Difference between revisions of "2020 AMC 12B Problems/Problem 10"
Awesome1st (talk | contribs) |
Argonauts16 (talk | contribs) (→Solution 2(Coordinate Bash)) |
||
Line 26: | Line 26: | ||
We plug this back into the linear equation to find <math>y=\frac{3}{10}</math>, and so <math>P=\left(-\frac{2}{5}, \frac{3}{10}\right)</math>. Finally, we use distance formula on <math>A</math> and <math>P</math> to get <math>AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\mathbf{(B) } \frac{\sqrt{5}}{10}}</math>. | We plug this back into the linear equation to find <math>y=\frac{3}{10}</math>, and so <math>P=\left(-\frac{2}{5}, \frac{3}{10}\right)</math>. Finally, we use distance formula on <math>A</math> and <math>P</math> to get <math>AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\mathbf{(B) } \frac{\sqrt{5}}{10}}</math>. | ||
+ | |||
+ | ~Argonauts16 | ||
==Solution 3(Power of a Point)== | ==Solution 3(Power of a Point)== |
Revision as of 16:52, 9 February 2020
Contents
Problem
In unit square the inscribed circle
intersects
at
and
intersects
at a point
different from
What is
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
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
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.