2006 SMT/Geometry Problems/Problem 4

Problem

The distance $AB$ is $l$ . Find the area of the locus of points $X$ such that $15^\circ\le\angle AXB\le30^\circ$ and $X$ is on the same side of line $AB$ as a given point $C$ .

Solution

$[asy] draw((-1,0)--(1,0)); draw(circle((0,2+sqrt(3)),2sqrt(2+sqrt(3)))); draw(circle((0,sqrt(3)),2)); label("$A$",(-1,0),SW); label("$B$",(1,0),SE); draw((0,0)--(0,2+sqrt(3)+2sqrt(2+sqrt(3)))); dot((0,2+sqrt(3))); dot((0,sqrt(3))); draw((0,sqrt(3))--(-1,0)); draw((0,2+sqrt(3))--(-1,0)); label("$X$",(0,sqrt(3)),E); label("$Y$",(0,2+sqrt(3)),NE); label("$M$",(0,0),NE); label("$Z$",(0,2+sqrt(3)+2sqrt(2+sqrt(3))),N); [/asy]$

Consider the locus of points $L$ such that $\angle ALB=15^\circ$ . As per the above diagram, let circle $Y$ be a circle and $Z$ a point on that circle such that $\angle AZB=15^\circ$ . Any point on that circle will intercept the same arc, and clearly any point not on that circle will have $\angle AZB$ either less than or greater than $15^\circ$ . Therefore, this circle $Y$ is the desired locus. We define circle $X$ the same way, except such that $\angle ALB=30^\circ$ .

Since $\angle AZB=15^\circ$ is an inscribed angle, we have $\angle AYB=30^\circ$ . Therefore, $Y$ must lie on circle $X$ . We now want the area between circles $X$ and $Y$ . Let $M$ be the midpoint of $AB$ , so that $AM=\frac{l}{2}$ .

First, let's try to find the radii of the circles. $\angle AYB=30^\circ\implies \angle AYM=15^\circ$ . From the half-angle identities, we have $\tan15^\circ=2-\sqrt{3}$ , so $\frac{\frac{l}{2}}{YM}=2-\sqrt{3}$ , and $YM=\frac{l(2+\sqrt{3})}{2}$ . Therefore, from the Pythagorean Theorem on $\triangle AMY$ , we have $AY=l\sqrt{2+\sqrt{3}}$ .

We do the same thing on circle $X$ . However, notice that $\triangle AXM$ is a $30-60-90$ right triangle, so we quickly find that $AX=2\left(\frac{l}{2}\right)=l$ .

Therefore, we have the radii of the circles. We now just need to find the area between them. We can do this by finding the area of sector $AYZ$ , subtract the area of the portion of circle $X$ cut off by chord $AY$ , and multiply the resulting difference by $2$ . Since $\angle AYM=15^\circ$ , we have $\angle AYZ=165^\circ$ , so the area of sector $AYZ$ is $\left(\frac{165}{360}\right)(\pi)\left(l\sqrt{2+\sqrt{3}}\right)^2=\frac{l^2(22\pi+11\pi\sqrt{3})}{24}$ .

We now need to find the area of circle $X$ cut off by chord $AY$ . We have $\angle AXY=180^\circ-30^\circ=150^\circ$ and $XA=XY=l$ . Therefore, the area of sector $AXY$ is $\left(\frac{150}{360}\right)(\pi)(l)^2=\frac{5\pi l^2}{12}$ , and the area of $\triangle AXY$ is $\frac{1}{2}(l)(l)\sin150^\circ=\frac{l^2}{4}$ . Thus, the area of circle $X$ cut off by chord $AY$ is $\frac{l^2(5\pi-3)}{12}$ .

Now, we subtract this from the area of sector $AYZ$ to get $\frac{l^2(22\pi+11\pi\sqrt{3})}{24}-\frac{l^2(10\pi-6)}{24}=l^2\left(\frac{12\pi+11\pi\sqrt{3}+6}{24}\right)$ , and we multiply this by $2$ to get the total area, which is $\boxed{l^2\left(\frac{12\pi+11\pi\sqrt{3}+6}{12}\right)}$ .

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2006 SMT/Geometry Problems/Problem 4

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