2020 AIME II Problems/Problem 13
Contents
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 .
Solution 1
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 (Complex Bash)
Suppose that the circle intersects , , , , and at , , , , and respectively. Then , , , , and . So , , , , and . Then , so . Then we can solve for each individually. , , , , and . To find the radius, we notice that , or . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, is real. Expanding, we get: , then . On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0. . . must equal 4, as r cannot be negative or be approximately equal to 1. Thus, the area of is
-nihao4112
Solution 3 (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 .
~Lopkiloinm
Solution 4 (Guess)
Because the AIME answers have to be a whole number it would meant the radius of the circle have to be a whole number, thus by drawing the diagram and experimenting, we can safely say the radius is 4 and the answer is 60
(Edit: While the guess would be technically correct, the assumption that the radius would have to be a whole number for the ans to be a whole number is wrong)
By EtherealMidnight
(Edit: I think that will actually work because the area of is equal to the semi-perimeter times the radius. By a simple calculation, we know that the semi-perimeter is an integer so the radius should also be an integer)
By YBSuburbanTea
...the radius could be a fraction with denominator 3, 5, or 15, and the area of the pentagon would still be an integer. - GeometryJake
Solution 5 (Official MAA 1)
Let be the inscribed circle, be its center, and be its radius. The area of is equal to its semiperimeter, times , so the problem is reduced to finding . Let be the length of the tangent segment from to , and analogously define , , , and . Then , , and , with a total of . Hence , , and . It follows that and . Let be the point where is tangent to . Then . The sum of the internal angles in polygons and are equal, so , which implies that must be . Therefore points , , and are collinear. Because , it follows thatAnother expression for can be found as follows. Note that and , so and Applying the Law of Cosines to and gives and Hence
yielding equivalently Substituting gives the quadratic equation , with solutions , and . The solution corresponds to a five-pointed star, which is not convex. Indeed, if , then , , and are less than implying that , , and are acute, which cannot happen in a convex pentagon. Thus and . The requested area is .
Solution 6 (Official MAA 2)
Define , , , , , and as in Solution 5. Then, as in Solution 5, , , , , and . Let , , and . It follows that , so . Thus , , and . By the Tangent Addition Formula, and Therefore which simplifies to . Then the solution proceeds as in Solution 5.
Solution 7 (Official MAA 3)
Define , , , , , and as in Solution 5. Note that Hence Therefore Simplifying this equation gives the same quadratic equation in as in Solution 5.
Solution 8 (The same circle)
Notation shown on diagram. As in soluti45on 5, we get and so on.
Let cross at cross at $356y^&*(REI^W$ (Error compiling LaTeX. Unknown error_msg)EU%W^&J OP(_{:>)}|?+_{"\angle BAQ = \angle EAQ \implies DFusnriyuvuti ou4y5ouy' = x + 2, EF' = x.\triangle AFF's = 2x + 11.\omegar =\sqrt{\frac{s-FF’}{s}}(s-AF) = \sqrt{\frac{3}{2x +11}}(x+4). \triangle BCFs = x + 4.\omegar = \sqrt{\frac{s(s-BF)(s-CF)}{s-BC}} = \sqrt{ \frac{(x+4)\cdot 2 \cdot 4}{x - 2}}.$
It is the same radius, therefore
Then the solution proceeds as in Solution 5.
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1
https://youtu.be/bz5N-jI2e0U?t=327
Video Solution 2
Video Solution 3
https://youtu.be/kn3c2LStiHA (solve in 5 minutes)
~MathProblemSolvingSkills.com
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.