Difference between revisions of "1999 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | [[Point]] <math>P_{}</math> is located inside [[ | + | [[Point]] <math>P_{}</math> is located inside [[triangle]] <math>ABC</math> so that [[angle]]s <math>PAB, PBC,</math> and <math>PCA</math> are all congruent. The sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relatively [[prime]] positive integers. Find <math>m+n.</math> |
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | <center><asy> | |
− | <!--This image does | + | real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ |
+ | pathpen = black +linewidth(0.65); pointpen = black; | ||
+ | pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); | ||
+ | D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); | ||
+ | |||
+ | /* constructing P, C is there as check */ | ||
+ | pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); | ||
+ | D(A--MP("P",P,NW)--B);D(P--C); | ||
+ | D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); | ||
+ | MP("13",(A+B)/2,S);MP("14",(A+C)/2,NW);MP("15",(C+B)/2,NE); | ||
+ | </asy></center> | ||
+ | <!--This image does exist now: [[Image:1999_AIME-14.png]]--> | ||
=== Solution 1 === | === Solution 1 === | ||
− | Drop [[perpendicular]]s from <math>P</math> to the three sides of <math>\triangle ABC</math> and let them meet <math>\overline{AB}, \overline{BC},</math> and <math>\overline{CA}</math> at <math>D, E,</math> and <math>F</math> respectively. Let <math>BE = x, CF = y,</math> and <math>AD = z</math>. We have that | + | Drop [[perpendicular]]s from <math>P</math> to the three sides of <math>\triangle ABC</math> and let them meet <math>\overline{AB}, \overline{BC},</math> and <math>\overline{CA}</math> at <math>D, E,</math> and <math>F</math> respectively. |
+ | |||
+ | <center><asy> | ||
+ | import olympiad; | ||
+ | real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other ways to construct Brocard */ | ||
+ | pathpen = black +linewidth(0.65); pointpen = black; | ||
+ | pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14)); | ||
+ | D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); | ||
+ | |||
+ | /* constructing P, C is there as check */ | ||
+ | pair Aa=A+(B-A)*dir(theta),Ba=B+(C-B)*dir(theta),Ca=C+(A-C)*dir(theta), P=IP(A--Aa,B--Ba); | ||
+ | D(A--MP("P",P,SSW)--B);D(P--C); | ||
+ | D(anglemark(B,A,P,30));D(anglemark(C,B,P,30));D(anglemark(A,C,P,30)); | ||
+ | MP("13",(A+B)/2,S);MP("14",(A+C)/2,NW);MP("15",(C+B)/2,NE); | ||
+ | |||
+ | /* constructing D,E,F as foot of perps from P */ | ||
+ | pair D=foot(P,A,B),E=foot(P,B,C),F=foot(P,C,A); | ||
+ | D(MP("D",D,NE)--P--MP("E",E,SSW),dashed);D(P--MP("F",F),dashed); | ||
+ | D(rightanglemark(P,E,C,15));D(rightanglemark(P,F,C,15));D(rightanglemark(P,D,A,15)); | ||
+ | </asy></center> | ||
+ | |||
+ | Let <math>BE = x, CF = y,</math> and <math>AD = z</math>. We have that | ||
<cmath> | <cmath> | ||
\begin{align*}DP&=z\tan\theta\ | \begin{align*}DP&=z\tan\theta\ |
Revision as of 11:18, 26 April 2008
Problem
Point is located inside triangle so that angles and are all congruent. The sides of the triangle have lengths and and the tangent of angle is where and are relatively prime positive integers. Find
Contents
[hide]Solution
Solution 1
Drop perpendiculars from to the three sides of and let them meet and at and respectively.
Let and . We have that We can then use the tool of calculating area in two ways On the other hand, We still need though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: Adding gives Recall that we found that . Plugging in , we get , giving us for an answer.
Solution 2
Let , , , , , and .
So by the Law of Cosines, we have: Adding these equations and rearranging, we have: Now , by Heron's formula.
Now the area of a triangle, , where and are sides on either side of an angle, . So, Adding these equations yields: Dividing by , we have: Thus, .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |