Difference between revisions of "2020 AIME II Problems/Problem 13"
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==Solution 2 (Guess)== | ==Solution 2 (Guess)== | ||
− | The area of a regular pentagon with side lengths <math>s</math> is <math>\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}</math>. <math>5-2\sqrt{5}</math> is slightly greater than <math>\frac{1}{2}</math> given that <math>2\sqrt{5}</math> is slightly less than <math>\frac{9}{2}</math>. <math>4\sqrt{5-2\sqrt{5}}</math> is then slightly greater than <math>2\sqrt{2}</math>. We will approximate that to be <math>2.9</math>. The area is now roughly <math>\frac{180}{2.9}</math>, but because the pentagon is not regular we can say the area is <math>\frac{180}{3}</math> which is <math>60</math> and since <math>60</math> is a multiple of the semiperimeter <math>15</math>, we can safely say that the answer is most likely <math>60</math>. | + | This pentagon is very close to a regular pentagon with side lengths <math>6</math>. The area of a regular pentagon with side lengths <math>s</math> is <math>\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}</math>. <math>5-2\sqrt{5}</math> is slightly greater than <math>\frac{1}{2}</math> given that <math>2\sqrt{5}</math> is slightly less than <math>\frac{9}{2}</math>. <math>4\sqrt{5-2\sqrt{5}}</math> is then slightly greater than <math>2\sqrt{2}</math>. We will approximate that to be <math>2.9</math>. The area is now roughly <math>\frac{180}{2.9}</math>, but because the pentagon is not regular we can say the area is <math>\frac{180}{3}</math> which is <math>60</math> and since <math>60</math> is a multiple of the semiperimeter <math>15</math>, we can safely say that the answer is most likely <math>60</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 14:53, 8 June 2020
Problem
Convex pentagon has side lengths
,
, and
. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of
.
Solutions (Misplaced problem?)
Assume the incircle touches ,
,
,
,
at
respectively. Then let
,
,
. So we have
,
and
=7, solve it we have
,
,
. Let the center of the incircle be
, by SAS we can proof triangle
is congruent to triangle
, and triangle
is congruent to triangle
. Then we have
,
. Extend
, cross ray
at
, ray
at
, then by AAS we have triangle
is congruent to triangle
. Thus
. Let
, then
. So by law of cosine in triangle
and triangle
we can obtain
, solved it gives us
, which yield triangle
to be a triangle with side length 15, 15, 24, draw a height from
to
divides it into two triangles with side lengths 9, 12, 15, so the area of triangle
is 108. Triangle
is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is
.
-Fanyuchen20020715
Solution 2 (Guess)
This pentagon is very close to a regular pentagon with side lengths . The area of a regular pentagon with side lengths
is
.
is slightly greater than
given that
is slightly less than
.
is then slightly greater than
. We will approximate that to be
. The area is now roughly
, but because the pentagon is not regular we can say the area is
which is
and since
is a multiple of the semiperimeter
, we can safely say that the answer is most likely
.
Video Solution
https://youtu.be/bz5N-jI2e0U?t=327
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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