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

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== Problem ==
 
== Problem ==
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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 integers. Let <math> s </math> be the sum of all possible perimeters of <math> \triangle ACD. </math> Find <math> s. </math>
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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>
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== Solution 1 (Pythagorean Theorem) ==
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<center><asy>
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size(220);
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pointpen = black; pathpen = black + linewidth(0.7);
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pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8);
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D(D(MP("A",A))--D(MP("C",C))--D(MP("D",D,NE))--cycle);
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D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7));
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MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2);
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</asy></center> <!-- Asymptote replacement for Image:2003_I_AIME-7.png by azjps -->
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Denote the height of <math>\triangle ACD</math> as <math>h</math>, <math>x = AD = CD</math>, and <math>y = BD</math>. Using the [[Pythagorean theorem]], we find that <math>h^2 = y^2 - 6^2</math> and <math>h^2 = x^2 - 15^2</math>. Thus, <math>y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189</math>. The LHS is [[difference of squares]], so <math>(x + y)(x - y) = 189</math>. As both <math>x,\ y</math> are integers, <math>x+y,\ x-y</math> must be integral divisors of <math>189</math>.
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The pairs of divisors of <math>189</math> are <math>(1,189)\ (3,63)\ (7,27)\ (9,21)</math>. This yields the four potential sets for <math>(x,y)</math> as <math>(95,94)\ (33,30)\ (17,10)\ (15,6)</math>. The last is not a possibility since it simply [[degenerate]]s into a [[line]]. The sum of the three possible perimeters of <math>
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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>.
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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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The solution follows as above.
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== Solution 3 (Law of Cosines) ==
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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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-intelligence_20
  
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== Solution ==
 
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{{solution}}
 
 
== See also ==
 
== See also ==
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* [[2003 AIME I Problems/Problem 6 | Previous problem]]
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*[[Stewart's Theorem]]
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* [[2003 AIME I Problems/Problem 8 | Next problem]]
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{{AIME box|year=2003|n=I|num-b=6|num-a=8}}
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* [[2003 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

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
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All AIME Problems and Solutions

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