Difference between revisions of "2020 AMC 12B Problems/Problem 10"
Argonauts16 (talk | contribs) (→Solution 3(Power of a Point)) |
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~FANYUCHEN20020715 | ~FANYUCHEN20020715 | ||
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+ | ==Solution 5 (Similar Triangles)== | ||
+ | Call the midpoint of <math>\overline{AB}</math> point <math>N</math>. Draw in <math>\overline{NM}</math> and <math>\overline{NP}</math>. Note that <math>\angle{NPM}=90^{\circ}</math> due to Thales's Theorem. | ||
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); | ||
+ | draw(circle((0.5,0.5),0.5)); | ||
+ | draw((0,0)--(0,0.5)--(1,0.5)--cycle); | ||
+ | label("A",(0,0),SW); | ||
+ | label("B",(0,1),NW); | ||
+ | label("C",(1,1),NE); | ||
+ | label("D",(1,0),SE); | ||
+ | label("M",(1,0.5),E); | ||
+ | label("P",(0.2,0.1),S); | ||
+ | label("N",(0,0.5),W); | ||
+ | draw((0,0.5)--(0.2,0.1)); | ||
+ | markscalefactor=0.007; | ||
+ | draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); | ||
+ | </asy> | ||
+ | Now we just need to find the length by setting up proportions on similar triangles. | ||
+ | <cmath>\triangle APN\sim\triangle ANM\Rightarrow\frac{AP}{AN}=\frac{AN}{AM}\Rightarrow\frac{AP}{\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}}</cmath> | ||
+ | <cmath>AP=\boxed{\textbf{(B) }\frac{\sqrt{5}}{10}}</cmath> | ||
+ | ~QIDb602 | ||
==Video Solution== | ==Video Solution== |
Revision as of 15:04, 24 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
~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.
Now we just need to find the length by setting up proportions on similar triangles.
~QIDb602
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.