Difference between revisions of "1986 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | + | <center><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); | ||
+ | D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); | ||
+ | dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s)); | ||
+ | D(D--Ea);D(Da--F);D(Fa--E); | ||
+ | MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); | ||
+ | </asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --> | ||
− | Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. | + | Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s. |
Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. | Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. | ||
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Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]: | Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]: | ||
− | < | + | <center><math>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></center> |
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>. | ||
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Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>[BEPD'] = c</math>, <math>[D'PD] = d</math>, <math>[DAF'P] = e</math> and <math>[F'D'P] = f</math> | Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>[BEPD'] = c</math>, <math>[D'PD] = d</math>, <math>[DAF'P] = e</math> and <math>[F'D'P] = f</math> | ||
− | + | 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 <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get: | Let the length of the segment be <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get: | ||
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{{AIME box|year=1986|num-b=8|num-a=10}} | {{AIME box|year=1986|num-b=8|num-a=10}} | ||
− | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 10:29, 25 April 2008
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
[hide]Solution
Solution 1
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 2
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.
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 |