Difference between revisions of "1999 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
+ | [[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__ | ||
== Solution == | == Solution == | ||
+ | <center><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("15",(A+C)/2,NW);MP("14",(C+B)/2,NE); | ||
+ | </asy></center> | ||
+ | <!--This image does exist now: [[Image:1999_AIME-14.png]]--> | ||
+ | |||
+ | === 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. | ||
+ | |||
+ | <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("15",(A+C)/2,NW);MP("14",(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> | ||
+ | \begin{align*}DP&=z\tan\theta\ | ||
+ | EP&=x\tan\theta\ | ||
+ | FP&=y\tan\theta\end{align*} | ||
+ | </cmath> | ||
+ | We can then use the tool of calculating area in two ways | ||
+ | <cmath> | ||
+ | \begin{align*}[ABC]&=[PAB]+[PBC]+[PCA]\ | ||
+ | &=\frac{1}{2}(13)(z\tan\theta)+\frac{1}{2}(14)(x\tan\theta)+\frac{1}{2}(15)(y\tan\theta)\ | ||
+ | &=\frac{1}{2}\tan\theta(13z+14x+15y)\end{align*} | ||
+ | </cmath> | ||
+ | On the other hand, | ||
+ | <cmath> | ||
+ | \begin{align*}[ABC]&=\sqrt{s(s-a)(s-b)(s-c)}\ | ||
+ | &=\sqrt{21\cdot6\cdot7\cdot8}\ | ||
+ | &=84\end{align*} | ||
+ | </cmath> | ||
+ | We still need <math>13z+14x+15y</math> though. We have all these [[right triangle]]s and we haven't even touched [[Pythagorean theorem|Pythagoras]]. So we give it a shot: | ||
+ | <cmath> | ||
+ | \begin{align}x^2+x^2\tan^2\theta&=z^2\tan^2\theta+(13-z)^2\ | ||
+ | z^2+z^2\tan^2\theta&=y^2\tan^2\theta+(15-y)^2\ | ||
+ | y^2+y^2\tan^2\theta&=x^2\tan^2\theta+(14-x)^2\end{align} | ||
+ | </cmath> | ||
+ | Adding <math>(1) + (2) + (3)</math> gives | ||
+ | <cmath> | ||
+ | \begin{align*}x^2+y^2+z^2&=(14-x)^2+(15-y)^2+(13-z)^2\ | ||
+ | \Rightarrow13z+14x+15y&=295\end{align*} | ||
+ | </cmath> | ||
+ | Recall that we found that <math>[ABC]=\frac{1}{2}\tan\theta(13z+14x+15y)=84</math>. Plugging in <math>13z+14x+15y=295</math>, we get <math>\tan\theta=\frac{168}{295}</math>, giving us <math>\boxed{463}</math> for an answer. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>, <math>PA=x</math>, <math>PB=y</math>, and <math>PC=z</math>. | ||
+ | |||
+ | So by the [[Law of Cosines]], we have: | ||
+ | <cmath> | ||
+ | \begin{align*}x^2 &= z^2 + b^2 - 2bz\cos{\theta}\ | ||
+ | y^2 &= x^2 + c^2 - 2cx\cos{\theta}\ | ||
+ | z^2 &= y^2 + a^2 - 2ay\cos{\theta}\end{align*} | ||
+ | </cmath> | ||
+ | Adding these equations and rearranging, we have: | ||
+ | <cmath> | ||
+ | a^2 + b^2 + c^2 = (2bz + 2cx + 2ay)\cos{\theta}\qquad(1) | ||
+ | </cmath> | ||
+ | Now <math>[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84</math>, by [[Heron's formula]]. | ||
+ | |||
+ | Now the area of a triangle, <math>[A] = \frac {mn\sin{\beta}}{2}</math>, where <math>m</math> and <math>n</math> are sides on either side of an angle, <math>\beta</math>. So, | ||
+ | <cmath> | ||
+ | \begin{align*}[CAP] &= \frac {bz\sin{\theta}}{2}\ | ||
+ | [ABP] &= \frac {cx\sin{\theta}}{2}\ | ||
+ | [BCP] &= \frac {ay\sin{\theta}}{2}\end{align*} | ||
+ | </cmath> | ||
+ | Adding these equations yields: | ||
+ | <cmath>\begin{align*}[ABC]= 84 &= \frac {(bz + cx + ay)\sin{\theta}}{2}\ | ||
+ | \Rightarrow 168&= (bz + cx + ay)\sin{\theta}\qquad (2)\end{align*} | ||
+ | </cmath> | ||
+ | Dividing <math>(2)</math> by <math>(1)</math>, we have: | ||
+ | <cmath> | ||
+ | \begin{align*}\frac {168}{a^2 + b^2 + c^2} &= \frac {(bz + cx + ay)\sin{\theta}}{(2bz + 2cx + 2ay)\cos{\theta}}\ | ||
+ | \Rightarrow \tan{\theta} = \frac {336}{a^2 + b^2 + c^2} &= \frac {336}{14^2 + 15^2 + 13^2} = \frac {336}{590} = \frac {168}{295}\end{align*} | ||
