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

Revision as of 22:17, 10 December 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: $\triangle ABT,$ with $45$ , $30$ , and $105$ degrees, and $\triangle ATC$ , with $30$ , $75$ , $75$ degrees. If we can sum these two triangles' areas, we have our answer.

Let's take care of $\triangle ATC$ first. We see that $\triangle ATC$ is a isosceles triangle, with $AT = AC = 24$ . Because the area of a triangle is $\frac{1}{2}ab\sin C$ , we have $\frac{1}{2}\cdot 24^2\cdot\frac{1}{2}$ , which is equal to $144.$

Now, on to $\triangle ABT$ . Draw the altitude from angle $\angle T$ to $AB$ , and call the point of intersection $D$ . This splits $\triangle ABT$ into $2$ triangles, one with $30-60-90$ ( $\triangle ADT$ ), and another with $45-45-90$ ( $\triangle BDT$ ). Now, because we know that $AT$ is $24$ , we have by special right triangle ratios. The area of $\triangle ADT$ is $\frac{12\sqrt{3}\cdot 12}{2}$ , and the area of $\triangle BDT$ is $\frac{12\cdot 12}{2}$ , which adds to $72\sqrt{3} + 72$ .

Adding this to $\triangle ATC$ we get a total sum of $216 + 72\sqrt{3}.$ Then, $a + b + c$ would be $216 + 72 + 3 = \boxed{291}.$

~MathCosine

Video Solution by OmegaLearn

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

~ pi_is_3.14

@@ Line 69: / Line 69: @@
 After drawing line AT, we see that we have two triangles: <math>\triangle ABT,</math> with <math>45</math>, <math>30</math>, and <math>105</math> degrees, and <math>\triangle ATC</math>, with <math>30</math>, <math>75</math>, <math>75</math> degrees. If we can sum these two triangles' areas, we have our answer.
-Let's take care of <math>\triangle ATC</math> first. We see that <math>\triangle ATC</math> is a isosceles triangle, with <math>AT = AC = 24</math>. Because the area of a triangle is <math>\frac{1}{2}ab\sinC</math>, we have <math>\frac{1}{2}\cdot 24^2\cdot\frac{1}{2}</math>, which is equal to <math>144.</math>
+Let's take care of <math>\triangle ATC</math> first. We see that <math>\triangle ATC</math> is a isosceles triangle, with <math>AT = AC = 24</math>. Because the area of a triangle is <math>\frac{1}{2}ab\sin C</math>, we have <math>\frac{1}{2}\cdot 24^2\cdot\frac{1}{2}</math>, which is equal to <math>144.</math>
 Now, on to <math>\triangle ABT</math>. Draw the altitude from angle <math>\angle T</math> to <math>AB</math>, and call the point of intersection <math>D</math>. This splits <math>\triangle ABT</math> into <math>2</math> triangles, one with <math>30-60-90</math> (<math>\triangle ADT</math>), and another with <math>45-45-90</math> (<math>\triangle BDT</math>). Now, because we know that <math>AT</math> is <math>24</math>, we have by special right triangle ratios. The area of <math>\triangle ADT</math> is <math>\frac{12\sqrt{3}\cdot 12}{2}</math>, and the area of <math>\triangle BDT</math> is <math>\frac{12\cdot 12}{2}</math>, which adds to  <math>72\sqrt{3} + 72</math>.

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