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

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[[Point]] <math> B </math> is on <math> \overline{AC} </math> with <math> AB = 9 </math> and <math> BC = 21. </math> Point <math> D </math> is not on <math> \overline{AC} </math> so that <math> AD = CD, </math> and <math> AD </math> and <math> BD </math> are [[integer]]s. Let <math> s </math> be the sum of all possible [[perimeter]]s of <math> \triangle ACD</math>. Find <math> s. </math>
 
[[Point]] <math> B </math> is on <math> \overline{AC} </math> with <math> AB = 9 </math> and <math> BC = 21. </math> Point <math> D </math> is not on <math> \overline{AC} </math> so that <math> AD = CD, </math> and <math> AD </math> and <math> BD </math> are [[integer]]s. Let <math> s </math> be the sum of all possible [[perimeter]]s of <math> \triangle ACD</math>. Find <math> s. </math>
  
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== Solution ==
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== Solution 1 (Pythagorean Theorem) ==
 
<center><asy>
 
<center><asy>
 
size(220);
 
size(220);
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\triangle ACD</math> is equal to <math>3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}</math>.
 
\triangle ACD</math> is equal to <math>3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}</math>.
  
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== Solution 2 ==
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== Solution 2 (Stewart's Theorem) ==
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Let <math>AD=c</math> and <math>BD=d</math>, then by [[Stewart's Theorem]] we have:
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<math>30d^2+21*9*30=9c^2+21c^2=30c^2</math>. After simplifying:
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<math>d^2-c^2=189</math>.
  
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Using [[Stewart's Theorem]], letting the side length be c, and the cevian be d, then we have
 
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<math>30d^2+21*9*30=9c^2+21c^2=30c^2</math>. Dividing both sides by thirty leaves
 
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<math>d^2+189=c^2</math>.
 
 
The solution follows as above.
 
The solution follows as above.
  
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== Solution 3 (Law of Cosines) ==
  
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== Solution 3 (LAW OF COSINES BASH) ==
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Drop an altitude from point <math>D</math> to side <math>AC</math>. Let the intersection point be <math>E</math>. Since triangle <math>ADC</math> is isosceles, AE is half of <math>AC</math>, or <math>15</math>. Then, label side AD as <math>x</math>. Since <math>AED</math> is a right triangle, you can figure out <math>\cos A</math> with adjacent divided by hypotenuse, which in this case is <math>AE</math> divided by <math>x</math>, or <math>\frac{15}{x}</math>. Now we apply law of cosines. Label <math>BD</math> as <math>y</math>. Applying law  of cosines,  
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<math>y^2 = x^2+9^2- 2 \cdot x \cdot 9 \cdot \cos A</math>. Since <math>\cos A</math> is equal to <math>\frac{15}{x}</math>, <math>y^2 = x^2+9^2- 2 \cdot x \cdot 9 \cdot \frac{15}{x}</math>, which can be simplified to <math>x^2-y^2=189</math>. The solution proceeds as the first solution does.
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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,  
 
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<math>y^2 = x^2+9^2- 2 \times x \times 9 \times cos(A)</math> Since cosine A is equal to 15/x, <math>y^2 = x^2+9^2- 2 \times x \times 9 \times 15/x</math>, which can be simplified to <math>y^2 = x^2 -189</math> then continue with the first solution
 
  
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u guys can edit the formatting if youd like to
 
 
-intelligence_20
 
-intelligence_20
  

Latest revision as of 17:13, 13 February 2021

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 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:

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

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

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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