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Euqivalence

In trapezoid $ABCD$ , $overline{AB}$ and $overline{CD}$ are perpendicular to $overline{AD}$ , with $AB+CD=BC$ , $AB<CD$ , and $AD=7$ . What is $ABcdot CD$ ?

$extbf{(A)} 12 qquad extbf{(B)} 12.25 qquad extbf{(C)} 12.5 qquad extbf{(D)} 12.75 qquad extbf{(E)} 13$

10.

In $riangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$ ?

$mathrm{(A)} 3 qquadmathrm{(B)} 2sqrt{3} qquadmathrm{(C)} 4 qquadmathrm{(D)} 5 qquadmathrm{(...$

11.

Square $ABCD$ has side length $2$ . A semicircle with diameter $overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $overline{AD}$ at $E$ . What is the length of $overline{CE}$ ?

$mathrm{(A) } frac{2+sqrt{5}}{2} qquad mathrm{(B) } sqrt{5} qquad mathrm{(C) } sqrt{6} qquad mathrm{(D) } ...$

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

12.

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

$mathrm{(A) } frac{1}{4}x^2qquad mathrm{(B) } frac{2}{5}x^2qquad mathrm{(C) } frac{1}{2}x^2qquad mathrm{(D) ...$

13.

A circle of radius $1$ is surrounded by $4$ circles of radius $r$ as shown. What is $r$ ?

unitsize(3mm);defaultpen(linewidth(.8pt)+fontsize(7pt));dotfactor=4;real r1=1, r2=1+sqrt(2);pair A=(0,0), B=(1+sqrt(2),1+sqrt...

$extbf{(A) } sqrt{2} qquad extbf{(B) } 1+sqrt{2} qquad extbf{(C) } sqrt{6} qquad extbf{(D) } 3 qquad extbf{(E) } ...$

14.

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle with radius 3. What is the area of the triangle?

$mathrm{(A)} 8.64qquad mathrm{(B)} 12qquad mathrm{(C)} 5piqquad mathrm{(D)} 17.28qquad mathrm{(E)} 18$

''>''

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle with radius 3. What is the area of the triangle?

$mathrm{(A)} 8.64qquad mathrm{(B)} 12qquad mathrm{(C)} 5piqquad mathrm{(D)} 17.28qquad mathrm{(E)} 18$

15.

Points $A$ and $C$ lie on a circle centered at $O$ , each of $overline{BA}$ and $overline{BC}$ are tangent to the circle, and $riangle ABC$ is equilateral. The circle intersects $overline{BO}$ at $D$ . What is $frac{BD}{BO}$ ?

$ext{(A) } frac {sqrt2}{3}qquad ext{(B) } frac {1}{2}qquad ext{(C) } frac {sqrt3}{3}qquad ext{(D) } frac {sqrt2...$

16.

Triangle $ABC$ has a right angle at $B$ , $AB=1$ , and $BC=2$ . The bisector of $angle BAC$ meets $overline{BC}$ at $D$ . What is $BD$ ?

$ext{(A) } frac {sqrt3 - 1}{2}qquad ext{(B) } frac {sqrt5 - 1}{2}qquad ext{(C) } frac {sqrt5 + 1}{2}qquad ext{(D...$

17.

A right triangle has perimeter $32$ and area $20$ . What is the length of its hypotenuse?

$mathrm{(A)} frac{57}{4}qquadmathrm{(B)} frac{59}{4}qquadmathrm{(C)} frac{61}{4}qquadmathrm{(D)} frac{63}{4}qq...$

1 Problem
2 Solution
3 See also

18.

19.

Nondegenerate $riangle ABC$ has integer side lengths, $overline{BD}$ is an angle bisector, $AD = 3$ , and $DC=8$ . What is the smallest possible value of the perimeter?

$extbf{(A)} 30 qquad extbf{(B)} 33 qquad extbf{(C)} 35 qquad extbf{(D)} 36 qquad extbf{(E)} 37$

20.

A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?

$extbf{(A)} 4+4sqrt{2} qquad extbf{(B)} 2+4sqrt{2}+2sqrt{3} qquad extbf{(C)} 2+3sqrt{2}+3sqrt{3} qquad extb...$

21.

Point B is due east of point A. Point C is due north of point B. The distance between points A and C is $10sqrt 2$ , and $angle BAC= 45^circ$ . Point D is 20 meters due north of point C. The distance AD is between which two integers?

