Difference between revisions of "2009 AIME I Problems/Problem 5"
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<cmath>LP=\boxed {072}</cmath> | <cmath>LP=\boxed {072}</cmath> | ||
− | ==Solution 2== | + | ==Solution 2 (Mass Points)== |
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: | 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> | <cmath>\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL</cmath> | ||
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By the SAS congruence, <math>\triangle MKA</math> = <math>\triangle PKC</math>. So, <math>MA</math> = <math>CP</math> = <math>180</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> | 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== | ==Video Solution== | ||
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~IceMatrix | ~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 == |
Latest revision as of 19:52, 7 March 2024
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
.
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]](http://latex.artofproblemsolving.com/5/b/d/5bd5ad7e87b56ebff30bc90b55c084d90c0f16c9.png)
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]](http://latex.artofproblemsolving.com/5/b/d/5bd5ad7e87b56ebff30bc90b55c084d90c0f16c9.png)
Since is the midpoint of
and
, quadrilateral
is a parallelogram, which implies
and
is similar to
Thus,
Now let's apply the angle bisector theorem.
Solution 2 (Mass Points)
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
.
By the definition of mass points,
By vertical angles, angle
angle
.
Also, it is given that
and
.
By the SAS congruence, =
. So,
=
=
.
Since
,
Solution 3 (Law of Cosines Bash)
Using the diagram from solution , we can also utilize the fact that
forms a parallelogram. Because of that, we know that
.
Applying the angle bisector theorem to , we get that
So, we can let
and
.
Now, apply law of cosines on and
If we let , then the law of cosines gives the following system of equations:
Bashing those out, we get that and
Since , we can use the double angle formula to calculate that
Now, apply Law of Cosines on to find
.
We get:
Bashing gives
From the angle bisector theorem on , we know that
So,
and
Now, we apply Law of Cosines on and
in order to solve for the length of
.
We get the following system:
The first equation gives or
and the second gives
or
.
The only value that satisfies both equations is , and since
, we have
Video Solution
~IceMatrix
Video Solution
~Shreyas S
Solution 4(Area Ratios)
Note that we are given that , that
Note then that
by vertical angles. From this, we have
This means that
is 180. Applying angle bisector theorem on
gives
Applying it on
yields
Now we can proceed with area ratios. Suppose the area of
This means that
Continuing on
we have
Since
Area ratios on
yield
Now, suppose
We have that the ratio of areas of
and
is
and is also
and equating these gives
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.