Difference between revisions of "2010 AIME II Problems/Problem 12"
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− | == Problem | + | == Problem == |
− | Two | + | Two non[[congruent]] integer-sided [[isosceles triangle]]s have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is <math>8: 7</math>. Find the minimum possible value of their common [[perimeter]]. |
− | + | == Solution 1== | |
− | == Solution == | + | Let the first triangle have side lengths <math>a</math>, <math>a</math>, <math>14c</math>, and the second triangle have side lengths <math>b</math>, <math>b</math>, <math>16c</math>, where <math>a, b, 2c \in \mathbb{Z}</math>. |
− | |||
− | Let the first triangle | ||
− | |||
− | and the second triangle | ||
− | |||
− | where <math>a, b, 2c \in \mathbb{Z}</math>. | ||
<br/> | <br/> | ||
− | Equal perimeter: | + | Equal perimeter: |
<center> | <center> | ||
Line 26: | Line 20: | ||
<center> | <center> | ||
− | <math>\begin{array}{ | + | <math>\begin{array}{cccl} |
7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ | 7c(\sqrt{a^2-(7c)^2})&=&8c(\sqrt{b^2-(8c)^2})&{}\\ | ||
− | 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{b+8c)(b-8c)})&{}\\ | + | 7(\sqrt{(a+7c)(a-7c)})&=&8(\sqrt{(b+8c)(b-8c)})&{}\\ |
− | 7(\sqrt{(a-7c)})&=&8(\sqrt(b-8c)})&\text{(Note that} a+7c=b+8c)\\ | + | 7(\sqrt{(a-7c)})&=&8(\sqrt{(b-8c)})&\text{(Note that } a+7c=b+8c)\\ |
49a-343c&=&64b-512c&{}\\ | 49a-343c&=&64b-512c&{}\\ | ||
49a+169c&=&64b&{}\\ | 49a+169c&=&64b&{}\\ | ||
− | 49a+169(a-b)&=&64b&\text{(Note that} c=a-b)\\ | + | 49a+169(a-b)&=&64b&\text{(Note that } c=a-b)\\ |
218a&=&233b&{}\\ | 218a&=&233b&{}\\ | ||
\end{array}</math> | \end{array}</math> | ||
</center> | </center> | ||
− | Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a= | + | Since <math>a</math> and <math>b</math> are integer, the minimum occurs when <math>a=233</math>, <math>b=218</math>, and <math>c=15</math>. Hence, the perimeter is <math>2a+14c=2(233)+14(15)=\boxed{676}</math>. |
+ | |||
+ | == Solution 2== | ||
+ | Let <math>s</math> be the semiperimeter of the two triangles. Also, let the base of the longer triangle be <math>16x</math> and the base of the shorter triangle be <math>14x</math> for some arbitrary factor <math>x</math>. Then, the dimensions of the two triangles must be <math>s-8x,s-8x,16x</math> and <math>s-7x,s-7x,14x</math>. By Heron's Formula, we have | ||
− | + | <center> | |
+ | <cmath>\sqrt{s(8x)(8x)(s-16x)}=\sqrt{s(7x)(7x)(s-14x)}</cmath> | ||
+ | <cmath>8\sqrt{s-16x}=7\sqrt{s-14x}</cmath> | ||
+ | <cmath>64s-1024x=49s-686x</cmath> | ||
+ | <cmath>15s=338x</cmath> | ||
+ | </center> | ||
+ | Since <math>15</math> and <math>338</math> are coprime, to minimize, we must have <math>s=338</math> and <math>x=15</math>. However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by <math>2</math>, which gives us a final answer of <math>\boxed{676}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=11|num-a=13|n=II}} | {{AIME box|year=2010|num-b=11|num-a=13|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 17:03, 9 August 2018
Contents
Problem
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is . Find the minimum possible value of their common perimeter.
Solution 1
Let the first triangle have side lengths , , , and the second triangle have side lengths , , , where .
Equal perimeter:
Equal Area:
Since and are integer, the minimum occurs when , , and . Hence, the perimeter is .
Solution 2
Let be the semiperimeter of the two triangles. Also, let the base of the longer triangle be and the base of the shorter triangle be for some arbitrary factor . Then, the dimensions of the two triangles must be and . By Heron's Formula, we have
Since and are coprime, to minimize, we must have and . However, we want the minimum perimeter. This means that we must multiply our minimum semiperimeter by , which gives us a final answer of .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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