from: category_eng

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$ , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

$extbf{(A)} frac{8}3qquad extbf{(B)} frac{30}{11}qquad extbf{(C)} 3qquad extbf{(D)} frac{25}{8}qquad extbf{(E...$

In rectangle $ABCD, AB=5$ and $BC=3$ . Points $F$ and $G$ are on $overline{CD}$ so that $DF=1$ and $GC=2$ . Lines $AF$ and $BG$ intersect at $E$ . Find the area of $riangle AEB$ .

$extbf{(A) } 10 qquad extbf{(B) } frac{21}{2} qquad extbf{(C) } 12 qquad extbf{(D) } frac{25}{2} qquad extbf{(E) }...$

Points $A,B,C,D,E$ and $F$ lie, in that order, on $overline{AF}$ , dividing it into five segments, each of length 1. Point $G$ is not on line $AF$ . Point $H$ lies on $overline{GD}$ , and point $J$ lies on $overline{GF}$ . The line segments $overline{HC}, overline{JE},$ and $overline{AG}$ are parallel. Find $HC/JE$ .

$ext{(A)} 5/4 qquad ext{(B)} 4/3 qquad ext{(C)} 3/2 qquad ext{(D)} 5/3 qquad ext{(E)} 2$

Points $E$ and $F$ are located on square $ABCD$ so that $riangle BEF$ is equilateral. What is the ratio of the area of $riangle DEF$ to that of $riangle ABE$ ?

$mathrm{(A) } frac{4}{3} qquad mathrm{(B) } frac{3}{2} qquad mathrm{(C) } sqrt{3} qquad mathrm{(D) } 2 qqua...$

Right $riangle ABC$ has $AB=3, BC=4,$ and $AC=5.$ Square $XYZW$ is inscribed in $riangle ABC$ with $X$ and $Y$ on $overline{AC}, W$ on $overline{AB},$ and $Z$ on $overline{BC}.$ What is the side length of the square?

$extbf{(A) } frac{3}{2} qquad extbf{(B) } frac{60}{37} qquad extbf{(C) } frac{12}{7} qquad extbf{(D) } frac{23}{13...$

A circle of radius 1 is tangent to a circle of radius 2. The sides of $riangle ABC$ are tangent to the circles as shown, and the sides $overline{AB}$ and $overline{AC}$ are congruent. What is the area of $riangle ABC$ ?

size(200); pathpen = linewidth(0.7); pointpen = black;real t=2^0.5;D((0,0)--(4*t,0)--(2*t,8)--cycle);D(CR((2*t,2),2));D(CR((2...

$mathrm{(A) } frac{35}{2}qquadmathrm{(B) } 15sqrt{2}qquadmathrm{(C) } frac{64}{3}qquadmathrm{(D) } 16sqrt{2...$

''>''

$mathrm{(A) } frac{35}{2}qquadmathrm{(B) } 15sqrt{2}qquadmathrm{(C) } frac{64}{3}qquadmathrm{(D) } 16sqrt{2...$

Consider the $12$ -sided polygon $ABCDEFGHIJKL$ , as shown. Each of its sides has length $4$ , and each two consecutive sides form a right angle. Suppose that $overline{AG}$ and $overline{CH}$ meet at $M$ . What is the area of quadrilateral $ABCM$ ?

$ext{(A)} frac {44}{3}qquad ext{(B)} 16 qquad ext{(C)} frac {88}{5}qquad ext{(D)} 20 qquad ext{(E)} frac...$

Rectangle $ABCD$ has $AB=8$ and $BC=6$ . Point $M$ is the midpoint of diagonal $overline{AC}$ , and $E$ is on $AB$ with $overline{ME}perpoverline{AC}$ . What is the area of $riangle AME$ ?

$ext{(A) } frac{65}{8}qquad ext{(B) } frac{25}{3}qquad ext{(C) } 9qquad ext{(D) } frac{75}{8}qquad ext{(E) } fra...$

10.

Points $A$ and $B$ lie on a circle centered at $O$ , and $angle AOB = 60^circ$ . A second circle is internally tangent to the first and tangent to both $overline{OA}$ and $overline{OB}$ . What is the ratio of the area of the smaller circle to that of the larger circle?

$mathrm{(A)} frac{1}{16}qquadmathrm{(B)} frac{1}{9}qquadmathrm{(C)} frac{1}{8}qquadmathrm{(D)} frac{1}{6}qquad...$

11.

Trapezoid $ABCD$ has bases $overline{AB}$ and $overline{CD}$ and diagonals intersecting at $K$ . Suppose that $AB = 9$ , $DC = 12$ , and the area of $riangle AKD$ is $24$ . What is the area of trapezoid $ABCD$ ?

$mathrm{(A)} 92qquadmathrm{(B)} 94qquadmathrm{(C)} 96qquadmathrm{(D)} 98 qquadmathrm{(E)} 100$

12.

