Difference between revisions of "1986 AIME Problems/Problem 9"
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__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | === Solution 1 === | + | ===Solution 1 (mass points) === |
+ | <asy> | ||
+ | size(200); | ||
+ | pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); | ||
+ | pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425)); | ||
+ | /* construct remaining points */ | ||
+ | pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); | ||
+ | pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); | ||
+ | /* Construct P */ | ||
+ | pair P = IP(D--Ea,E--Fa); dot(MP("P",P,NE)); | ||
+ | pair X = IP(L(A,P,4), B--C); dot(MP("X",X,NW)); | ||
+ | pair Y = IP(L(B,P,4), C--A); dot(MP("Y",Y,NE)); | ||
+ | pair Z = IP(L(C,P,4), A--B); dot(MP("Z",Z,N)); | ||
+ | |||
+ | D(A--X); D(B--Y); D(C--Z); | ||
+ | D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); | ||
+ | MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); | ||
+ | </asy> | ||
+ | |||
+ | Construct cevians <math>AX</math>, <math>BY</math> and <math>CZ</math> through <math>P</math>. Place masses of <math>x,y,z</math> on <math>A</math>, <math>B</math> and <math>C</math> respectively; then <math>P</math> has mass <math>x+y+z</math>. | ||
+ | |||
+ | Notice that <math>Z</math> has mass <math>x+y</math>. On the other hand, by similar triangles, <math>\frac{CP}{CZ} = \frac{d}{AB}</math>. Hence by mass points we find that <cmath> \frac{x+y}{x+y+z} = \frac{d}{AB} </cmath> Similarly, we obtain <cmath> \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} </cmath> Summing these three equations yields <cmath> \frac{d}{AB} + \frac{d}{BC} + \frac{d}{CA} = \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} = \frac{2x+2y+2z}{x+y+z} = 2 </cmath> | ||
+ | |||
+ | Hence, <cmath> d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \boxed{306} </cmath> | ||
+ | |||
+ | === Solution 2 === | ||
<center><asy> | <center><asy> | ||
size(200); | size(200); | ||
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Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>. | Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>. | ||
− | === Solution | + | === Solution 3 === |
Define the points the same as above. | Define the points the same as above. | ||
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<math>\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math> | <math>\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math> | ||
answer follows after some hideous computation. | answer follows after some hideous computation. | ||
− | ===Solution | + | |
− | Refer to the diagram; let <math>a^2=[E'EP]</math>, <math>b^2=[D'DP]</math>, and <math>c^2=[F'FP]</math>. Now, note that <math>[E'BD]</math>, <math>[D'DP]</math>, and <math>[E'EP]</math> are similar, so through some similarities we find that <math>\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2</math>. Similarly, we find that <math>[D'AF]=(b+c)^2</math> and <math>[F'CE]=(c+a)^2</math>, so <math>[ABC]=(a+b+c)^2</math>. Now, again from similarity, it follows that <math>\frac{d}{510}=\frac{a+b}{a+b+c}</math>, <math>\frac{d}{450}=\frac{b+c}{a+b+c}</math>, and <math>\frac{d}{425}=\frac{c+a}{a+b+c}</math>, so adding these together, simplifying, and solving gives <math>d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}</math> | + | ===Solution 4=== |
+ | Refer to the diagram in solution 2; let <math>a^2=[E'EP]</math>, <math>b^2=[D'DP]</math>, and <math>c^2=[F'FP]</math>. Now, note that <math>[E'BD]</math>, <math>[D'DP]</math>, and <math>[E'EP]</math> are similar, so through some similarities we find that <math>\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2</math>. Similarly, we find that <math>[D'AF]=(b+c)^2</math> and <math>[F'CE]=(c+a)^2</math>, so <math>[ABC]=(a+b+c)^2</math>. Now, again from similarity, it follows that <math>\frac{d}{510}=\frac{a+b}{a+b+c}</math>, <math>\frac{d}{450}=\frac{b+c}{a+b+c}</math>, and <math>\frac{d}{425}=\frac{c+a}{a+b+c}</math>, so adding these together, simplifying, and solving gives <math>d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}</math> | ||
<math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>. | <math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>. | ||
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== See also == | == See also == |
Revision as of 01:08, 5 April 2012
Problem
In , , , and . An interior point is then drawn, and segments are drawn through parallel to the sides of the triangle. If these three segments are of an equal length , find .
Contents
Solution
Solution 1 (mass points)
Construct cevians , and through . Place masses of on , and respectively; then has mass .
Notice that has mass . On the other hand, by similar triangles, . Hence by mass points we find that Similarly, we obtain Summing these three equations yields
Hence,
Solution 2
Let the points at which the segments hit the triangle be called as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar (). The remaining three sections are parallelograms.
Since is a parallelogram, we find , and similarly . So . Thus . By the same logic, .
Since , we have the proportion:
Doing the same with , we find that . Now, .
Solution 3
Define the points the same as above.
Let , , , , and
The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.
Let the length of the segment be and the area of the triangle be , using the theorem, we get:
, , adding all these together and using we get
Using corresponding angles from parallel lines, it is easy to show that , since and are parallelograms, it is easy to show that
Now we have the side length ratio, so we have the area ratio , by symmetry, we have and
Substituting these into our initial equation, we have answer follows after some hideous computation.
Solution 4
Refer to the diagram in solution 2; let , , and . Now, note that , , and are similar, so through some similarities we find that . Similarly, we find that and , so . Now, again from similarity, it follows that , , and , so adding these together, simplifying, and solving gives .
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |