AIME 1999 · 第 12 题

Solution

Solution 1

Let $Q$ be the tangency point on $\overline{AC}$, and $R$ on $\overline{BC}$. By the Two Tangent Theorem, $AP = AQ = 23$, $BP = BR = 27$, and $CQ = CR = x$. Using $rs = A$, where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$, we get $(21)(50 + x) = A$. By Heron's formula, $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$. Equating and squaring both sides,

$$\begin{aligned} [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2} \end{aligned}$$ We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$.

Solution 2

Let the incenter be denoted $I$. It is commonly known that the incenter is the intersection of the angle bisectors of a triangle. So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$

We have that

$$\begin{aligned} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\ \tan \gamma & = & \frac {21}x. \end{aligned}$$ So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have

$$\begin{aligned} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\ & = & \frac 1{\tan(\alpha + \beta)} \\ \Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{aligned}$$ Doing the algebra, we get $x = \frac {245}2.$

The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$

Solution 3

Let unknown side has length as $x$, Assume three sides of triangles are $a,b,c$, the area of the triangle is $S$.

Note that $r=\frac{2S}{a+b+c}=21,S=1050+21x$

$\tan\angle{\frac{B}{2}}=\frac{7}{9}, \tan\angle{B}=\frac{63}{16}$. Use trig identity, knowing that $1+\cot^2\angle{B}=\csc^2\angle{B}$, getting that $\sin\angle{B}=\frac{63}{65}$

Now equation $(x+27)*50*\frac{63}{65}*\frac{1}{2}=1050+21x; x=\frac{245}{2}$, the final answer is $245+100=345$ ~bluesoul

Solution 4 (Trig Bash)

Let $CR = CQ = x$. Using the inradius formula for the area of a triangle, we have

$$[ABC] = 1050 + 21x,$$ where $[ABC]$ denotes the area of $\triangle ABC$.

Next, we find $\sin \angle CBA$. Note that

$$\angle CBA = \angle RBP = 2\angle OBP.$$ First, we find the hypotenuse $OB$ using the Pythagorean Theorem:

$$OB^2 = 27^2 + 21^2 = 1170 \quad \Rightarrow \quad OB = 3\sqrt{130}.$$ Let $\angle OBP = \theta$. Then

$$\sin \theta = \frac{21}{3\sqrt{130}}, \qquad \cos \theta = \frac{27}{3\sqrt{130}}.$$ Using the double-angle identity,

$$\sin(2\theta) = 2\sin \theta \cos \theta= 2\left(\frac{21}{3\sqrt{130}}\right)\left(\frac{27}{3\sqrt{130}}\right)= \frac{1134}{1170}= \frac{63}{65}.$$ Now we compute the area of $\triangle ABC$ using the formula $\tfrac{1}{2}ab\sin C$:

$$[ABC] = \frac{1}{2}(50)(27+x)\left(\frac{63}{65}\right)= \frac{315}{13}(27+x).$$ Equating the two expressions for the area,

$$1050 + 21x = \frac{315}{13}(27+x).$$ Solving,

$$\frac{13}{15}x + \frac{130}{3} = 27 + x,$$ $$\frac{49}{3} = \frac{2}{15}x,$$ so

$$x = \frac{245}{2}.$$ Finally, we compute the perimeter of $\triangle ABC$:

$$2(23+27) + 2\left(\frac{245}{2}\right) = \boxed{345}.$$ ~Voidling