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

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==Problem==
  
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In triangle <math>ABC</math>, angles <math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The bisector of angle <math>A</math> intersects <math>\overline{BC}</math> at <math>T</math>, and <math>AT=24</math>. The area of triangle <math>ABC</math> can be written in the form <math>a+b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>.
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==Solution==
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After chasing angles, <math>\angle ATC=75^{\circ}</math> and <math>\angle TCA=75^{\circ}</math>, meaning <math>\triangle TAC</math> is an isosceles triangle and <math>AC=24</math>.
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Using law of sines on <math>\triangle ABC</math>, we can create the following equation:
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<math>\frac{24}{\sin(\angle ABC)}</math> <math>=</math> <math>\frac{BC}{\sin(\angle BAC)}</math>
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<math>\angle ABC=45^{\circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqrt{6}</math>.
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We can then use the Law of Sines area formula <math>\frac{1}{2} \cdot BC \cdot AC \cdot \sin(\angle BCA)</math> to find the area of the triangle.
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<math>\sin(75)</math> can be found through the sin addition formula.
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<math>\sin(75)</math> <math>=</math> <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math>
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Therefore, the area of the triangle is <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math> <math>\cdot</math> <math>24</math> <math>\cdot</math> <math>12\sqrt{6}</math> <math>\cdot</math> <math>\frac{1}{2}</math>
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<math>72\sqrt{3} + 216</math>
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<math>72 + 3 + 216 =</math> <math>\boxed{291}</math>
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==Solution 2 (no trig)==
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First, draw a good diagram.
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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>. 
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<center><asy>
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size(200);
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defaultpen(linewidth(0.4)+fontsize(8));
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pair A,B,C,D,T,F;
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A = origin;
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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);
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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;
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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);
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</asy></center>
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By 30-60-90 triangles, <math>AD=12</math> and <math>CD=12\sqrt{3}</math>. 
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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>.
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-youyanli
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==Solution 3(Speedy and Simple)==
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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.
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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>
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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>.
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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>
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~MathCosine
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==Solution 4 (very fast)==
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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>
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<cmath>=(12+12\sqrt{3})(6)(\sqrt{3})</cmath>
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<cmath>=72\sqrt{3}+72\times 3</cmath>
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<cmath>=216 + 72\sqrt{3}</cmath>
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<cmath>\Longrightarrow 216+72+3=\boxed{291}</cmath>.
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~martianrunner
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== Video Solution by OmegaLearn ==
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https://youtu.be/BIyhEjVp0iM?t=526
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~ pi_is_3.14
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==See also==
 +
{{AIME box|year=2001|n=I|num-b=3|num-a=5}}
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 +
{{MAA Notice}}

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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