Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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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>. | ||
==Solution 6(bash)== | ==Solution 6(bash)== | ||
− | Let <math>CI=a</math>, <math>BE=b</math>. Because opposite angles in a cyclic quadrilateral are supplementary, we have <math>\angle EMI=90^{\circ}</math>. By the law of cosines, we have <math>MI^2=a^2+\frac{9}{4}-\frac{3}{2}a</math>, and <math>ME^2=b^2+\frac{9}{4}-\frac{3}{2}b</math>. Notice that <math>EI=2MO</math>, where <math>O</math> is the origin of the circle mentioned in the problem. Thus <math>\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}</math>. By the Pythagorean Theorem, we have <math>ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8</math>. By the Pythagorean Theorem, we have <math>AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2 | + | Let <math>CI=a</math>, <math>BE=b</math>. Because opposite angles in a cyclic quadrilateral are supplementary, we have <math>\angle EMI=90^{\circ}</math>. By the law of cosines, we have <math>MI^2=a^2+\frac{9}{4}-\frac{3}{2}a</math>, and <math>ME^2=b^2+\frac{9}{4}-\frac{3}{2}b</math>. Notice that <math>EI=2MO</math>, where <math>O</math> is the origin of the circle mentioned in the problem. Thus <math>\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}</math>. By the Pythagorean Theorem, we have <math>ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8</math>. By the Pythagorean Theorem, we have <math>AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8</math>. Thus we have <math>18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}</math> <math>=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a</math>. We know that <math>(3-a)^2+(3-b)^2=8\implies (3-a)^2+a^2=8\implies 2a^2-6a+9=8\implies 2a^2-6a+1=0\implies a=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}</math>. We take the smaller solution because we have <math>AI>AE\implies 3-AI<3-AE\implies CI<CE</math>, and we want <math>CI</math>, not <math>CE</math>, thus <math>CI=\frac{3-\sqrt{7}}{2}</math>. Thus our final answer is <math>3+7+2=\boxed{12\textbf{(D)}}</math> |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:40, 6 April 2020
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
?
Diagram
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
.
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
.
Solution 6(bash)
Let ,
. Because opposite angles in a cyclic quadrilateral are supplementary, we have
. By the law of cosines, we have
, and
. Notice that
, where
is the origin of the circle mentioned in the problem. Thus
. By the Pythagorean Theorem, we have
. By the Pythagorean Theorem, we have
. Thus we have
. We know that
. We take the smaller solution because we have
, and we want
, not
, thus
. Thus our 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.