Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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− | == Solution == | + | == Solution 1== |
Observe that <math>\triangle{EMI}</math> is isosceles right (<math>M</math> is the midpoint of diameter arc <math>EI</math>), so <math>MI=2,MC=\frac{3}{\sqrt{2}}</math>. With <math>\angle{MCI}=45^\circ</math>, we can use Law of Cosines to determine that <math>CI=\frac{3\pm\sqrt{7}}{2}</math>. The same calculations hold for <math>BE</math> also, and since <math>CI<BE</math>, we deduce that <math>CI</math> is the smaller root, giving the answer of <math>\boxed{12}</math>. (trumpeter) | Observe that <math>\triangle{EMI}</math> is isosceles right (<math>M</math> is the midpoint of diameter arc <math>EI</math>), so <math>MI=2,MC=\frac{3}{\sqrt{2}}</math>. With <math>\angle{MCI}=45^\circ</math>, we can use Law of Cosines to determine that <math>CI=\frac{3\pm\sqrt{7}}{2}</math>. The same calculations hold for <math>BE</math> also, and since <math>CI<BE</math>, we deduce that <math>CI</math> is the smaller root, giving the answer of <math>\boxed{12}</math>. (trumpeter) | ||
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Now let <math>AE=a</math> and <math>AI=b</math>. By Ptolemy's Theorem, in cyclic quadrilateral <math>AIME</math>, we have <math>2a+2b=6</math>. By Pythagorean Theorem, we have <math>a^2+b^2=8</math>. One can solve the simultaneous system and find <math>b=\frac{3+\sqrt{7}}{2}</math>. Then by deducting the length of <math>\overline{AI}</math> from 3 we get <math>CI=\frac{3-\sqrt{7}}{2}</math>, giving the answer of <math>\boxed{12}</math>. (Surefire2019) | Now let <math>AE=a</math> and <math>AI=b</math>. By Ptolemy's Theorem, in cyclic quadrilateral <math>AIME</math>, we have <math>2a+2b=6</math>. By Pythagorean Theorem, we have <math>a^2+b^2=8</math>. One can solve the simultaneous system and find <math>b=\frac{3+\sqrt{7}}{2}</math>. Then by deducting the length of <math>\overline{AI}</math> from 3 we get <math>CI=\frac{3-\sqrt{7}}{2}</math>, giving the answer of <math>\boxed{12}</math>. (Surefire2019) | ||
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+ | == Solution 3 (More Elementary) == | ||
+ | |||
+ | Like above, notice that <math>\triangle{EMI}</math> is isosceles and right, which means that <math>\dfrac{ME \cdot MI}{2} = 2</math>, so <math>MI^2=4</math> and <math>MI = 2</math>. Then construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math> as well as <math>\overline{MI}</math>. It's clear that <math>MG^2+GI^2 = MI^2</math> by Pythagorean, so knowing that <math>MG = \dfrac{AB}{2} = \dfrac{3}{2}</math> allows one to solve to get <math>GI = \dfrac{\sqrt{7}}{2}</math>. By just looking at the diagram, <math>CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}</math>. The answer is thus <math>3+7+2=12</math>. | ||
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==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 20:49, 6 May 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
.
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.