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

Revision as of 17:22, 7 August 2010

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 $LP$ .

Solution

$[asy] import markers; defaultpen(fontsize(8)); size(300); pair A=(0,0), B=(30*sqrt(331),0), C, K, L, M, P; C = intersectionpoints(Circle(A,450), Circle(B,300))[0]; K = midpoint(A--C); L = (3*B+2*A)/5; P = extension(B,K,C,L); M = 2*K-P; draw(A--B--C--cycle); draw(C--L);draw(B--M--A); markangle(n=1,radius=15,A,C,L,marker(markinterval(stickframe(n=1),true))); markangle(n=1,radius=15,L,C,B,marker(markinterval(stickframe(n=1),true))); dot(A^^B^^C^^K^^L^^M^^P); label("$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("$72$",(L+P)/2,(-1,0));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 lets apply the angle bisector theorem.

$\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}$

$\frac {AM}{LP}=\frac {AL}{LB}+1=\frac {5}{2}$

$\frac {180}{LP}=\frac {5}{2}$

$LP=\boxed {072}$

2009 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2009 AIME I Problems/Problem 5"

Revision as of 17:22, 7 August 2010

Problem

Solution

See also

@@ Line 4: / Line 4: @@
 == Solution ==
-{{image}}
+<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("$72$",(L+P)/2,(-1,0));label("$300$",(B+C)/2,(1,1));
+</asy></center>
-Since <math>K</math> is the midpoint of <math>\overline{PM}, \overline{AC}</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>
-Thus, <math>AK=CK,PK=MK</math> and the opposite angles are congruent.
-Therefore, <math>\bigtriangleup{AMK}</math> is congruent to <math>\bigtriangleup{CPK}</math> because of SAS
-<math>\angle{KMA}</math> is congruent to <math>\angle{KPC}</math> because of CPCTC
-That shows <math>\overline{AM}</math> is parallel to <math>\overline{CP}</math> (also <math>CL</math>)
-That makes <math>\bigtriangleup{AMB}</math> similar to <math>\bigtriangleup{LPB}</math>
 Thus,