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

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

@@ Line 1: / Line 1: @@
 == Problem ==
-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>
+[[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>
-== Solution ==
+== Solution 1 (Pythagorean Theorem) ==
+<center><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></center> <!-- Asymptote replacement for Image:2003_I_AIME-7.png by azjps -->
+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>.
+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>
+\triangle ACD</math> is equal to <math>3(AC) + 2(x_1 + x_2 + x_3) = 90 + 2(95 + 33 + 17) = \boxed{380}</math>.
+== Solution 2 (Stewart's Theorem) ==
+Let <math>AD=c</math> and <math>BD=d</math>, then by [[Stewart's Theorem]] we have:
+<math>30d^2+21*9*30=9c^2+21c^2=30c^2</math>. After simplifying:
+<math>d^2-c^2=189</math>.
+The solution follows as above.
+== Solution 3 (Law of Cosines) ==
+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,
+<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.
+-intelligence_20
 == See also ==
-* [[2003 AIME I Problems]]
+*[[Stewart's Theorem]]
+{{AIME box|year=2003|n=I|num-b=6|num-a=8}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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