Difference between revisions of "2020 AIME II Problems/Problem 13"
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==Solution 2 (Guess)== | ==Solution 2 (Guess)== | ||
− | 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 | + | 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 actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the <math>2.9</math> into <math>3</math> thus turning the area into <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 15:02, 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 actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the
into
thus turning the area into
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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