+ | </cmath> | ||
+ | Thus, <math>m + n = 168 + 295 = \boxed{463}</math>. | ||
+ | |||
+ | Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by | ||
+ | <cmath>cot(\theta)=\frac{a^2+b^2+c^2}{4\Delta}</cmath> | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Let <math>\angle{PAB} = \angle{PBC} = \angle{PCA} = x.</math> Then, using Law of Cosines on the three triangles containing vertex <math>P,</math> we have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | b^2 &= a^2 + 169 - 26a \cos x \ | ||
+ | c^2 &= b^2 + 196 - 28b \cos x \ | ||
+ | a^2 &= c^2 + 225 - 30c \cos x. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Add the three equations up and rearrange to obtain <cmath>(13a + 14b + 15c) \cos x = 295.</cmath> Also, using <math>[ABC] = \frac{1}{2}ab \sin \angle C</math> we have <cmath>[ABC] = [APB] + [BPC] + [CPA] = \dfrac{\sin x}{2}(13a + 14b + 15c) = 84 \iff (13a + 14b + 15c) \sin x = 168.</cmath> Divide the two equations to obtain <math>\tan x = \frac{168}{295} \iff \boxed{463}.~\square</math> | ||
+ | |||
+ | |||
+ | === Solution 4 (Law of sines) === | ||
+ | Firstly, denote angles <math>ABC</math>, <math>BCA</math>, and <math>CAB</math> as <math>B</math>, <math>A</math>, and <math>C</math> respectively. Let <math>\angle{PAB}=x</math>. | ||
+ | Notice that by angle chasing that <math>\angle{BPC}=180-C</math> and <math>\angle{BPA}=180-B</math>. | ||
+ | Using the nice properties of the 13-14-15 triangle, we have <math>\sin B = \frac{12}{13}</math> and <math>\sin C = \frac{4}{5}</math>. <math>\cos C</math> is easily computed, so we have <math>\cos C=\frac{3}{5}</math>. | ||
+ | |||
+ | Using Law of Sines, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{BP}{\sin x} &= \frac{13}{\sin (180 - B)} \ | ||
+ | \frac{BP}{\sin (C - x)} &= \frac{14}{\sin (180 - C)} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | hence, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{BP}{\sin x} &= \frac{13}{\sin B} \ | ||
+ | \frac{BP}{\sin (C - x)} &= \frac{14}{\sin C} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Now, computation carries the rest. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{13 \sin x}{\sin B} &= \frac{14 \sin (C-x)}{\sin C} \ | ||
+ | \frac{169 \sin x}{12} &= \frac{210 \sin (C-x)}{12} \ | ||
+ | 169 \sin x &= 210 (\sin C \cos x - \cos C \sin x) \ | ||
+ | 169 \sin x &= 210 (\frac{4}{5} \cos x - \frac{3}{5} \sin x) \ | ||
+ | 169 \sin x &= 168 \cos x - 126 \sin x \ | ||
+ | 295 \sin x &= 168 \cos x \ | ||
+ | \tan x &= \frac{168}{295} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Extracting yields <math>168 + 295 = \boxed{463}</math>. | ||
== See also == | == See also == | ||
− | * [[1999 | + | *[[Brocard point]] |
+ | {{AIME box|year=1999|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:25, 22 December 2022
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, .
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
Solution 3
Let Then, using Law of Cosines on the three triangles containing vertex we have Add the three equations up and rearrange to obtain Also, using we have Divide the two equations to obtain
Solution 4 (Law of sines)
Firstly, denote angles , , and as , , and respectively. Let . Notice that by angle chasing that and . Using the nice properties of the 13-14-15 triangle, we have and . is easily computed, so we have .
Using Law of Sines, hence, Now, computation carries the rest. Extracting yields .
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 |
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