2003 AIME I Problems/Problem 7


Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21.$ Point $D$ is not on $\overline{AC}$ so that $AD = CD,$ and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s.$

Solution 1 (Pythagorean Theorem)

[asy] size(220); pointpen = black; pathpen = black + linewidth(0.7); pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle); D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7)); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2); [/asy]

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 $3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}$.

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:


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.


See also

2003 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AIME Problems and Solutions

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