AIME 2003 I · 第 7 题

Solution 1 (Pythagorean Theorem)

Denote the height of $\triangle ACD$ as $h$, $x = AD = CD$, and $y = BD$. Using the Pythagorean theorem, we find that $h^2 = y^2 - 6^2$ and $h^2 = x^2 - 15^2$. Thus, $y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189$. The LHS is difference of squares, so $(x + y)(x - y) = 189$. As both $x,\ y$ are integers, $x+y,\ x-y$ must be integral divisors of $189$.

The pairs of divisors of $189$ are $(1,189)\ (3,63)\ (7,27)\ (9,21)$. This yields the four potential sets for $(x,y)$ as $(95,94)\ (33,30)\ (17,10)\ (15,6)$. The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of $\triangle ACD$ is equal to $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}$.

Note

One can also conclude $(15,6)$ doesn't work as it violates the triangle inequality.

~Pinotation

Solution 2 (Stewart's Theorem)

Let $AD=c$ and $BD=d$, then by Stewart's Theorem we have:

$30d^2+21*9*30=9c^2+21c^2=30c^2$. After simplifying:

$d^2-c^2=189$.

The solution follows as above.

Solution 3 (Law of Cosines)

Drop an altitude from point $D$ to side $AC$. Let the intersection point be $E$. Since triangle $ADC$ is isosceles, AE is half of $AC$, or $15$. Then, label side AD as $x$. Since $AED$ is a right triangle, you can figure out $\cos A$ with adjacent divided by hypotenuse, which in this case is $AE$ divided by $x$, or $\frac{15}{x}$. Now we apply law of cosines. Label $BD$ as $y$. Applying law of cosines, $y^2 = x^2+9^2- 2 \cdot x \cdot 9 \cdot \cos A$. Since $\cos A$ is equal to $\frac{15}{x}$, $y^2 = x^2+9^2- 2 \cdot x \cdot 9 \cdot \frac{15}{x}$, which can be simplified to $x^2-y^2=189$. The solution proceeds as the first solution does.

-intelligence_20