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Length

Problem 19

Three semicircles of radius $1$ are constructed on diameter $overline{AB}$ of a semicircle of radius $2$ . The centers of the small semicircles divide $overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

$extbf{(A) } pi - sqrt{3} qquad extbf{(B) } pi - sqrt{2} qquad extbf{(C) } frac{pi + sqrt{2}}{2} qquad extbf{(D...$

A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle?

$(mathrm {A}) 200+25pi quad (mathrm {B}) 100+75pi quad (mathrm {C}) 75+100pi quad (mathrm {D}) 100+100pi quad (ma...$

10.

A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

$mathrm{(A) } frac{3sqrt{5}}{2} qquad mathrm{(B) } frac{7}{2} qquad mathrm{(C) } sqrt{15} qquad mathrm{(D) ...$

11.

Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

$ext{(A)} pi qquad ext{(B)} 1.5pi qquad ext{(C)} 2pi qquad ext{(D)} 3pi qquad ext{(E)} 3.5pi$

''>''

$ext{(A)} pi qquad ext{(B)} 1.5pi qquad ext{(C)} 2pi qquad ext{(D)} 3pi qquad ext{(E)} 3.5pi$

12.

An $8$ -foot by $10$ -foot floor is tiles with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?

$mathrm{(A)} 80-20pi qquadmathrm{(B)} 60-10pi qquadmathrm{(C)} 80-10pi qquadmathrm{(D)} 60+10pi ...$

13.

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

$mathrm{(A) } frac{3sqrt{2}}{pi}qquad mathrm{(B) } frac{3sqrt{3}}{pi}qquad mathrm{(C) } sqrt{3}qquad math...$

14.

A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$ , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.

$mathrm{(A) } frac{1}{6}pi-frac{sqrt{3}}{4}qquad mathrm{(B) } frac{sqrt{3}}{4}-frac{1}{12}piqquad mathrm{(C) ...$

15.

Circles $A$ , $B$ , and $C$ are externally tangent to each other and internally tangent to circle $D$ . Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$ . What is the radius of circle $B$ ?

$mathrm{(A) } frac{2}{3} qquad mathrm{(B) } frac{sqrt{3}}{2} qquad mathrm{(C) } frac{7}{8} qquad mathrm{(D) ...$

16.

17.

An equilateral triangle has side length $6$ . What is the area of the region containing all points that are outside the triangle but not more than $3$ units from a point of the triangle?

$mathrm{(A)} 36+24sqrt{3}qquadmathrm{(B)} 54+9piqquadmathrm{(C)} 54+18sqrt{3}+6piqquadmathrm{(D)} left(2sqrt{...$

18.

A square of side length $1$ and a circle of radius $dfrac{sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?

$extbf{(A)} dfrac{pi}{3}-1 qquad extbf{(B)} dfrac{2pi}{9}-dfrac{sqrt{3}}{3} qquad extbf{(C)} dfrac{pi}{18} ...$

19.

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

$mathrm{(A)} 3-2sqrt2qquadmathrm{(B)} 2-sqrt2qquadmathrm{(C)} 4(3-2sqrt2)qquadmathrm{(D)} frac12(3-sqrt2)qqua...$

1.

2.

3.

4. 7/9	Understanding

5. 8.75

6. 12	Complex

7. 8

By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$ .
This gives the area of the white region as $frac{5}{6}pi+frac{2cdotsqrt3}{4}=frac{5}{6}pi+frac{sqrt3}{2}$ .
The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2pi-frac{5}{6}pi+frac{sqrt3}{2}=frac{7}{6}pi-frac{sqrt3}{2}$ .

Thus the answer is $oxed{ extbf{(E)} frac{7}{6}pi-frac{sqrt3}{2}}$ .

The area of the circle is $S_{igcirc}=100pi$ , the area of the square is $S_{square}=100$ .

Exactly $1/4$ of the circle lies inside the square. Thus the total area is $dfrac34 S_{igcirc} + S_{square} = oxed{100+75pi} Longrightarrow mathrm{(B)}$ .

10.

This is obviously a right triangle. Pick a coordinate system so that the right angle is at $(0,0)$ and the other two vertices are at $(12,0)$ and $(0,5)$ .

As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at $(6,2.5)$ .

The radius $r$ of the inscribed circle can be computed using the well-known identity $frac{rP}2=S$ , where $S$ is the area of the triangle and $P$ its perimeter. In our case, $S=5cdot 12/2=30$ and $P=5+12+13=30$ , thus $r=2$ . As the inscribed circle touches both legs, its center must be at $(r,r)=(2,2)$ .

The distance of these two points is then $sqrt{ (6-2)^2 + (2.5-2)^2 } = sqrt{16.25} = sqrt{frac{65}4} = oxed{frac{sqrt{65}}2}$ .

11.

The outer circle has radius $1+1+1=3$ , and thus area $9pi$ . The little circles have area $pi$ each; since there are 7, their total area is $7pi$ . Thus, our answer is $9pi-7pi=oxed{2piRightarrow ext{(C)}}$ .

12.

There are 80 tiles. Each tile has $[mbox{square} - 4 cdot (mbox{quarter circle})]$ shaded. Thus:

$egin{align*}mbox{shaded area} &= 80 left( 1 - 4 cdot dfrac{1}{4} cdot pi cdot left(dfrac{1}{2} ight)^2 ight)...$

13.

