Difference between revisions of "2020 AMC 12B Problems/Problem 10"
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==Problem== | ==Problem== | ||
− | |||
In unit square <math>ABCD,</math> the inscribed circle <math>\omega</math> intersects <math>\overline{CD}</math> at <math>M,</math> and <math>\overline{AM}</math> intersects <math>\omega</math> at a point <math>P</math> different from <math>M.</math> What is <math>AP?</math> | In unit square <math>ABCD,</math> the inscribed circle <math>\omega</math> intersects <math>\overline{CD}</math> at <math>M,</math> and <math>\overline{AM}</math> intersects <math>\omega</math> at a point <math>P</math> different from <math>M.</math> What is <math>AP?</math> | ||
<math>\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}</math> | <math>\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | [[File:2020 AMC 12B Problem 12 Diagram.png]] | ||
+ | |||
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
==Solution 1 (Angle Chasing/Trig)== | ==Solution 1 (Angle Chasing/Trig)== | ||
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~awesome1st | ~awesome1st | ||
− | ==Solution 2(Coordinate Bash)== | + | |
+ | ==Solution 2 (Coordinate Bash)== | ||
Place circle <math>\omega</math> in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for <math>\omega</math> as <math>x^2+y^2=\frac{1}{4}</math>, because it is not translated and the radius is <math>\frac{1}{2}</math>. | Place circle <math>\omega</math> in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for <math>\omega</math> as <math>x^2+y^2=\frac{1}{4}</math>, because it is not translated and the radius is <math>\frac{1}{2}</math>. | ||
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~Argonauts16 | ~Argonauts16 | ||
− | ==Solution 3(Power of a Point)== | + | ==Solution 3 (Power of a Point)== |
− | |||
Let circle <math>\omega</math> intersect <math>\overline{AB}</math> at point <math>N</math>. By Power of a Point, we have <math>AN^2=AP\cdot AM</math>. We know <math>AN=\frac{1}{2}</math> because <math>N</math> is the midpoint of <math>\overline{AB}</math>, and we can easily find <math>AM</math> by the Pythagorean Theorem, which gives us <math>AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}</math>. Our equation is now <math>\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}</math>, or <math>AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}</math>, thus our answer is | Let circle <math>\omega</math> intersect <math>\overline{AB}</math> at point <math>N</math>. By Power of a Point, we have <math>AN^2=AP\cdot AM</math>. We know <math>AN=\frac{1}{2}</math> because <math>N</math> is the midpoint of <math>\overline{AB}</math>, and we can easily find <math>AM</math> by the Pythagorean Theorem, which gives us <math>AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}</math>. Our equation is now <math>\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}</math>, or <math>AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}</math>, thus our answer is | ||
<math>\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.</math> | <math>\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.</math> | ||
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~QIDb602 | ~QIDb602 | ||
− | ==Solution 6 | + | ==Solution 6 (Intersecting Chords)== |
Label the midpoint of <math>AD</math> as <math>N</math>, and let <math>Q</math> be the intersection of <math>ON</math> and <math>AM</math> | Label the midpoint of <math>AD</math> as <math>N</math>, and let <math>Q</math> be the intersection of <math>ON</math> and <math>AM</math> | ||
Then <math>MQ=AQ=\frac{1}{2}AM = \frac{\sqrt{5}}{4}</math> and <math>NQ=\frac{1}{4}</math> | Then <math>MQ=AQ=\frac{1}{2}AM = \frac{\sqrt{5}}{4}</math> and <math>NQ=\frac{1}{4}</math> |
Revision as of 08:59, 22 April 2021
Contents
Problem
In unit square the inscribed circle
intersects
at
and
intersects
at a point
different from
What is
Diagram
File:2020 AMC 12B Problem 12 Diagram.png
~MRENTHUSIASM (by Geometry Expressions)
Solution 1 (Angle Chasing/Trig)
Let be the center of the circle and the point of tangency between
and
be represented by
. We know that
. Consider the right triangle
. Let
.
Since is tangent to
at
,
. Now, consider
. This triangle is iscoceles because
and
are both radii of
. Therefore,
.
We can now use Law of Cosines on to find the length of
and subtract it from the length of
to find
. Since
and
, the double angle formula tells us that
. We have
By Pythagorean theorem, we find that
~awesome1st
Solution 2 (Coordinate Bash)
Place circle in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for
as
, because it is not translated and the radius is
.
We have and
. The slope of the line passing through these two points is
, and the
-intercept is simply
. This gives us the line passing through both points as
.
We substitute this into the equation for the circle to get , or
. Simplifying gives
. The roots of this quadratic are
and
, but if
we get point
, so we only want
.
We plug this back into the linear equation to find , and so
. Finally, we use distance formula on
and
to get
.
~Argonauts16
Solution 3 (Power of a Point)
Let circle intersect
at point
. By Power of a Point, we have
. We know
because
is the midpoint of
, and we can easily find
by the Pythagorean Theorem, which gives us
. Our equation is now
, or
, thus our answer is
~Argonauts16
Solution 4
Take as the center and draw segment
perpendicular to
,
, link
. Then we have
. So
. Since
, we have
. As a result,
Thus
. Since
, we have
. Put
.
~FANYUCHEN20020715
Solution 5 (Similar Triangles)
Call the midpoint of point
. Draw in
and
. Note that
due to Thales's Theorem.
Using the Pythagorean theorem,
. Now we just need to find
using similar triangles.
~QIDb602
Solution 6 (Intersecting Chords)
Label the midpoint of as
, and let
be the intersection of
and
Then
and
Then and
~ERMSCoach
Video Solution
~IceMatrix
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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.