Difference between revisions of "1986 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
+ | === Simple Solution === | ||
+ | <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); | ||
+ | /*P copied from above solution*/ | ||
+ | pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); | ||
+ | </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. 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. | ||
+ | |||
+ | By similar triangles, <math>CF'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>FA=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Since <math>E'D=CA-FF'</math>, we have <math>900-\frac{33}{17}d=d</math>, so <math>d=\boxed{306}</math>. | ||
+ | |||
===Solution 1 === | ===Solution 1 === | ||
<asy> | <asy> |
Revision as of 10:07, 3 August 2018
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
Simple Solution
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.
By similar triangles, and . Since , we have , so .
Solution 1
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 and the 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 .
Solution 5
Refer to the diagram from Solution 2. Notice that because , , and are parallelograms, , , and .
Let . Then, because , , so . Simplifying the LHS and cross-multiplying, we have . From the same triangles, we can find that .
is also similar to . Since , . We now have , and . Cross multiplying, we have . Using the previous equation to substitute for , we have: This is a linear equation in one variable, and we can solve to get
- I did not show the multiplication in the last equation because most of it cancels out when solving.
(Note: I chose to be only because that is what I had written when originally solving. The solution would work with other choices for .)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.