Triangle $ABC$ has a right angle at $B$ . Point $D$ is the foot of the altitude from $B$ , $AD=3$ , and $DC=4$ . What is the area of $riangle ABC$ ?

$mathrm{(A)} 4sqrt3qquadmathrm{(B)} 7sqrt3qquadmathrm{(C)} 21qquadmathrm{(D)} 14sqrt3 qquadmathrm{(E)} 42$

13.

' The area of $riangle$ $EBD$ is one third of the area of $3-4-5$ $riangle$ $ABC$ . Segment $DE$ is perpendicular to segment $AB$ . What is $BD$ ?

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$extbf{(A)} frac{4}{3} qquad extbf{(B)} sqrt{5} qquad extbf{(C)} frac{9}{4} qquad extbf{(D)} frac{4sqrt{3}}{3...$

14.

Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C. What is BC?

$extbf{(A)} 4qquad extbf{(B)} 4.8qquad extbf{(C)} 10.2qquad extbf{(D)} 12qquad extbf{(E)} 14.4$

1. 21/4	Complex

Let the diameter of the cylinder be $2r$ . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $frac{12-2r}{12}=frac{2r}{10}$ which we solve to find $r=frac{30}{11}$ . Our answer is $oxed{ extbf{(B)} frac{30}{11}}$ .

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$riangle EFG sim riangle EAB$ because $FG parallel AB.$ The ratio of $riangle EFG$ to $riangle EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $riangle EAB,$

$egin{align*}frac{2}{5} &= frac{h-3}{h}2h &= 5h-153h &= 15h &= 5$

The height is $5$ so the area of $riangle EAB$ is $frac{1}{2}(5)(5) = oxed{ extbf{(D)} frac{25}{2}}$ .

4.

As $overline{JE}$ is parallel to $overline{AG}$ , angles FHD and FGA are congruent. Also, angle F is clearly congruent to itself. From SSS similarity, $riangle AGF sim riangle EJF$ ; hence $frac {AG}{JE} =5$ . Similarly, $frac {AG}{HC} = 3$ . Thus, $frac {HC}{JE} = oxed{frac {5}{3}Rightarrow ext{(D)}}$ .

5.

Since triangle $BEF$ is equilateral, $EA=FC$ , and $EAB$ and $FCB$ are $SAS$ congruent. Thus, triangle $DEF$ is an isosceles right triangle. So we let $DE=x$ . Thus $EF=EB=FB=xsqrt{2}$ . If we go angle chasing, we find out that $angle AEB=75^{circ}$ , Thus $angle ABE=15^{circ}$ . $frac{AE}{EB}=sin{15^{circ}}=frac{sqrt{6}-sqrt{2}}{4}$ . Thus $frac{AE}{xsqrt{2}}=frac{sqrt{6}-sqrt{2}}{4}$ , or $AE=frac{x(sqrt{3}-1)}{2}$ . Thus $AB=frac{x(sqrt{3}+1)}{2}$ , and $[AEB]=frac{x^2}{4}$ , and $[DEF]=frac{x^2}{2}$ . Thus the ratio of the areas is $2$ . $mathrm{(D)}$

There are many similar triangles in the diagram, but we will only be using $riangle WBZ sim riangle ABC.$ If $h$ is the altitude from $B$ to $AC$ and $s$ is the sidelength of the square, then $h-s$ is the altitude from $B$ to $WZ.$ By similar triangles,
$egin{align*}frac{h-s}{s}&=frac{h}{5}5h-5s&=hs5h&=s(h+5)s&=frac{5h}{h+5}end{align*}$

Find the length of the altitude of $riangle ABC.$ Since it is a right triangle, the area of $riangle ABC$ is $frac{1}{2}(3)(4) = 6.$

The area can also be expressed as $frac{1}{2}(5)(h),$ so $frac{5}{2}h=6 longrightarrow h=2.4.$

Substitute back into $s.$

$s=frac{5h}{h+5} = frac{12}{7.4} = oxed{mathrm{(B) } frac{60}{37}}$

Note that $riangle ADO_1 sim riangle AEO_2 sim riangle AFC$ . Using the first pair of similar triangles, we write the proportion:

$frac{AO_1}{AO_2} = frac{DO_1}{EO_2} Longrightarrow frac{AO_1}{AO_1 + 3} = frac{1}{2} Longrightarrow AO_1 = 3$

By the Pythagorean Theorem we have that $AD = sqrt{3^2-1^2} = sqrt{8}$ .

Now using $riangle ADO_1 sim riangle AFC$ ,

$frac{AD}{AF} = frac{DO_1}{CF} Longrightarrow frac{2sqrt{2}}{8} = frac{1}{CF} Longrightarrow CF = 2sqrt{2}$

The area of the triangle is $frac{1}{2}cdot AF cdot BC = frac{1}{2}cdot AF cdot (2cdot CF) = AF cdot CF = 8left(2sqrt{2} ight) = 16sqrt{2} m...$ .

Solution 1

We can obtain the solution by calculating the area of rectangle $ABGH$ minus the combined area of triangles $riangle AHG$ and $riangle CGM$ .

