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> | ||
− | == Solution == | + | == Solution 1 (Pythagorean Theorem) == |
<center><asy> | <center><asy> | ||
size(220); | size(220); | ||
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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> | 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{ | + | \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 == | + | == 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: | ||
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+ | <math>d^2-c^2=189</math>. | ||
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The solution follows as above. | 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. | ||
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+ | -intelligence_20 | ||
== See also == | == See also == |
Latest revision as of 17:13, 13 February 2021
Contents
Problem
Point is on with and Point is not on so that and and are integers. Let be the sum of all possible perimeters of . Find
Solution 1 (Pythagorean Theorem)
Denote the height of as , , and . Using the Pythagorean theorem, we find that and . Thus, . The LHS is difference of squares, so . As both are integers, must be integral divisors of .
The pairs of divisors of are . This yields the four potential sets for as . The last is not a possibility since it simply degenerates into a line. The sum of the three possible perimeters of is equal to .
Solution 2 (Stewart's Theorem)
Let and , then by Stewart's Theorem we have:
. After simplifying:
.
The solution follows as above.
Solution 3 (Law of Cosines)
Drop an altitude from point to side . Let the intersection point be . Since triangle is isosceles, AE is half of , or . Then, label side AD as . Since is a right triangle, you can figure out with adjacent divided by hypotenuse, which in this case is divided by , or . Now we apply law of cosines. Label as . Applying law of cosines, . Since is equal to , , which can be simplified to . 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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