Difference between revisions of "1986 AIME Problems/Problem 9"
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<center><asy> | <center><asy> | ||
size(200); | size(200); | ||
− | pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); | + | pathpen = black; |
− | pair C=(0,0),A=(510,0) | + | pointpen = black + linewidth(0.6); |
− | / | + | pen s = fontsize(10); |
− | pair Da=IP( | + | |
− | pair D=IP(Ea--(Ea+A-C),A--B) | + | // Define points |
− | + | pair C = (0,0), A = (510,0); | |
− | 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)); | + | pair B = IP(circle(C,450),circle(A,425)); |
− | + | ||
− | + | // Construct remaining points | |
− | / | + | pair Da = IP(circle(A,289),A--B); |
− | pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); | + | pair E = IP(circle(C,324),B--C); |
+ | pair Ea = IP(circle(B,270),B--C); | ||
+ | pair D = IP(Ea--(Ea+A-C),A--B); | ||
+ | pair F = IP(Da--(Da+C-B),A--C); | ||
+ | pair Fa = IP(E--(E+A-B),A--C); | ||
+ | |||
+ | // Draw the main triangle | ||
+ | draw(A--B--C--cycle); | ||
+ | dot(MP("A",A,s)); | ||
+ | dot(MP("B",B,N,s)); | ||
+ | dot(MP("C",C,s)); | ||
+ | |||
+ | // Mark and draw the other points | ||
+ | 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)); | ||
+ | |||
+ | // Draw connecting lines | ||
+ | draw(D--Ea); | ||
+ | draw(Da--F); | ||
+ | draw(Fa--E); | ||
+ | |||
+ | // Label distances | ||
+ | label("450", (B+C)/2, NW); | ||
+ | label("425", (A+B)/2, NE); | ||
+ | label("510", (A+C)/2, S); | ||
+ | |||
+ | // Additional point P | ||
+ | pair P = IP(D--Ea, E--Fa); | ||
+ | dot(MP("P",P,N)); | ||
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --> | </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 | + | 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 DD'P \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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=== Solution 6 === | === Solution 6 === | ||
− | <asy> | + | <center><asy> |
size(200); | size(200); | ||
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); | pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10); | ||
Line 112: | Line 144: | ||
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); | 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); | 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); | 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); | MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2); | ||
− | </asy> | + | /*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 --> | ||
Refer to the diagram above. Notice that because <math>CE'PF</math>, <math>AF'PD</math>, and <math>BD'PE</math> are parallelograms, <math>\overline{DD'} = 425-d</math>, <math>\overline{EE'} = 450-d</math>, and <math>\overline{FF'} = 510-d</math>. | Refer to the diagram above. Notice that because <math>CE'PF</math>, <math>AF'PD</math>, and <math>BD'PE</math> are parallelograms, <math>\overline{DD'} = 425-d</math>, <math>\overline{EE'} = 450-d</math>, and <math>\overline{FF'} = 510-d</math>. | ||
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(Note: I chose <math>F'P</math> to be <math>x</math> only because that is what I had written when originally solving. The solution would work with other choices for <math>x</math>.) | (Note: I chose <math>F'P</math> to be <math>x</math> only because that is what I had written when originally solving. The solution would work with other choices for <math>x</math>.) | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/FWmrHV1dWPM?t=396 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Revision as of 12:08, 29 July 2024
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.
By similar triangles, and . Since , we have , so .
Solution 2
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 3
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 4
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 5
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 6
Refer to the diagram above. 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 .)
Video Solution by OmegaLearn
https://youtu.be/FWmrHV1dWPM?t=396
~ pi_is_3.14
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.