Answer: C

Solution:

Simply solving directly: $$\begin{align*} &\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} \\ &= \dfrac{12}9 - \dfrac 9{12} \\ &= \dfrac {16}{12} - \dfrac 9{12} \\ &=\dfrac 7{12}.\end{align*}$$

Thus, the correct answer is C.

Answer: E

Solution:

Her current average is $$\dfrac{90+80+70+60+85}5 = 77,$$ and the sum of her scores is $385.$ The desired average is then $80,$ so the sum of scores required is $80\cdot 6.$ Therefore, the answer is $$80\cdot 6-385 = 95.$$

Thus, the correct answer is E.

Answer: A

Solution:

The smallest possible dimensions are $1.5\times 2.5,$ so the area is $$1.5\cdot 2.5=3.75.$$

Thus, the correct answer is A.

Answer: C

Solution:

The amount they each would have to pay is $\dfrac{A+B}2,$ and LeRoy paid $A.$ Thus, he has to pay $$\dfrac{A+B}2-A = \dfrac{B-A}2$$ more.

Thus, the correct answer is C.

Answer: E

Solution:

The number $161$ is equal to $7\cdot 23.$ There are no other pairs of numbers that multiply to $161$ besides $1\cdot 161,$ so $23$ is the only two digit factor. Thus, $23$ is the number reversed, so he mean to get $32\cdot 7$ which is $224.$

Thus, the correct answer is E.

Answer: A

Solution:

Let $c$ be the total amount of candies.

After day one, he used $\frac c3 +2$ of his candies, so he had $\frac {2c}3-2$ left.

After day two, he used $$\dfrac { \dfrac {2c}3-2}3 +4 = \dfrac {2c}9+ \dfrac{10}3$$ of his candies, so he had $\frac {2c}9-\frac{14}3$ left.

He had $8$ candies after this, so $$\dfrac {4c}9-\dfrac{16}3 =8.$$ This makes $c=30.$

Thus, the correct answer is A.

Answer: B

Solution:

The two angles add to $\frac 65 \cdot 90 = 108.$ This makes the other angle $180-108=72.$ Then, if the larger of the two angles is $x$ then the smaller of them is $x-30$ so their sum is $2x-30=108,$ making $x=69.$

This means no angle is larger than $72,$ making the largest eqaul to $72.$

Thus, the correct answer is B.

Answer: B

Solution:

We know that over $80$ degrees and sunny combinded implies crowded, so not crowded implies that the combination of over $80$ degrees and sunny is not true. This immeadiately eliminated choice C.

We have no more information, so if it was below $80$ degrees, we don't know if it is crowded. Thus, A, D and E are eliminated.

Thus, the correct answer is B.

Answer: D

Solution:

By angle angle similarity, we have $BDE \sim BCA .$

Then, since the ratio of the areas is $\frac 13,$ the ratio of the sidelengths is $\frac{1}{\sqrt 3}.$

As such, $$\dfrac{BD}{BC} = \dfrac{BD}4 = \dfrac{1}{\sqrt 3},$$ making $$BD = \dfrac 4{ \sqrt 3} = \dfrac{4\sqrt{3}}{3} .$$

Thus, the correct answer is D.

Answer: B

Solution:

The largest number is $10^{10}.$ The rest of the number have a sum of $$S=\sum_{i=0}^9 10^i.$$ Then, $10S = \sum_{i=0}^9 10^{i+1},$ making $9S = 10^{10}-1.$ This means that $\dfrac{10^{10}}{9S}$ is close to one, so the ratio between $10^{10}$ and the sum is close to $9.$

Thus, the correct answer is B.

Answer: D

Solution:

It isn't nessicarily true for $n\geq 6$ as we could have $5$ people born in the first $4$ months and $4$ people born in the subsequent months.

However, one month must be greater than or equal to $5$ as the average of the number of people born in each month is $\frac{52}{12}$ which is greater than $4,$ and some month must be above average.

Thus, the correct answer is D.

Answer: A

Solution:

Let the radius of the inside loop be $r$ and let the straights have length $s.$ Then, the distance he walks is on the inside is then $2\pi r+ 2s.$ Then, the radius of the outside is $r+6,$ so the distance he walks is on the inside is then $$2\pi (r+6)+ 2s = 2\pi r + 12 \pi + 2s.$$ Therefore, he walks $12 \pi$ more meters in $36$ seconds.

