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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\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>. | ||
− | == 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: | ||
+ | |||
+ | <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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− | <math>y^2 = x^2+9^2- 2 \ | ||
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-intelligence_20 | -intelligence_20 | ||
Latest revision as of 18:13, 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)
![[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]](http://latex.artofproblemsolving.com/1/5/b/15bb134453eeb3db27cdd27ad65a8176fe0393b3.png)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.