Difference between revisions of "2016 AIME II Problems/Problem 5"
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Solution by brightaz | Solution by brightaz | ||
+ | |||
+ | == Solution 3 == | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | |||
+ | // Setup | ||
+ | pair A, B; | ||
+ | pair C0, C1, C2, C3, C4, C5, C6, C7, C8; | ||
+ | A = (5, 0); | ||
+ | B = (0, 3); | ||
+ | C0 = (0, 0); | ||
+ | C1 = foot(C0, A, B); | ||
+ | C2 = foot(C1, C0, B); | ||
+ | C3 = foot(C2, C1, B); | ||
+ | C4 = foot(C3, C2, B); | ||
+ | C5 = foot(C4, C3, B); | ||
+ | C6 = foot(C5, C4, B); | ||
+ | C7 = foot(C6, C5, B); | ||
+ | C8 = foot(C7, C6, B); | ||
+ | |||
+ | // Labels | ||
+ | label("$A$", A, SE); | ||
+ | label("$B$", B, NW); | ||
+ | label("$C_0$", C0, SW); | ||
+ | label("$C_1$", C1, NE); | ||
+ | label("$C_2$", C2, W); | ||
+ | label("$C_3$", C3, NE); | ||
+ | label("$C_4$", C4, W); | ||
+ | |||
+ | // Drawings | ||
+ | draw(A--B--C0--cycle); | ||
+ | draw(C0--C1--C2, red); | ||
+ | draw(C2--C3--C4--C5--C6--C7--C8); | ||
+ | draw(C0--C2, green); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>a = BC_0</math>, <math>b = AC_0 </math>, and <math>c = AB</math>. | ||
+ | Note that the total length of the red segments in the figure above is equal to the length of the green segment times <math>\frac{a+c}{b}</math>. | ||
+ | |||
+ | The desired sum is equal to the total length of the infinite path <math>C_0 C_1 C_2 C_3 \cdots</math>, shown in red in the figure below. | ||
+ | Since each of the triangles <math>\triangle C_0 C_1 C_2, \triangle C_2 C_3 C_4, \dots</math> on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the green segment times <math>\frac{a+c}{b}</math>. | ||
+ | In other words, we have that <math>a\left(\frac{a+c}{b}\right) = 6p</math>. | ||
+ | |||
+ | Guessing and checking Pythagorean triples reveals that <math>a = 84</math>, <math>b=13</math>, <math>c = 85</math>, and <math>p = a + b + c = \boxed{182}</math> satisfies this equation. | ||
+ | |||
+ | <asy> | ||
+ | size(10cm); | ||
+ | |||
+ | // Setup | ||
+ | pair A, B; | ||
+ | pair C0, C1, C2, C3, C4, C5, C6, C7, C8; | ||
+ | A = (5, 0); | ||
+ | B = (0, 3); | ||
+ | C0 = (0, 0); | ||
+ | C1 = foot(C0, A, B); | ||
+ | C2 = foot(C1, C0, B); | ||
+ | C3 = foot(C2, C1, B); | ||
+ | C4 = foot(C3, C2, B); | ||
+ | C5 = foot(C4, C3, B); | ||
+ | C6 = foot(C5, C4, B); | ||
+ | C7 = foot(C6, C5, B); | ||
+ | C8 = foot(C7, C6, B); | ||
+ | |||
+ | // Labels | ||
+ | label("$A$", A, SE); | ||
+ | label("$B$", B, NW); | ||
+ | label("$C_0$", C0, SW); | ||
+ | label("$a$", (B+C0)/2, W); | ||
+ | label("$b$", (A+C0)/2, S); | ||
+ | label("$c$", (A+B)/2, NE); | ||
+ | |||
+ | // Drawings | ||
+ | draw(A--B--C0--cycle); | ||
+ | draw(C0--C1--C2--C3--C4--C5--C6--C7--C8, red); | ||
+ | draw(C0--B, green); | ||
+ | </asy> | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=4|num-a=6}} | {{AIME box|year=2016|n=II|num-b=4|num-a=6}} |
Revision as of 19:27, 26 May 2017
Triangle has a right angle at . Its side lengths are pariwise relatively prime positive integers, and its perimeter is . Let be the foot of the altitude to , and for , let be the foot of the altitude to in . The sum . Find .
Contents
Solution 1
Note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is for reach height, so by the geometric series formula, we have . Multiplying by the denominator and expanding, the equation becomes . Cancelling and multiplying by yields , so and . Checking for Pythagorean triples gives and , so
Solution modified/fixed from Shaddoll's solution.
Solution 2
We start by splitting the sum of all into two parts: those where is odd and those where is even.
First consider the sum of the lengths of the segments for which is odd for each . The perimeters of these triangles can be expressed using and ratios that result because of similar triangles. Considering triangles where is odd, we find that the perimeter for each such is . Thus,
.
Simplifying,
. (1)
Continuing with a similar process for the sum of the lengths of the segments for which is even,
.
Simplifying,
. (2)
Adding (1) and (2) together, we find that
.
Setting , , and , we can now proceed as in Shaddoll's solution, and our answer is .
Solution by brightaz
Solution 3
Let , , and . Note that the total length of the red segments in the figure above is equal to the length of the green segment times .
The desired sum is equal to the total length of the infinite path , shown in red in the figure below. Since each of the triangles on the left are similar, it follows that the total length of the red segments in the figure below is equal to the length of the green segment times . In other words, we have that .
Guessing and checking Pythagorean triples reveals that , , , and satisfies this equation.
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |