Difference between revisions of "2009 AIME I Problems/Problem 5"
IMOJonathan (talk | contribs) (This proposes a second solution.) |
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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 == | + | == Solution 1== |
<center><asy> | <center><asy> | ||
import markers; | import markers; | ||
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<cmath>LP=\boxed {072}</cmath> | <cmath>LP=\boxed {072}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | <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("$y$",(P+C)/2,(-1,0));label("$x$",(A+K)/2,(0,2));label("$x$",(K+C)/2,(0,2)); | ||
+ | label("$2y/5$",(L+P)/2,(-1,0));label("$180$",(B+C)/2,(1,1)); | ||
+ | </asy></center> | ||
+ | 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}\implies3BL=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>. | ||
+ | <cmath>\n</cmath> | ||
+ | By the definition of mass points, <cmath>\frac{LP}{CP}=\frac{2}{5}\impliesLP=\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>. | ||
+ | <cmath>\n</cmath>By the SAS congruence, triangle <math>MKA</math> = triangle <math>PKC</math>. So, <math>MA</math> = <math>CP</math> = 180. | ||
+ | Since <math>LP=\frac{2}{5}CP</math>, <math>LP = \frac{2}{5}180 = \boxed{072}</math> | ||
+ | |||
== See also == | == See also == |
Revision as of 21:49, 30 August 2014
Contents
Problem
Triangle has and . Points and are located on and respectively so that , and is the angle bisector of angle . Let be the point of intersection of and , and let be the point on line for which is the midpoint of . If , find .
Solution 1
Since is the midpoint of and , quadrilateral is a parallelogram, which implies and is similar to
Thus,
Now lets apply the angle bisector theorem.
Solution 2
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: So, we can weight as and as and as . Since is the midpoint of and , the weight of is equal to the weight of , which equals . Also, since the weight of is and is , we can weight as .
\[\n\] (Error compiling LaTeX. Unknown error_msg)
By the definition of mass points,
\[\frac{LP}{CP}=\frac{2}{5}\impliesLP=\frac{2}{5}CP\] (Error compiling LaTeX. Unknown error_msg)
By vertical angles, angle angle . Also, it is given that and .
\[\n\] (Error compiling LaTeX. Unknown error_msg)
By the SAS congruence, triangle = triangle . So, = = 180.
Since ,
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 |
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