area Ratios) ==
 +
Note that we are given that <math>\overline{MK} = \overline{KP}</math>, that <math>\overline{AK} = \overline{CK}.</math> Note then that <math>\angle MKA = \angle CKB</math> by vertical angles. From this, we have <math>\triangle MKA \cong PKC.</math> This means that <math>\overline{CP}</math> is 180. Applying angle bisector theorem on <math>\triangle ACB</math> gives <math>\frac{\overline{AL}}{\overline{LB}} = \frac{450}{300} = \frac{3}{2}.</math> Applying it on <math>\triangle KCB</math> yields
 +
<cmath>\frac{\overline{KP}}{\overline{PB}} = \frac{225}{300} = \frac{3}{4}</cmath>
 +
Now we can proceed with area ratios. Suppose the area of <math>\triangle ACB = A.</math> This means that <cmath>[\triangle AKL] = \left(\frac{225}{225+225}\right)\left(\frac{3}{5}\right)A = \frac{3}{10}A</cmath>
 +
Continuing on <math>\triangle LPB</math> we have <cmath>[\triangle LPB] = \left(\frac{2}{2+3}\right)\left(\frac{1}{2}\right)\left(\frac{4}{4+3}\right) = \frac{4}{35}A</cmath>
 +
Since <math>\overline{AK}=\overline{KC}</math> <math>[\triangle KPL] = [\triangle AKB]-[\triangle AKL] - [\triangle LPB] = \frac{1}{2}A - \frac{3}{10}A - \frac{4}{35}A = \frac{3}{35}A.</math>
 +
Area ratios on <math>\triangle KCP</math> yield <math>[\triangle KCP] = \left(\frac{1}{2}\right)\left(\frac{3}{3+4}\right) = \frac{3}{14}.</math> Now, suppose <math>\overline{LP} = x.</math> We have that the ratio of areas of <math>\triangle LKP</math> and <math>\triangle PKC</math> is <math>\frac{x}{180}</math>
 +
and is also <math>\frac{\frac{3}{35}}{\frac{3}{14}}</math> and equating these gives <cmath>x = \boxed{72}</cmath>
  
 
== 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}}

Latest revision as of 18:52, 7 March 2024

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 (Mass Points)

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

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 \[x = \boxed{72}\]

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_5&oldid=216576"