We know that triangles $riangle AMH$ and $riangle CGM$ are similar because $overline{AH} parallel overline{CG}$ . Also, since $frac{AH}{CG} = frac{3}{2}$ , the ratio of the distance from $M$ to $overline{AH}$ to the distance from $M$ to $overline{CG}$ is also $frac{3}{2}$ . Solving with the fact that the distance from $overline{AH}$ to $overline{CG}$ is 4, we see that the distance from $M$ to $overline{CG}$ is $frac{8}{5}$ .

The area of $riangle AHG$ is simply $frac{1}{2} cdot 4 cdot 12 = 24$ , the area of $riangle CGM$ is $frac{1}{2} cdot frac{8}{5} cdot 8 = frac{32}{5}$ , and the area of rectangle $ABGH$ is $4 cdot 12 = 48$ .

Taking the area of rectangle $ABGH$ and subtracting the combined area of $riangle AHG$ and $riangle CGM$ yields $48 - (24 + frac{32}{5}) = oxed{frac{88}{5}} ext{(C)}$ .

Solution 2

Extend $AB$ and $CH$ and call their intersection $N$ .

The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$ , hence $BN=2$ and thus $AN=6$ . The area of the triangle $BCN$ is $frac{2cdot 4}2 = 4$ .

The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$ .

Draw altitudes from $M$ onto $AN$ and $GH$ , let their feet be $M_1$ and $M_2$ . We get that $MM_1 : MM_2 = 3:2$ . Hence $MM1 = frac 35 cdot 12 = frac {36}5$ .

Then the area of $AMN$ is $frac 12 cdot AN cdot MM_1 = frac{108}5$ , and the area of $ABCM$ can be obtained by subtracting the area of $BCN$ , which is $4$ . Hence the answer is $frac{108}5 - 4 = oxed{frac{88}5}$ .

By the Pythagorean theorem we have $AC=10$ , hence $AM=5$ .

The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar.

The ratio of their sides is $frac{AM}{AB} = frac 58$ , hence the ratio of their areas is $left( frac 58 ight)^2 = frac{25}{64}$ .

And as the area of triangle $ABC$ is $frac{6cdot 8}2 = 24$ , the area of triangle $AME$ is $24cdot frac{25}{64} = oxed{ frac{75}8 }$ .

10.

Let $P$ be the center of the small circle with radius $r$ , and let $Q$ be the point where the small circle is tangent to $OA$ . Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$ .

Then $PQO$ is a right triangle, and a $30-60-90$ triangle at that. Therefore, $OP=2PQ$ .

Since $OP=OC-PC=OC-r=R-r$ , we have $R-r=2PQ$ , or $R-r=2r$ , or $frac{1}{3}=frac{r}{R}$ .

Then the ratio of areas will be $frac{1}{3}$ squared, or $frac{1}{9}Rightarrow oxed{ ext{B}}$ .

11.

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Since $overline{AB} parallel overline{DC}$ it follows that $riangle ABK sim riangle CDK$ . Thus $frac{KA}{KC} = frac{KB}{KD} = frac{AB}{DC} = frac{3}{4}$ .

We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since $riangle AKB, riangle AKD$ share a common altitude to $overline{BD}$ , it follows that (we let $[ riangle ldots]$ denote the area of the triangle) $frac{[ riangle AKB]}{[ riangle AKD]} = frac{KB}{KD} = frac{3}{4}$ , so $[ riangle AKB] = frac{3}{4}(24) = 18$ . Similarly, we find $[ riangle DKC] = frac{4}{3}(24) = 32$ and $[ riangle BKC] = 24$ .

Therefore, the area of $ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98 mathrm{(D)}$ .

12.

It is a well-known fact that in any right triangle $ABC$ with the right angle at $B$ and $D$ the foot of the altitude from $B$ onto $AC$ we have $BD^2 = ADcdot CD$ . (See below for a proof.) Then $BD = sqrt{ 3cdot 4 } = 2sqrt 3$ , and the area of the triangle $ABC$ is $frac{ACcdot BD}2 = 7sqrt3Rightarrowoxed{ ext{(B)}}$ .

Proof: Consider the Pythagorean theorem for each of the triangles $ABC$ , $ABD$ , and $CBD$ . We get:

Substituting equations 2 and 3 into the left hand side of equation 1, we get $BD^2 = AD cdot DC$ .

Alternatively, note that $riangle ABD sim riangle BCD Longrightarrow frac{AD}{BD} = frac{BD}{CD}$ . $lacksquare$

13.

14.

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Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ . $AB=8$ . Also, let $BC=x$ . As $angle ADC$ and $angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $angle ACD$ , $riangle ADC sim riangle BCE$ . From this we can get a proportion.

$frac{BC}{AC}=frac{BE}{AD} ightarrow frac{x}{x+8}=frac{3}{5} ightarrow 5x=3x+24 ightarrow x=oxed{ extbf{(D)} 12}$