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
![[asy] 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]](http://latex.artofproblemsolving.com/5/e/7/5e76f7278340e0e455b61e12aad27c54018c3d18.png)
Solution 1
Drop perpendiculars from to the three sides of
and let them meet
and
at
and
respectively.
![[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]](http://latex.artofproblemsolving.com/e/e/b/eebcd8926b0fcb54f3ffc9dfb257f2ab733331d2.png)
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 |