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

(→‎Solution 2 (no trig))
(→‎Solution 4 (very fast))
 
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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 AC must equal 24.
+
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 of it a point D on side AB. 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);
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C = scale(24)*dir(60);
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B = extension(C,T,A,(1,0));
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F = foot(T,A,B);
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D = foot(C,A,B);
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draw(A--B--C--A--T, black+0.8);
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draw(C--D, dashed);
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label(rotate(degrees(T-A))*"$24$", A--T, N);
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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 12+<math>12\sqrt{3}</math>, and the altitude CD is <math>12\sqrt{3}</math>.  We can easily find that the area triangle ABC is 216+<math>72\sqrt{3}</math>, so a+b+c=<math>\boxed{291}</math>.
+
label("$12\sqrt 3$", C--D, E);
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label("$12\sqrt 3$", D--B, S);
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label("$12$", A--D, S);
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pen p = fontsize(8)+red;
 +
MA("45^\circ", C,B,A,2);
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MA("30^\circ", B,A,T,2.5);
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MA("30^\circ", T,A,C,3.5);
 +
 
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dot("$A$", A, SW);
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dot("$B$", B, SE);
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dot("$C$", C, N);
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dot("$T$", T, NE);
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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
 +
 +
==Solution 3(Speedy and Simple)==
 +
 +
After drawing line <math>AT</math>, 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\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>.
 +
 +
Adding this to <math>\triangle ATC</math> we get a total sum of <math>216 + 72\sqrt{3}.</math> Thus, <math>a + b + c</math> would be <math>216 + 72 + 3 = \boxed{291}.</math>
 +
 +
~MathCosine
 +
 +
==Solution 4 (very fast)==
 +
Recall the triangle area via sine formula <math>\frac{ab\sin{C}}{2}</math>. We notice that they have given almost all we need to use this, since <math>AC=24</math> by properties of isosceles triangles and <math>\angle A</math> itself equals <math>60</math>. So, we are trying to find <math>AB</math>. This is very trivial, as when we drop an altitude from <math>T</math> to <math>AB</math> (let the intersecting point be <math>U</math>), <math>AU=12\sqrt{3}</math> and <math>BU=12</math> by <math>30-60-90</math> and <math>45-45-90</math> triangles respectively. Thus the answer is just <cmath>\frac{(12+12\sqrt{3})(24)\sin{60}}{2}</cmath>
 +
<cmath>=(12+12\sqrt{3})(6)(\sqrt{3})</cmath>
 +
<cmath>=72\sqrt{3}+72\times 3</cmath>
 +
<cmath>=216 + 72\sqrt{3}</cmath>
 +
<cmath>\Longrightarrow 216+72+3=\boxed{291}</cmath>.
 +
 +
~martianrunner
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/BIyhEjVp0iM?t=526
 +
 +
~ pi_is_3.14
  
 
==See also==
 
==See also==

Latest revision as of 21:08, 23 June 2025

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 $216 + 72\sqrt{3}.$ Thus, $a + b + c$ would be $216 + 72 + 3 = \boxed{291}.$

~MathCosine

Solution 4 (very fast)

Recall the triangle area via sine formula $\frac{ab\sin{C}}{2}$. We notice that they have given almost all we need to use this, since $AC=24$ by properties of isosceles triangles and $\angle A$ itself equals $60$. So, we are trying to find $AB$. This is very trivial, as when we drop an altitude from $T$ to $AB$ (let the intersecting point be $U$), $AU=12\sqrt{3}$ and $BU=12$ by $30-60-90$ and $45-45-90$ triangles respectively. Thus the answer is just \[\frac{(12+12\sqrt{3})(24)\sin{60}}{2}\] \[=(12+12\sqrt{3})(6)(\sqrt{3})\] \[=72\sqrt{3}+72\times 3\] \[=216 + 72\sqrt{3}\] \[\Longrightarrow 216+72+3=\boxed{291}\].

~martianrunner

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

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