Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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== Solution 5 (Quick) == | == Solution 5 (Quick) == | ||
From <math>AIME</math> cyclic we get <math>\angle{MEI} = \angle{MAI} = 45^\circ</math> and <math>\angle{MIE} = \angle{MAE} = 45^\circ</math>, so <math>\triangle{EMI}</math> is an isosceles right triangle. | From <math>AIME</math> cyclic we get <math>\angle{MEI} = \angle{MAI} = 45^\circ</math> and <math>\angle{MIE} = \angle{MAE} = 45^\circ</math>, so <math>\triangle{EMI}</math> is an isosceles right triangle. | ||
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From <math>[EMI]=2</math> we get <math>EM=MI=2</math>. | From <math>[EMI]=2</math> we get <math>EM=MI=2</math>. | ||
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Notice <math>\triangle{AEM} \cong \triangle{CIM}</math>, because <math>\angle{AEM}=180-\angle{AIM}=\angle{CIM}</math>, <math>EM=IM</math>, and <math>\angle{EAM}=\angle{ICM}=45^\circ</math>. | Notice <math>\triangle{AEM} \cong \triangle{CIM}</math>, because <math>\angle{AEM}=180-\angle{AIM}=\angle{CIM}</math>, <math>EM=IM</math>, and <math>\angle{EAM}=\angle{ICM}=45^\circ</math>. | ||
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Let <math>CI=AE=x</math>, so <math>AI=3-x</math>. | Let <math>CI=AE=x</math>, so <math>AI=3-x</math>. | ||
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By Pythagoras on <math>\triangle{EAI}</math> we have <math>x^2+(3-x)^2=EI^2=8</math>, and solve this to get <math>x=CI=\dfrac{3-\sqrt{7}}{2}</math> for a final answer of <math>3+7+2=\boxed{12}</math>. | By Pythagoras on <math>\triangle{EAI}</math> we have <math>x^2+(3-x)^2=EI^2=8</math>, and solve this to get <math>x=CI=\dfrac{3-\sqrt{7}}{2}</math> for a final answer of <math>3+7+2=\boxed{12}</math>. | ||
Revision as of 02:33, 24 December 2018
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
. Since
is cyclic,
(which is opposite of another right angle) must be a right angle; therefore,
. Compute the dot product to arrive at the relation
. We can set up another equation involving the area of
using the Shoelace Theorem. This is
. Multiplying, substituting
for
, and simplifying, we get
. Thus,
. But
, meaning
, and the final answer is
.
Solution 5 (Quick)
From cyclic we get
and
, so
is an isosceles right triangle.
From we get
.
Notice , because
,
, and
.
Let , so
.
By Pythagoras on we have
, and solve this to get
for a final answer of
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.