AIME 2001 I · 第 4 题

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 $D$.

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. Therefore, 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