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Difference between revisions of "2003 AIME I Problems/Problem 7"

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== Solution 3 ==
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== Solution 3 (LAW OF COSINES BASH) ==
  
 
drop an altitude from angle D to side AC. let the intersection point be E. since the triangle is isosceles, AE is half of AC, which is 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, which is just 15/x. now law of cosines. label bd as y. so, by law  of cosines,  
 
drop an altitude from angle D to side AC. let the intersection point be E. since the triangle is isosceles, AE is half of AC, which is 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, which is just 15/x. now law of cosines. label bd as y. so, by law  of cosines,  

Revision as of 11:16, 21 March 2020

Problem

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

[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

Using Stewart's Theorem, letting the side length be c, and the cevian be d, then we have $30d^2+21*9*30=9c^2+21c^2=30c^2$. Dividing both sides by thirty leaves $d^2+189=c^2$. The solution follows as above.


Solution 3 (LAW OF COSINES BASH)

drop an altitude from angle D to side AC. let the intersection point be E. since the triangle is isosceles, AE is half of AC, which is 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, which is just 15/x. now law of cosines. label bd as y. so, by law of cosines, $y^2 = x^2+9^2- 2 \times x \times 9 \times cos(A)$ Since cosine A is equal to 15/x, $y^2 = x^2+9^2- 2 \times x \times 9 \times 15/x$, which can be simplified to $y^2 = x^2 -189$ then continue with the first solution


u guys can edit the formatting if youd like to -intelligence_20

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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