Let $s$ be the length of a side of the equilateral triangle and let $r$ be the radius of the circle.

In a circle with a radius $r$ the side of an inscribed equilateral triangle is $rsqrt{3}$ .

So $s=rsqrt{3}$ .

The perimeter of the triangle is $3s=3rsqrt{3}$

The area of the circle is $pi r^{2}$

So:
$pi r^{2} = 3rsqrt{3}$

$pi r=3sqrt{3}$

$r=frac{3sqrt{3}}{pi} Rightarrow B$

14.

Let $[X]$ denote the area of region $X$ in the figure above.

The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$ .

The area of the smaller semicircle is $[A+B] = frac{1}{2}picdot(frac{1}{2})^{2}=frac{1}{8}pi$ .

Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures $60^circ$ .

The area of the $60^circ$ sector of the larger semicircle is $[B+C] = frac{60}{360}picdot(frac{2}{2})^{2}=frac{1}{6}pi$ .

The area of the triangle is $[C] = frac{1^{2}sqrt{3}}{4}=frac{sqrt{3}}{4}$ .

So the shaded area is $[A] = [A+B]-[B+C]+[C] = left(frac{1}{8}pi ight)-left(frac{1}{6}pi ight)+left(frac{sqrt{3}}{4} ight)=oxed{mathr...$ .

15.

Let $O_{i}$ be the center of circle $i$ for all $i in {A,B,C,D}$ and let $E$ be the tangent point of $B,C$ . Since the radius of $D$ is the diameter of $A$ , the radius of $D$ is $2$ . Let the radius of $B,C$ be $r$ and let $O_{D}E = x$ . If we connect $O_{A},O_{B},O_{C}$ , we get an isosceles triangle with lengths $1 + r, 2r$ . Then right triangle $O_{D}O_{B}E$ has legs $r, x$ and hypotenuse $2-r$ . Solving for $x$ , we get $x^2 = (2-r)^2 - r^2 Longrightarrow x = sqrt{4-4r}$ .

Also, right triangle $O_{A}O_{B}E$ has legs $r, 1+x$ , and hypotenuse $1+r$ . Solving,

$egin{eqnarray*}r^2 + (1+sqrt{4-4r})^2 &=& (1+r)^21+4-4r+2sqrt{4-4r}&=& 2r + 11-r &=& left(f...$

So the answer is $mathrm{(D)}$ .

16.	Complex

17.

The region described contains three rectangles of dimensions $3 imes 6$ , and three $120^{circ}$ degree arcs of circles of radius $3$ . Thus the answer is $3(3 imes 6) + 3 left( frac{120^{circ}}{360^{circ}} imes 3^2 pi<br /> ight) = 54 + 9pi Longrightarrow mathrm{(B)}.$

18.

The radius of circle is $frac{sqrt{3}}{3} = sqrt{frac{1}{3}}$ . Half the diagonal of the square is $frac{sqrt{1^2+1^2}}{2} = frac{sqrt{2}}{2} = sqrt{frac12}$ . We can see that the circle passes outside the square, but the square is NOT completely contained in the circle Therefore the picture will look something like this:

unitsize(5cm);defaultpen(linewidth(.8pt)+fontsize(10pt));dotfactor=3;real r=sqrt(1/3);pair O=(0,0);pair W=(0.5,0.5), X=(0.5,-...

Then we proceed to find: 4 * (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).

First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits $AB$ in half. Let this half-length be $a$ . Also note that $OX=frac12$ because it is half the sidelength of the square. Because this is a right triangle, we can use the Pythagorean Theorem to solve for $a.$

$a^2+left( frac12 ight) ^2 = left( frac{sqrt{3}}{3} ight) ^2$

Solving, $a= frac{sqrt{3}}{6}$ and $2a=frac{sqrt{3}}{3}$ . Since $AB=AO=BO$ , $riangle AOB$ is an equilateral triangle and the central angle is $60^{circ}$ . Therefore the sector has an area $pi left( frac{sqrt{3}}{3} ight) ^2 left( frac{60}{360} ight) = frac{pi}{18}$ .

Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is

$frac{s^2sqrt{3}}{4} = frac{frac13 sqrt{3}}{4} = frac{sqrt{3}}{12}$

Putting it together, we get the answer to be $4 left( frac{pi}{18}-frac{sqrt{3}}{12} ight)= oxed{ extbf{(B)} frac{2pi}{9}-frac{sqrt{3}}{3}}$

19.

Draw some of the radii of the small circles as in the picture below.

Out of symmetry, the quadrilateral in the center must be a square. Its side is obviously $2r$ , and therefore its diagonal is $2rsqrt{2}$ . We can now compute the length of the vertical diameter of the large circle as $2r + 2rsqrt{2}$ . Hence $2R=2r + 2rsqrt{2}$ , and thus $R=r+rsqrt{2}=r(1+sqrt{2})$ .

Then the area of the large circle is $L = pi R^2 = pi r^2 (1+sqrt 2)^2 = pi r^2 (3+2sqrt 2)$ .
The area of four small circles is $S = 4pi r^2$ . Hence their ratio is:

$egin{align*}frac SL & = frac{4pi r^2}{pi r^2 (3+2sqrt 2)} & = frac 4{3+2sqrt 2} & = frac 4{3+2sqrt...$