AIME 2002 II · 第 14 题

Solution 1

Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have

$$\frac{19}{AM} = \frac{152-2AM-19+19}{152} = \frac{152-2AM}{152}$$ Solving, $AM = 38$. So the ratio of the side lengths of the triangles is 2. Therefore,

$$\frac{PB+38}{OP}= 2 \text{ and } \frac{OP+19}{PB} = 2$$ so $2OP = PB+38$ and $2PB = OP+19.$ Substituting for $PB$, we see that $4OP-76 = OP+19$, so $OP = \frac{95}3$ and the answer is $\boxed{098}$.

Solution 2

Reflect triangle $PAM$ across line $AP$, creating an isoceles triangle. Let $x$ be the distance from the top of the circle to point $P$, with $x + 38$ as $AP$. Given the perimeter is 152, subtracting the altitude yields the semiperimeter $s$ of the isoceles triangle, as $114 - x$. The area of the isoceles triangle is:

$[PAM] = r \cdot s$ $[PAM] = 19 \cdot (114 - x)$

Now use similarity, draw perpendicular from $O$ to $PM$, name the new point $D$. Triangle $PDO$ is similar to triangle $PAM$, by AA Similarity. Equating the legs, we get:

$\frac{\sqrt{x}}{19} = \frac{\sqrt{x + 38}}{AM}$

Solving for $AM$, it yields $19 \cdot \sqrt{\frac{x + 38}{x}}$.

$19 \cdot (114 - x) = AM \cdot AP = 19 \cdot (x + 38) \cdot \sqrt{\frac{x + 38}{x}}$

The $x^3$ cancels, yielding a quadratic. Solving yields $x = \frac{38}{3}$. Add $19$ to find $OP$, yielding $\frac{95}{3}$ or $\boxed{098}$.

Solution 3

Let the foot of the perpendicular from $O$ to $PM$ be $D;$ now $OD=19.$ Also let $AM=x$ and $PM=y.$ This means that $OP=\frac{y}{x}\cdot 19$, since $O$ is on the angle bisector of $\angle M.$

We have that $\tan(\angle AMO)=\frac{19}{x},$ so

$$\tan(\angle M)=\tan (2\cdot \angle AMO)=\frac{38x}{x^{2}-361}.$$ However $\tan(\angle M)=\frac{PA}{AM}=\frac{PO+OA}{AM}=\frac{\frac{y}{x}\cdot 19 + 19}{x}$, so

$$\frac{38x}{x^{2}-361}=19\cdot \frac{\frac{y}{x}+1}{x}$$ $$\frac{2x^{2}}{x^{2}-361}=\frac{y}{x}+1$$ $$\frac{x^{2}+361}{x^{2}-361}=\frac{y}{x}.$$ $$x\cdot \frac{x^{2}+361}{x^{2}-361}=y$$ We now use the fact that the perimeter of $\triangle PAM$ is $152$:

$$PO+OA+AM+MP=152$$ $$\frac{y}{x}\cdot 19+19+x+y=152$$ $$19\left(\frac{x^{2}+361}{x^{2}-361}\right)+x\cdot \left(\frac{x^{2}+361}{x^{2}-361}\right)+x+19=152$$ $$(x+19)\left(\frac{x^{2}+361}{x^{2}-361}+\frac{x^{2}-361}{x^{2}-361}\right)=152$$ $$\frac{2x^{2}}{x-19}=152$$ $$x^{2}-76x+19\cdot 76=0.$$ This quadratic factors as $(x-38)^{2}=0,$ so $x=38$, and

$$\frac{y}{x}=\frac{38^{2}+361}{38^{2}-361}=\frac{5}{3}$$ $$OP=\frac{y}{x}\cdot 19=\frac{95}{3}\to \boxed{98.}$$

Solution 4

Let $K$ be the foot of the altitude from $O$ to $PM,$ and notice how $OK$ is a radius of the circle. Also, we have that $\angle OKM=\angle OAM=90^\circ,$ $OK=OA=90,$ and $OM=OM,$ so $\triangle OKM$ is congruent to $\triangle OAM.$ This means that because $\angle PMO=\angle KMO=\angle AMO,$ $O$ is the foot of the angle bisector from $M.$

Then, let $k=\angle AMO=\angle PMO.$ We have that

$$\angle POK=180^\circ-\angle KOM-\angle AOM=2k=\angle PMA,$$ so $\triangle PKO$ is similar to $\triangle PAM.$ Let $a=AM=KM,$ and let $p=152$ be the perimeter of $\triangle PAM.$ By similarity, we have that the perimeter of $\triangle PKO$ is $\frac{19}{a}\cdot p,$ so

$$PK+PO=\dfrac{19}{a}\cdot p-19,$$ so

$$p=PK+PO+OA+AM+MK=\dfrac{19}{a}\cdot p-19+19+2a.$$ Solving for $p,$ we have that

$$p=\dfrac{2a}{1-\frac{19}{a}}$$ and using $p=152$ we find that

$$a^2-76a+38^2=(a-38)^2=0$$ so $a=38.$

Finally, let $b=PO.$ We have by the angle bisector theorem that $PM=2b,$ so using the perimeter information one more time we get $3b+57=152$ and solving gives us

$$b=\dfrac{95}{3}\implies \boxed{098}.$$ ~BS2012

Solution 5

Let $X$ be the point of tangency from the circle to $PM$. Letting $AM=b$, we have $AM=XM=b$. We also know that triangle $POX$ is similar to triangle $PMA$. Taking advantage of the similarity between the sum of the shortest leg and hypotenuse of the two triangles, we have $AP+PM=\frac{b}{19} \cdot (XP+PO)$ $\implies$ $152-b=\frac{b}{19} \cdot (133-2b)$. If we solve this equation, we obtain $b=38$. Now letting $OP=a$, we know that $PM=2a$, so $3a+57=152$, or $OP=a= \frac{95}{3}$, so our desired answer is $\boxed{098}$. ~Irfans123