Difference between revisions of "2001 AIME I Problems/Problem 4"
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==Solution 2 (no trig)== | ==Solution 2 (no trig)== | ||
− | + | First, draw a good diagram. | |
− | We realize that | + | We realize that <math>\angle C = 75^\circ</math>, and <math>\angle CAT = 30^\circ</math>. Therefore, <math>\angle CTA = 75^\circ</math> as well, making <math>\triangle CAT</math> an isosceles triangle. <math>AT</math> and <math>AC</math> are congruent, so <math>AC=24</math>. We now drop an altitude from <math>C</math>, and call the foot this altitude point <math>D</math>. |
+ | <center><asy> | ||
+ | size(200); | ||
+ | defaultpen(linewidth(0.4)+fontsize(8)); | ||
− | + | pair A,B,C,D,T,F; | |
+ | A = origin; | ||
+ | T = scale(24)*dir(30); | ||
+ | C = scale(24)*dir(60); | ||
+ | B = extension(C,T,A,(1,0)); | ||
+ | F = foot(T,A,B); | ||
+ | D = foot(C,A,B); | ||
+ | draw(A--B--C--A--T, black+0.8); | ||
+ | draw(C--D, dashed); | ||
+ | label(rotate(degrees(T-A))*"$24$", A--T, N); | ||
+ | label(rotate(degrees(C-A))*"$24$", A--C, 2*NW); | ||
− | We also notice that CDB is an isosceles right triangle. CD is congruent to BD, which makes | + | label("$12\sqrt 3$", C--D, E); |
+ | label("$12\sqrt 3$", D--B, S); | ||
+ | label("$12$", A--D, S); | ||
+ | pen p = fontsize(8)+red; | ||
+ | MA("45^\circ", C,B,A,2); | ||
+ | MA("30^\circ", B,A,T,2.5); | ||
+ | MA("30^\circ", T,A,C,3.5); | ||
+ | |||
+ | dot("$A$", A, SW); | ||
+ | dot("$B$", B, SE); | ||
+ | dot("$C$", C, N); | ||
+ | dot("$T$", T, NE); | ||
+ | dot("$D$", D, S); | ||
+ | </asy></center> | ||
+ | By 30-60-90 triangles, <math>AD=12</math> and <math>CD=12\sqrt{3}</math>. | ||
+ | |||
+ | We also notice that <math>\triangle CDB</math> is an isosceles right triangle. <math>CD</math> is congruent to <math>BD</math>, which makes <math>BD=12\sqrt{3}</math>. The base <math>AB</math> is <math>12+12\sqrt{3}</math>, and the altitude <math>CD=12\sqrt{3}</math>. We can easily find that the area of triangle <math>ABC</math> is <math>216+72\sqrt{3}</math>, so <math>a+b+c=\boxed{291}</math>. | ||
+ | |||
+ | -youyanli | ||
==See also== | ==See also== |
Latest revision as of 22:56, 24 June 2021
Problem
In triangle , angles and measure degrees and degrees, respectively. The bisector of angle intersects at , and . The area of triangle can be written in the form , where , , and are positive integers, and is not divisible by the square of any prime. Find .
Solution
After chasing angles, and , meaning is an isosceles triangle and .
Using law of sines on , we can create the following equation:
and , so .
We can then use the Law of Sines area formula to find the area of the triangle.
can be found through the sin addition formula.
Therefore, the area of the triangle is
Solution 2 (no trig)
First, draw a good diagram.
We realize that , and . Therefore, as well, making an isosceles triangle. and are congruent, so . We now drop an altitude from , and call the foot this altitude point .
By 30-60-90 triangles, and .
We also notice that is an isosceles right triangle. is congruent to , which makes . The base is , and the altitude . We can easily find that the area of triangle is , so .
-youyanli
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.