$extbf{(A)} 30 ext{and} 31 qquad extbf{(B)} 31 ext{and} 32 qquad extbf{(C)} 32 ext{and} 33 qquad extbf{...$

1.

2.

3.

4.

5. 4	Complex

6. 7 sqrt 2

7. 59	Complex

8. 15	Complex

If $AB=x$ and $CD=y$ , we have $BC=x+y$ .

By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49$

Solving the equation, we get $4xy=49 implies xy = oxed{ extbf{(B)} 12.25}$ .

10.

Draw height $CH$ . We have that $BH=1$ . From the Pythagorean Theorem, $CH=sqrt{48}$ . Since $CD=8$ , $HD=sqrt{8^2-48}=sqrt{16}=4$ , and $BD=HD-1$ , so $BD=3<br /> ightarrowoxed{A}$ .

11.

Solution 1

Let the point of tangency be $F$ . By the Two Tangent Theorem $BC = FC = 2$ and $AE = EF = x$ . Thus $DE = 2-x$ . The Pythagorean Theorem on $riangle CDE$ yields

$egin{align*}DE^2 + CD^2 &= CE^2(2-x)^2 + 2^2 &= (2+x)^2x^2 - 4x + 8 &= x^2 + 4x + 4x &= frac{1}{2}e...$

Hence $CE = FC + x = frac{5}{2} Rightarrow mathrm{(D)}$ .

Solution 2

Clearly, $EA = EF = BG$ . Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$ , $CE = 5/2$ .

12.

Let the width of the rectangle be $w$ . Then the length is $2w$

Using the Pythagorean Theorem:

$x^{2}=w^{2}+(2w)^{2}$

$x^{2}=5w^{2}$

$w=frac{x}{sqrt{5}}$

$2w=frac{2x}{sqrt{5}}$

So the area of the rectangle is $w cdot 2w = frac{x}{sqrt{5}} cdot frac{2x}{sqrt{5}} = frac{2}{5}x^{2} Longrightarrow mathrm{(B)}$

13.

Solution 1

You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get $r+1.$ You can also add the radius of two outer circles and use a $45-45-90$ triangle to get $frac{2r}{sqrt{2}} = rsqrt{2}.$ Since both representations are for the same thing, you can set them equal to each other.
$egin{align*}r+1&=rsqrt{2}1&=r(sqrt{2}-1)end{align*}$
$r = frac{1}{sqrt{2}-1} = frac{sqrt{2}+1}{2-1} = oxed{mathrm{(B) } 1 + sqrt{2}}$

Solution 2

You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in $2r+2$ . The two legs are each the length between two large, adjacent circles, thus $2r$ .
Using the Pythagorean Theorem:
$egin{align*}(2r+2)^2 = 2(2r)^24r^2+8r+4=8r^2r^2+2r+1=2r^2r^2-2r-1=0r=frac{2+sqrt{4-(-4)}}{2}=frac{2+sqrt{2}}{2}...$

14.

Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is $2r = 6$ . Then the other legs are $frac{24}5=4.8$ and $frac{18}5=3.6$ . The area is $frac{4.8 cdot 3.6}2 = 8.64 mathrm{(A)}$

15.

Solution 1

As $riangle ABC$ is equilateral, we have $angle BAC = angle BCA = 60^circ$ , hence $angle OAC = angle OCA = 30^circ$ . Then $angla AOC = 120^circ$ , and from symmetry we have $angle AOB = angle COB = 60^circ$ . Finally this gives us $angle ABO = angle CBO = 30^circ$ .

We know that $DO = AO$ , as $D$ lies on the circle. From $riangle ABO$ we also have $AO = BO sin 30^circ = frac{BO}2$ , Hence $DO = frac{BO}2$ , therefore $BD = BO - DO = frac{BO}2$ , and $frac{BD}{BO} = oxed{frac 12}$ .

Solution 2

As in the previous solution, we find out that $angle AOB = angle COB = 60^circ$ . Hence $riangle AOD$ and $riangle COD$ are both equilateral.

We then have $angle SCD = angle SAD = 30^circ$ , hence $D$ is the incenter of $riangle ABC$ , and as $riangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 cdot SD = BD$ , and as $SD = SO$ , we have $2cdot SD = SD + SO = OD$ , therefore $BD=OD$ , and as before we conclude that $frac{BD}{BO} = oxed{frac 12}$ .

16.

By the Pythagorean Theorem, $AC=sqrt5$ . The Angle Bisector Theorem now yields that

$frac{BD}{1}=frac{2-BD}{sqrt5}BDleft(1+frac{1}{sqrt5}<br /> ight)=frac{2}{sqrt5}BD(sqrt5+1)=2BD=frac{2}{sqrt5+1}=...$

17.

