2018 AMC 12A Problems/Problem 20
Contents
Problem
Triangle is an isosceles right triangle with
. Let
be the midpoint of hypotenuse
. Points
and
lie on sides
and
, respectively, so that
and
is a cyclic quadrilateral. Given that triangle
has area
, the length
can be written as
, where
,
, and
are positive integers and
is not divisible by the square of any prime. What is the value of
?
Solution 1
Observe that is isosceles right (
is the midpoint of diameter arc
), so
. With
, we can use Law of Cosines to determine that
. The same calculations hold for
also, and since
, we deduce that
is the smaller root, giving the answer of
. (trumpeter)
Solution 2 (Using Ptolemy)
We first claim that is isosceles and right.
Proof: Construct and
. Since
bisects
, one can deduce that
. Then by AAS it is clear that
and therefore
is isosceles. Since quadrilateral
is cyclic, one can deduce that
. Q.E.D.
Since the area of is 2, we can find that
,
Since is the mid-point of
, it is clear that
.
Now let and
. By Ptolemy's Theorem, in cyclic quadrilateral
, we have
. By Pythagorean Theorem, we have
. One can solve the simultaneous system and find
. Then by deducting the length of
from 3 we get
, giving the answer of
. (Surefire2019)
Solution 3 (More Elementary)
Like above, notice that is isosceles and right, which means that
, so
and
. Then construct
and
as well as
. It's clear that
by Pythagorean, so knowing that
allows one to solve to get
. By just looking at the diagram,
. The answer is thus
.
Solution 4 (Coordinate Geometry)
Let lie on
,
on
,
on
, and
on
.
is cyclic, so
is a right angle; therefore
. Multiply out and simplify to arrive at the relation
. We can set up another equation for the area of
using the Shoelace Theorem. This is
. Multiplying, substituting
for
and simplifying, we arrive at
. Thus, the solution set is
. But
, meaning
, and the final answer is
.
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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