AIME 2009 II · 第 10 题

AIME 2009 II — Problem 10

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Four lighthouses are located at points $A$ , $B$ , $C$ , and $D$ . The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$ , the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$ , and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$ . To an observer at $A$ , the angle determined by the lights at $B$ and $D$ and the angle determined by the lights at $C$ and $D$ are equal. To an observer at $C$ , the angle determined by the lights at $A$ and $B$ and the angle determined by the lights at $D$ and $B$ are equal. The number of kilometers from $A$ to $D$ is given by $\frac {p\sqrt{q}}{r}$ , where $p$ , $q$ , and $r$ are relatively prime positive integers, and $r$ is not divisible by the square of any prime. Find $p$ + $q$ + $r$ .

Diagram

$AIME diagram$

-asjpz

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=f3zEesJh4Ws

解析

Solution 1

Let $O$ be the intersection of $BC$ and $AD$ . By the Angle Bisector Theorem, $\frac {5}{BO}$ = $\frac {13}{CO}$ , so $BO$ = $5x$ and $CO$ = $13x$ , and $BO$ + $OC$ = $BC$ = $12$ , so $x$ = $\frac {2}{3}$ , and $OC$ = $\frac {26}{3}$ . Let $P$ be the foot of the altitude from $D$ to $OC$ . It can be seen that triangle $DOP$ is similar to triangle $AOB$ , and triangle $DPC$ is similar to triangle $ABC$ . If $DP$ = $15y$ , then $CP$ = $36y$ , $OP$ = $10y$ , and $OD$ = $5y\sqrt {13}$ . Since $OP$ + $CP$ = $46y$ = $\frac {26}{3}$ , $y$ = $\frac {13}{69}$ , and $AD$ = $\frac {60\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$ ). The answer is $60$ + $13$ + $23$ = $\boxed{096}$ .

Solution 2

Extend $AB$ and $CD$ to intersect at $P$ . Note that since $\angle ACB=\angle PCB$ and $\angle ABC=\angle PBC=90^{\circ}$ by ASA congruency we have $\triangle ABC\cong \triangle PBC$ . Therefore $AC=PC=13$ .

By the angle bisector theorem, $PD=\dfrac{130}{23}$ and $CD=\dfrac{169}{23}$ . Now we apply Stewart's theorem to find $AD$ :

$\begin{aligned}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\ 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\ 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\ AD^2&=130-\dfrac{130\cdot 169}{23^2}\\ AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\ AD^2&=\dfrac{130(23^2-169)}{23^2}\\ AD^2&=\dfrac{130(360)}{23^2}\\ AD&=\dfrac{60\sqrt{13}}{23}\\ \end{aligned}$ and our final answer is $60+13+23=\boxed{096}$ .

Solution 3

Notice that by extending $AB$ and $CD$ to meet at a point $E$ , $\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$ , $B=(12,0)$ , $E=(12,-5)$ and $A=(12,5)$ , and the equation of line $CD$ is $y=-\dfrac{5}{12}x$ . Let F be the intersection point of $AD$ and $BC$ , and by using the Angle Bisector Theorem: $\dfrac{BF}{AB}=\dfrac{FC}{AC}$ we have $FC=\dfrac{26}{3}$ . Then the equation of the line $AF$ through the points $(12,5)$ and $\left(\frac{26}{3},0\right)$ is $y=\frac32 x-13$ . Hence the intersection point of $AF$ and $CD$ is the point $D$ at the coordinates $\left(\dfrac{156}{23},-\dfrac{65}{23}\right)$ . Using the distance formula, $AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}$ for an answer of $60+13+23=\fbox{096}$ .

Solution 4

After drawing a good diagram, we reflect $ABC$ over the line $BC$ , forming a new point that we'll call $A'$ . Also, let the intersection of $AD$ and $BC$ be point $E$ . Point $D$ lies on line $A'C$ . Since line $AD$ bisects $\angle{CAB}$ , we can use the Angle Bisector Theorem. $AA'=10$ and $AC=13$ , so $\frac{CD}{A'D}=\frac{13}{10}$ . Letting the segments be $13x$ and $10x$ respectively, we now have $13x+10x=13$ . Therefore, $x=\frac{13}{23}$ . By the Pythagorean Theorem, $AE=\frac{5\sqrt{13}}{3}$ . Using the Angle Bisector Theorem on $\angle{ACD}$ , we have that $ED=\frac{5x\sqrt{13}}{3}$ . Substituting in $x=\frac{13}{23}$ , we have that $AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}$ , so the answer is $60+13+23=\boxed{096}$ .

-RootThreeOverTwo

Solution 5 (Angle Bisector Theorem + Law of Cosines)

Let $CD$ and $AB$ meet at $A'$ ; then $\overline{AD}$ is an angle bisector of isosceles $\triangle AA'C$ . Then by the Angle Bisector Theorem, $A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}$ , and $\cos \angle DA'A=\frac {5}{13}$ . By the Law of Cosines on $\triangle AA'D$ , we have

$AD^2=10^2+\left(\frac {130}{23}\right)^2-2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13}=\frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD=\frac {60\sqrt {13}}{23}$ and the answer is $60+13+23=\boxed{096}$ .

$AIME diagram$

-azjps

Solution 6 (Law of Sines)

$AIME diagram$

Using the law of sines on $\bigtriangleup ADE$ ,

$\begin{aligned} AD &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\ &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\ &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\ &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23} \end{aligned}$ $\therefore$ the answer is $60+13+23=\boxed{096}$ .

-m1sterzer0