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Difference between revisions of "2020 AMC 12B Problems/Problem 10"

(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...")
 
(→‎Solution)
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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>
  
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==Solution==
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==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>.  
  
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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>.
  
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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}}$.
+
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 $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$

Solution 1(Coordinate Bash)

Place circle $\omega$ in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for $\omega$ as $x^2+y^2=\frac{1}{4}$, because it is not translated and the radius is $\frac{1}{2}$.

We have $A=\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $M=\left(0, -\frac{1}{2}\right)$. The slope of the line passing through these two points is $\frac{\frac{1}{2}+\frac{1}{2}}{-\frac{1}{2}-0}=\frac{1}{-\frac{1}{2}}=-2$, and the $y$-intercept is simply $M$. This gives us the line passing through both points as $y=-2x-\frac{1}{2}$.

We substitute this into the equation for the circle to get $x^2+\left(-2x-\frac{1}{2}\right)^2=\frac{1}{4}$, or $x^2+4x^2+2x+\frac{1}{4}=\frac{1}{4}$. Simplifying gives $x(5x+2)=0$. The roots of this quadratic are $x=0$ and $x=-\frac{2}{5}$, but if $x=0$ we get point $M$, so we only want $x=-\frac{2}{5}$.

We plug this back into the linear equation to find $y=\frac{3}{10}$, and so $P=\left(-\frac{2}{5}, \frac{3}{10}\right)$. Finally, we use distance formula on $A$ and $P$ to get $AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\mathbf{(B) } \frac{\sqrt{5}}{10}}$.

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