Difference between revisions of "1995 AIME Problems/Problem 9"
(tex fixes) |
m (→Problem: image replace with asy) |
||
Line 2: | Line 2: | ||
Triangle <math>ABC</math> is [[isosceles triangle|isosceles]], with <math>AB=AC</math> and [[altitude]] <math>AM=11.</math> Suppose that there is a point <math>D</math> on <math>\overline{AM}</math> with <math>AD=10</math> and <math>\angle BDC=3\angle BAC.</math> Then the perimeter of <math>\triangle ABC</math> may be written in the form <math>a+\sqrt{b},</math> where <math>a</math> and <math>b</math> are integers. Find <math>a+b.</math> | Triangle <math>ABC</math> is [[isosceles triangle|isosceles]], with <math>AB=AC</math> and [[altitude]] <math>AM=11.</math> Suppose that there is a point <math>D</math> on <math>\overline{AM}</math> with <math>AD=10</math> and <math>\angle BDC=3\angle BAC.</math> Then the perimeter of <math>\triangle ABC</math> may be written in the form <math>a+\sqrt{b},</math> where <math>a</math> and <math>b</math> are integers. Find <math>a+b.</math> | ||
− | [[Image:AIME_1995_Problem_9.png|center]] | + | <asy> |
+ | import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; | ||
+ | draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); | ||
+ | dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);</asy> | ||
+ | <!-- [[Image:AIME_1995_Problem_9.png|center]] --> | ||
== Solution == | == Solution == |
Revision as of 00:08, 21 August 2012
Problem
Triangle is isosceles, with and altitude Suppose that there is a point on with and Then the perimeter of may be written in the form where and are integers. Find
Solution
Let , so . Then, . Expanding using the angle sum identity gives Thus, . Solving, we get . Hence, and by the Pythagorean Theorem. The total perimeter is . The answer is thus .
See also
1995 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 |