Difference between revisions of "2016 AIME II Problems/Problem 5"
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− | Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pairwise relatively prime positive integers, and its perimeter is <math>p</math>. Let <math>C_1</math> be the foot of the altitude to <math>\overline{AB}</math>, and for <math>n \geq 2</math>, let <math>C_n</math> be the foot of the altitude to <math>\overline{C_{n-2}B}</math> in <math>\triangle C_{n-2}C_{n-1}B</math>. The sum <math>\sum_{ | + | ==Problem== |
+ | Triangle <math>ABC_0</math> has a right angle at <math>C_0</math>. Its side lengths are pairwise relatively prime positive integers, and its perimeter is <math>p</math>. Let <math>C_1</math> be the foot of the altitude to <math>\overline{AB}</math>, and for <math>n \geq 2</math>, let <math>C_n</math> be the foot of the altitude to <math>\overline{C_{n-2}B}</math> in <math>\triangle C_{n-2}C_{n-1}B</math>. The sum <math>\sum_{n=2}^\infty C_{n-2}C_{n-1} = 6p</math>. Find <math>p</math>. | ||
==Solution 1== | ==Solution 1== | ||
+ | |||
+ | Do note that by counting the area in 2 ways, the first altitude is <math>x = \frac{ab}{c}</math>. By similar triangles, the common ratio is <math>\rho = \frac{a}{c}</math> for each height, so by the geometric series formula, we have | ||
+ | <cmath>\begin{align} | ||
+ | 6p=\frac{x}{1-\rho} = \frac{ab}{c-a}. | ||
+ | \end{align}</cmath> | ||
+ | Writing <math>p=a+b+c</math> and clearing denominators, we get <cmath>13a=6p .</cmath>Thus <math>p=13q</math>, <math>a=6q</math>, and <math>b+c=7q</math>, i.e. <math>c=7q-b</math>. Plugging these into <math>(1)</math>, we get <math>78q(q-b)=6bq</math>, i.e., <math>14b=13q</math>. Thus <math>q=14r</math> and <math>p=182r</math>, <math>b=13r</math>, <math>a=84r</math>, <math>c=85r</math>. Taking <math>r=1</math> (since <math>a,b,c</math> are relatively prime) we get <math>p=\boxed{182}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ab}{c}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+85=\boxed{182}</math> | Note that by counting the area in 2 ways, the first altitude is <math>\dfrac{ab}{c}</math>. By similar triangles, the common ratio is <math>\dfrac{a}{c}</math> for reach height, so by the geometric series formula, we have <math>6p=\dfrac{\dfrac{ab}{c}}{1-\dfrac{a}{c}}</math>. Multiplying by the denominator and expanding, the equation becomes <math>\dfrac{ab}{c}=6a+6b+6c-\dfrac{6a^2}{c}-\dfrac{6ab}{c}-6a</math>. Cancelling <math>6a</math> and multiplying by <math>c</math> yields <math>ab=6bc+6c^2-6a^2-6ab</math>, so <math>7ab = 6bc+6b^2</math> and <math>7a=6b+6c</math>. Checking for Pythagorean triples gives <math>13,84,</math> and <math>85</math>, so <math>p=13+84+85=\boxed{182}</math> | ||
+ | |||
+ | Note: a more rigorous solution instead of checking for triples would be to substitute <math>c = \sqrt{a^2 + b^2}</math> and heavily simplifying the equation. Eventually we are left with <math>13a = 84b,</math> and since <math>a, b</math> are relatively prime, we know <math>a = 84</math> and <math>b = 13.</math> From here we can note that <math>(13, 84, 85)</math> is a pythagorean triple or again use the Pythagorean Theorem to find <math>c = 85.</math> Thus, the answer is <math>\boxed{182}.</math> | ||
Solution modified/fixed from Shaddoll's solution. | Solution modified/fixed from Shaddoll's solution. | ||
− | ==Solution | + | ==Solution 3== |
We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> is odd and those where <math>n-2</math> is even. | We start by splitting the sum of all <math>C_{n-2}C_{n-1}</math> into two parts: those where <math>n-2</math> is odd and those where <math>n-2</math> is even. | ||
Line 36: | Line 47: | ||
Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution, and our answer is <math>p=13+84+85=\boxed{182}</math>. | Setting <math>a=C_{0}B</math>, <math>b=C_{0}A</math>, and <math>c=AB</math>, we can now proceed as in Shaddoll's solution, and our answer is <math>p=13+84+85=\boxed{182}</math>. | ||
− | + | == Solution 4== | |
− | |||
− | == Solution | ||
<asy> | <asy> | ||
size(10cm); | size(10cm); | ||
Line 65: | Line 74: | ||
label("$C_3$", C3, NE); | label("$C_3$", C3, NE); | ||
label("$C_4$", C4, W); | label("$C_4$", C4, W); | ||
+ | label("$a$", (B+C0)/2, W); | ||
+ | label("$b$", (A+C0)/2, S); | ||
+ | label("$c$", (A+B)/2, NE); | ||
+ | |||
// Drawings | // Drawings | ||
Line 114: | Line 127: | ||
</asy> | </asy> | ||
− | ==Solution | + | ==Solution 5== |
