Difference between revisions of "2020 AMC 12B Problems/Problem 10"
Argonauts16 (talk | contribs) (Created page with "==Problem 10== 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}</ma...") |
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<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> | ||
− | ==Solution== | + | ==Solution 1(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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We have <math>A=\left(-\frac{1}{2}, \frac{1}{2}\right)</math> and <math>M=\left(0, -\frac{1}{2}\right)</math>. The slope of the line passing through these two points is <math>\frac{\frac{1}{2}+\frac{1}{2}}{-\frac{1}{2}-0}=\frac{1}{-\frac{1}{2}}=-2</math>, and the <math>y</math>-intercept is simply <math>M</math>. This gives us the line passing through both points as <math>y=-2x-\frac{1}{2}</math>. | We have <math>A=\left(-\frac{1}{2}, \frac{1}{2}\right)</math> and <math>M=\left(0, -\frac{1}{2}\right)</math>. The slope of the line passing through these two points is <math>\frac{\frac{1}{2}+\frac{1}{2}}{-\frac{1}{2}-0}=\frac{1}{-\frac{1}{2}}=-2</math>, and the <math>y</math>-intercept is simply <math>M</math>. This gives us the line passing through both points as <math>y=-2x-\frac{1}{2}</math>. | ||
− | We substitute this into the equation for the circle to get <math>x^2+\left(-2x-\frac{1}{2}\right)^2=\frac{1}{4}</math>, or <math>x^2+4x^2+2x+\frac{1}{4}=\frac{1}{4}</math>. Simplifying gives <math>x(5x+2)=0</math>. The roots of this quadratic are <math>x=0</math> and <math>x=-\frac{2}{5}</math>, but if <math>x=0</math> we get point <math>M</math>, so we only want <math>x=-\frac{2}{5}. | + | We substitute this into the equation for the circle to get <math>x^2+\left(-2x-\frac{1}{2}\right)^2=\frac{1}{4}</math>, or <math>x^2+4x^2+2x+\frac{1}{4}=\frac{1}{4}</math>. Simplifying gives <math>x(5x+2)=0</math>. The roots of this quadratic are <math>x=0</math> and <math>x=-\frac{2}{5}</math>, but if <math>x=0</math> we get point <math>M</math>, so we only want <math>x=-\frac{2}{5}</math>. |
− | We plug this back into the linear equation to find < | + | 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>. |
Revision as of 21:26, 7 February 2020
Problem 10
In unit square the inscribed circle
intersects
at
and
intersects
at a point
different from
What is
Solution 1(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
.