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 (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 (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 (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 that
Another expression for
can be found as follows. Note that
and
, so
and
Applying the Law of Cosines to
and
gives
\[
AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}
\]and
\[
AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.
\]Hence
\[
12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},
\]yielding
\[
2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;
\]equivalently
\[
7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).
\]
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
.
Video Solution
https://youtu.be/bz5N-jI2e0U?t=327
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