Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (no trig))
(→‎Solution 3(Speedy and Simple))
(3 intermediate revisions by the same user not shown)
Line 64: Line 64:
  
 
-youyanli
 
-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 <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 gives <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
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Revision as of 16:16, 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 gives $72*{\sqrt{3}}+72}$ (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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_4&oldid=223917"