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

m (→‎Solution 2 (no trig))
m (→‎Solution 2 (no trig))
Line 28: Line 28:
 
First, draw a good diagram.
 
First, draw a good diagram.
  
−
We realize that the measure of angle C is 75 degrees.  The measure of angle CAT is 30 degrees.  Therefore, the measure of angle CTA must also be 75 degrees, making CAT an isosceles triangle.  AT and AC are congruent, so <math>AC=24</math>.
+
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));
  
−
We now drop an altitude from C, and call the foot this altitude point D. By 30-60-90 triangles, AD equals twelve and CD equals <math>12\sqrt{3}</math>. 
+
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 BD also equal to <math>12\sqrt{3}</math>.  The base AB is <math>12+12\sqrt{3}</math>, and the altitude CD is <math>12\sqrt{3}</math>.  We can easily find that the area triangle <math>ABC</math> is <math>216+72\sqrt{3}</math>, so <math>a+b+c=\boxed{291}</math>.
+
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
 
-youyanli

Revision as of 22:56, 24 June 2021

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

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

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