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

Revision as of 14:04, 24 February 2020

Problem

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 (Angle Chasing/Trig)

Let $O$ be the center of the circle and the point of tangency between $\omega$ and $\overline{AD}$ be represented by $K$ . We know that $\overline{AK} = \overline{KD} = \overline{DM} = \frac{1}{2}$ . Consider the right triangle $\bigtriangleup ADM$ . Let $\measuredangle AMD = \theta$ .

Since $\omega$ is tangent to $\overline{DC}$ at $M$ , $\measuredangle PMO = 90 - \theta$ . Now, consider $\bigtriangleup POM$ . This triangle is iscoceles because $\overline{PO}$ and $\overline{OM}$ are both radii of $\omega$ . Therefore, $\measuredangle POM = 180 - 2(90 - \theta) = 2\theta$ .

We can now use Law of Cosines on $\angle{POM}$ to find the length of ${PM}$ and subtract it from the length of ${AM}$ to find ${AP}$ . Since $\cos{\theta} = \frac{1}{\sqrt{5}}$ and $\sin{\theta} = \frac{2}{\sqrt{5}}$ , the double angle formula tells us that $\cos{2\theta} = -\frac{3}{5}$ . We have $PM^2 = \frac{1}{2} - \frac{1}{2}\cos{2\theta} \implies PM = \frac{2\sqrt{5}}{5}$ By Pythagorean theorem, we find that $AM = \frac{\sqrt{5}}{2} \implies \boxed{\textbf{(B) } \frac{\sqrt5}{10}}$

~awesome1st

Solution 2(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 $AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\mathbf{(B) } \frac{\sqrt{5}}{10}}$ .

~Argonauts16

Solution 3(Power of a Point)

Let circle $\omega$ intersect $\overline{AB}$ at point $N$ . By Power of a Point, we have $AN^2=AP\cdot AM$ . We know $AN=\frac{1}{2}$ because $N$ is the midpoint of $\overline{AB}$ , and we can easily find $AM$ by the Pythagorean Theorem, which gives us $AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}$ . Our equation is now $\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}$ , or $AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}$ , thus our answer is $\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.$

~Argonauts16

Solution 4

Take $O$ as the center and draw segment $ON$ perpendicular to $AM$ , $ON\cap AM=N$ , link $OM$ . Then we have $OM\parallel AD$ . So $\angle DAM=\angle OMA$ . Since $AD=2AM=2OM=1$ , we have $\cos\angle DAM=\cos\angle OMP=\frac{2}{\sqrt{5}}$ . As a result, $NM=OM\cos\angle OMP=\frac{1}{2}\cdot \frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}.$ Thus $PM=2NM=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$ . Since $AM=\frac{\sqrt{5}}{2}$ , we have $AP=AM-PM=\frac{\sqrt{5}}{10}$ . Put $\boxed{B}$ .

~FANYUCHEN20020715

Solution 5 (Similar Triangles)

Call the midpoint of $\overline{AB}$ point $N$ . Draw in $\overline{NM}$ and $\overline{NP}$ . Note that $\angle{NPM}=90^{\circ}$ 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. $\triangle APN\sim\triangle ANM\Rightarrow\frac{AP}{AN}=\frac{AN}{AM}\Rightarrow\frac{AP}{\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}}$ $AP=\boxed{\textbf{(B) }\frac{\sqrt{5}}{10}}$ ~QIDb602

Video Solution

https://youtu.be/6ujfjGLzVoE

~IceMatrix

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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