Solution 1

Let the legs of the triangle have lengths $a,b$ . Then, by the Pythagorean Theorem, the length of the hypotenuse is $sqrt{a^2+b^2}$ , and the area of the triangle is $frac 12 ab$ . So we have the two equations

$egin{align}a+b+sqrt{a^2+b^2} &= 32 frac{1}{2}ab &= 20end{align}$

Re-arranging the first equation and squaring,

$egin{align*}sqrt{a^2+b^2} &= 32-(a+b)a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2a^2 + b^2 + 64(a+b) &= a^2 + b^2...$

From $(2)$ we have $2ab = 80$ , so

$a+b &= frac{80 + 32^2}{64} = frac{69}{4}.$

The length of the hypotenuse is $p - a - b = 32 - frac{69}{4} = frac{59}{4} mathrm{(B)}$ .

Solution 2

From the formula $A = rs$ , where $A$ is the area of a triangle, $r$ is its inradius, and $s$ is the semiperimeter, we can find that $r = frac{20}{32/2} = frac{5}{4}$ . It is known that in a right triangle, $r = s - h$ , where $h$ is the hypotenuse, so $h = 16 - frac{5}{4} = frac{59}{4}$ .

Solution 3

From the problem, we know that

$egin{align*}a+b+c &= 32 2ab &= 80. end{align*}$

Subtracting $c$ from both sides of the first equation and squaring both sides, we get

$egin{align*}(a+b)^2 &= (32 - c)^2a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.end{align*}$

Now we substitute in $a^2 + b^2 = c^2$ as well as $2ab = 80$ into the equation to get

$egin{align*}80 &= 1024 - 64cc &= frac{944}{64}.end{align*}$

Further simplification yields the result of $frac{59}{4}$ .

Solution 4

Let $a$ and $b$ be the legs of the triangle, and $c$ the hypotenuse.

Since the area is 20, we have $frac{1}{2}ab = 20 => ab=40$ .

Since the perimeter is 32, we have $a + b + c = 32$ .

The Pythagorean Theorem gives $c^2 = a^2 + b^2$ .

This gives us three equations with three variables:

$ab = 40 a + b + c = 32 c^2 = a^2 + b^2$

Rewrite equation 3 as $c^2 = (a+b)^2 - 2ab$ .
Substitute in equations 1 and 2 to get $c^2 = (32-c)^2 - 80$ .

$c^2 = (32-c)^2 - 80 c^2 = 1024 - 64c + c^2 - 80 64c = 944 c = frac{944}{64} = frac{236}{16} = frac{59}{4}$ .

The answer is choice (B).

18.	Complex

19.

By the Angle Bisector Theorem, we know that $frac{AB}{3} = frac{BC}{8}$ . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then $AB + BC = AD + DC = AC$ , contradicting the Triangle Inequality. If we use the next lowest values ( $AB = 6$ and $BC = 16$ ), the Triangle Inequality is satisfied. Therefore, our answer is $6 + 16 + 3 + 8 = oxed{33}$ , or choice $extbf{(B)}$ .

20.

The distance of an interior diagonal in this cube is $sqrt{3}$ and the distance of a diagonal on one of the square faces is $sqrt{2}$ . It is not possible for the fly to travel any interior diagonal twice, as then it would visit a corner more than once. So, the final sum can have at most $4$ as the coefficient of $sqrt{3}$ . The other 4 paths taken can be across a diagonal on one of the faces (try to find a such path!), so the maximum distance traveled is $extbf{(D)} 4sqrt{2}+4sqrt{3}$ .

21.

By constructing the bisectors of each angle and the perpendicular radii of the incircle the triangle consists of 3 kites.

Hence $AD=AF$ and $BD=BE$ and $CE=CF$ . Let $AD = x, BD = y$ and $CE = z$ gives three equations:

$x+y = a-1$

$x+z = a$

$y+z = a+1$

(where $a = 2012$ for the first triangle.)

Solving gives:

$x= frac{a}{2} - 1$

$y = frac{a}{2}$

$z = frac{a}{2}+1$

Subbing in gives that $T_2$ has sides of $1005, 1006, 1007$ .

$T_3$ can easily be derivied from this as the sides still differ by 1 hence the above solutions still work (now with $a=1006$ ). All additional triangles will differ by one as the solutions above differ by one so this process can be repeated indefinately until the side lengths no longer form a triangle.