Difference between revisions of "2001 AIME I Problems/Problem 4"

Revision as of 16:15, 29 July 2024

Problem

In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT=24$ . The area of triangle $ABC$ can be written in the form $a+b\sqrt{c}$ , where $a$ , $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c$ .

Solution

After chasing angles, $\angle ATC=75^{\circ}$ and $\angle TCA=75^{\circ}$ , meaning $\triangle TAC$ is an isosceles triangle and $AC=24$ .

Using law of sines on $\triangle ABC$ , we can create the following equation:

$\frac{24}{\sin(\angle ABC)}$ $=$ $\frac{BC}{\sin(\angle BAC)}$

$\angle ABC=45^{\circ}$ and $\angle BAC=60^{\circ}$ , so $BC = 12\sqrt{6}$ .

We can then use the Law of Sines area formula $\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)$ to find the area of the triangle.

$\sin(75)$ can be found through the sin addition formula.

$\sin(75)$ $=$ $\frac{\sqrt{6} + \sqrt{2}}{4}$

Therefore, the area of the triangle is $\frac{\sqrt{6} + \sqrt{2}}{4}$ $\cdot$ $24$ $\cdot$ $12\sqrt{6}$ $\cdot$ $\frac{1}{2}$

$72\sqrt{3} + 216$

$72 + 3 + 216 =$ $\boxed{291}$

Solution 2 (no trig)

First, draw a good diagram.

We realize that $\angle C = 75^\circ$ , and $\angle CAT = 30^\circ$ . Therefore, $\angle CTA = 75^\circ$ as well, making $\triangle CAT$ an isosceles triangle. $AT$ and $AC$ are congruent, so $AC=24$ . We now drop an altitude from $C$ , and call the foot this altitude point $D$ .

$[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); 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]$

By 30-60-90 triangles, $AD=12$ and $CD=12\sqrt{3}$ .

We also notice that $\triangle CDB$ is an isosceles right triangle. $CD$ is congruent to $BD$ , which makes $BD=12\sqrt{3}$ . The base $AB$ is $12+12\sqrt{3}$ , and the altitude $CD=12\sqrt{3}$ . We can easily find that the area of triangle $ABC$ is $216+72\sqrt{3}$ , so $a+b+c=\boxed{291}$ .

-youyanli

Solution 3(Speedy and Simple)

After drawing line AT, we see that we have two triangles: (ABT) with 45, 30, and 105 degrees, and (ATC), with 30, 75, 75 degrees. If we can sum these two triangles' areas, we have our answer. Let's take care of (ATC) first. We see that ATC is a iscoceles triangle, with AT = AC = 24. Because the area of a triangle is $\frac{1}{2}absinC$ , we have $\frac{1}{2}24^2\frac{1}{2}$ , which is equal to 144. Now on to triangle (ABT). Draw the altitude from angle T to AB, and call the point of intersection D. This splits (ABT) into 2 triangles, one with 30-60-90 (ADT), and another with 45-45-90 (BDT). Now, because we know that AT is 24, we have by special right triangle ratios the area of ADT $\frac{12\sqrt{3}*12}{2}$ and the area of BDT $\frac{12*12}{2}$ , which give $72\sqrt{3}+72}$ (Error compiling LaTeX. Unknown error_msg). Adding this to ATC we have $216+72*\sqrt{3}$ , from which we sum to get 291.

~MathCosine

Video Solution by OmegaLearn

https://youtu.be/BIyhEjVp0iM?t=526

~ pi_is_3.14

@@ Line 67: / Line 67: @@
 ==Solution 3(Speedy and Simple)==
-After drawing line AT, we see that we have two triangles: (ABT) with 45, 30, and 105 degrees, and (ATC), with 30, 75, 75 degrees. If we can sum these two triangles' areas, we have our answer. Let's take care of (ATC) first. We see that ATC is a iscoceles triangle, with AT = AC = 24. Because the area of a triangle is <math>\frac{1}{2}absinC</math>, we have <math>\frac{1}{2}24^2\frac{1}{2}</math>, which is equal to 144. Now on to triangle (ABT). Draw the altitude from angle T to AB, and call the point of intersection D. This splits (ABT) into 2 triangles, one with 30-60-90 (ADT), and another with 45-45-90 (BDT). Now, because we know that AT is 24, we have by special right triangle ratios the area of ADT <math>\frac{12\sqrt{3}*12}{2}</math> and the area of BDT <math>\frac{12*12}{2}</math>, which give <math>72\sqrt{3}+72}. Adding this to ATC we have </math>216+72*\sqrt{3}$, from which we sum to get 291.
+After drawing line AT, we see that we have two triangles: (ABT) with 45, 30, and 105 degrees, and (ATC), with 30, 75, 75 degrees. If we can sum these two triangles' areas, we have our answer. Let's take care of (ATC) first. We see that ATC is a iscoceles triangle, with AT = AC = 24. Because the area of a triangle is <math>\frac{1}{2}absinC</math>, we have <math>\frac{1}{2}24^2\frac{1}{2}</math>, which is equal to 144. Now on to triangle (ABT). Draw the altitude from angle T to AB, and call the point of intersection D. This splits (ABT) into 2 triangles, one with 30-60-90 (ADT), and another with 45-45-90 (BDT). Now, because we know that AT is 24, we have by special right triangle ratios the area of ADT <math>\frac{12\sqrt{3}*12}{2}</math> and the area of BDT <math>\frac{12*12}{2}</math>, which give <math>72\sqrt{3}+72}</math>. Adding this to ATC we have <math>216+72*\sqrt{3}</math>, from which we sum to get 291.
 ~MathCosine

2001 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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