This solution proceeds from <math>\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)</math>. Note the general from for a primitive pythagorean triple, <math>m^2-n^2, 2mn, m^2+n^2</math> and after substitution, letting <math>a = m^2-n^2, b = 2mn, c = m^2+n^2</math> into the previous equation simplifies down very nicely into <math>m = 13n</math>. Thus <math>a = 168n^2, b = 26n^2, c = 170n^2</math>. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving <math>a = 84, b = 13, c = 85</math> yielding <math>p = a + b + c = \boxed{182}</math>. | This solution proceeds from <math>\frac{\frac{ab}{c}}{1-\frac{a}{c}} = \frac{\frac{ab}{c}}{\frac{c-a}{c}} = \frac{ab}{c-a} = 6(a+b+c)</math>. Note the general from for a primitive pythagorean triple, <math>m^2-n^2, 2mn, m^2+n^2</math> and after substitution, letting <math>a = m^2-n^2, b = 2mn, c = m^2+n^2</math> into the previous equation simplifies down very nicely into <math>m = 13n</math>. Thus <math>a = 168n^2, b = 26n^2, c = 170n^2</math>. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving <math>a = 84, b = 13, c = 85</math> yielding <math>p = a + b + c = \boxed{182}</math>. | ||
+ | |||
+ | Alternate way to find a, b, and c (from VarunGotem) : <math>7a=6b+6c</math> implies that <math>6b=7a-6c</math>. Because <math>a^2+b^2=c^2</math>, <math>36a^2+(6b)^2=36c^2</math>. Plugging in gives <math>85a^2-84ac+36c^2=36c^2</math>. Simplifying gives <math>85a-84c=0</math>, and since <math>a</math> and <math>c</math> are relatively prime, <math>a=84</math> and <math>c=85</math>. This means <math>b=13</math> and <math>p=13+84+85=\boxed{182}</math>. | ||
+ | |||
+ | ==Solution 6== | ||
+ | For this problem, first notice that its an infinite geometric series of <math>6(a+b+c)=\frac{ab}{c-b}</math> if <math>c</math> is the hypotenuse. WLOG <math>a<b</math>, we can generalize a pythagorean triple of <math>x^2-y^2, 2xy, x^2+y^2</math>. Let <math>b=2xy</math>, then this generalization gives <math>6(a+b+c)(c-b)=ab</math> | ||
+ | <cmath>(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2</cmath> | ||
+ | <cmath>(x+y)xy=6(x^2+xy)(x-y)</cmath> | ||
+ | <cmath>xy=6x(x-y)</cmath> | ||
+ | <cmath>7xy=6x^2</cmath> | ||
+ | <cmath>y=\frac{6}{7}x</cmath> | ||
+ | |||
+ | Now this is just clear. Let <math>x=7m</math> and <math>y=6m</math> for <math>m</math> to be a positive integer, the pythagorean triple is <math>13-84-85</math> which yields <math>\boxed{182}</math>. | ||
== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:42, 6 August 2023
Contents
Problem
Triangle has a right angle at
. Its side lengths are pairwise 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
.
Solution 1
Do note that by counting the area in 2 ways, the first altitude is . By similar triangles, the common ratio is
for each height, so by the geometric series formula, we have
Writing
and clearing denominators, we get
Thus
,
, and
, i.e.
. Plugging these into
, we get
, i.e.,
. Thus
and
,
,
,
. Taking
(since
are relatively prime) we get
.
Solution 2
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
Note: a more rigorous solution instead of checking for triples would be to substitute and heavily simplifying the equation. Eventually we are left with
and since
are relatively prime, we know
and
From here we can note that
is a pythagorean triple or again use the Pythagorean Theorem to find
Thus, the answer is
Solution modified/fixed from Shaddoll's solution.
Solution 3
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 4
Let ,
, and
.
Note that the total length of the red segments in the figure above is equal to the length of the blue 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 blue segment times
.
In other words, we have that
.
Guessing and checking Pythagorean triples reveals that ,
,
, and
satisfies this equation.
Solution 5
This solution proceeds from . Note the general from for a primitive pythagorean triple,
and after substitution, letting
into the previous equation simplifies down very nicely into
. Thus
. Since we know all three side lengths are relatively prime, we must divide each by 2 and let n = 1 giving
yielding
.
Alternate way to find a, b, and c (from VarunGotem) : implies that
. Because
,
. Plugging in gives
. Simplifying gives
, and since
and
are relatively prime,
and
. This means
and
.
Solution 6
For this problem, first notice that its an infinite geometric series of if
is the hypotenuse. WLOG
, we can generalize a pythagorean triple of
. Let
, then this generalization gives
Now this is just clear. Let and
for
to be a positive integer, the pythagorean triple is
which yields
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.