Since $d = vt$ where $v$ is speed, we have $12 \pi = 36v.$ Thus, $v = \dfrac \pi 3.$

Thus, the correct answer is A.

Answer: D

Solution:

There is $0$ probability that our number is $0,$ so we need to just find the probability that the product isn't less than $0.$ The number product is less than zero if one of the numbers is less than $0$ and one of them is greater than $0.$

First, there are $2$ ways to choose the designated lower number. Then, the probability that the designated lower number is less than $0$ is $\frac 23$ and the probability that the designated higher number is greater than $0$ is $\frac 13.$

This makes the probability that the product is less than $0$ equal to $$2 \cdot \dfrac 23 \cdot \dfrac 13 = \dfrac 49.$$ As such, the probability that the product is greater than $0$ equal to $\dfrac 59.$

Thus, the correct answer is D.

Answer: C

Solution:

Let the side lengths be $l,w.$ We wish to find $2(l+w).$ From the Pythagorean Theorem, we get $$25 = \sqrt{l^2+w^2}$$$$l^2+w^2 = 625$$ We also know $lw = 168.$

As such $$\begin{align*}l^2+2lw+ w^2 &= (l+w)^2 \\&= 961 \\&= 31^2.\end{align*}$$ This makes $l+w = 31,$ and as such, our answer is $31\cdot 2=62.$

Thus, the correct answer is C.

Answer: E

Solution:

In text 1, the left hand side equals $\dfrac{x+y+z}2$ and the right hand side equals $x+ \dfrac{y+z}2,$ so they aren't equal.

In text 2, the left hand side equals $x+ \dfrac{y+z}2$ and the right hand side equals $x+ \dfrac{y+z}2,$ so they are equal.

In text 3, the left hand side equals $\dfrac{2x+y+z}4$ and the right hand side equals $\dfrac{2x+y+z}4,$ so they are equal.

Thus, the correct answer is E.

Answer: A

Solution:

Let the side length be $1.$ Then, the area of the center is $1.$

Then, we must find the area of the octagon. It can be found as a square with $4$ isoceles right triangles taken out. The side length of this square is $$1 + 2\cdot \dfrac 1{\sqrt 2} = 1 + \sqrt 2 .$$ It has an area of $$(1+\sqrt 2)^2 = 3 + 2 \sqrt 2.$$

Then, the side length of the right triangles is $\dfrac {1}{\sqrt 2} ,$ making the area of one equal to $$\dfrac {\frac {1}{\sqrt 2}^2}{2} = \dfrac 14 .$$ This makes them have a total combined area of $1,$ so the area of the octagon is $2+ 2 \sqrt 2.$

Thus, the ratio is $$\begin{align*}&\dfrac 1{2+ 2 \sqrt 2} \\&= \dfrac {-2+ 2 \sqrt 2}{(2+ 2 \sqrt 2)(-2+ 2 \sqrt 2)}\\ &=\dfrac {2(\sqrt 2-1)}{4} \\&=\dfrac {\sqrt 2-1}{2}. \end{align*}$$

Thus, the correct answer is A.

Answer: C

Solution:

The angle is equal to the angle of the major arc $$\dfrac{\overset{\huge\frown}{BAD}}2 = 180^\circ-\dfrac{\overset{\huge\frown}{BCD}}2.$$

Then, since $AB || ED,$ $AE$ and $BD$ are equal making their arcs equal, making our answer equal to $180^\circ-\dfrac{\overset{\huge\frown}{AE}}2.$

Then, by the angle ratio, we have $\overset{\huge\frown}{AE} : \overset{\huge\frown}{ED} =5:4.$ Since the sum of the arcs is $180^\circ,$ the arc $\overset{\huge\frown}{AE}$ is eqaul to $100^\circ.$ Therefore, the answer is $$180^\circ-\dfrac{\overset{\huge\frown}{100^\circ}}2 = 130^\circ .$$

Thus, the correct answer is C.

Answer: E

Solution:

The angles $\angle AMD$ and $\angle MDC$ are equal since $AB \mid \mid DC.$

As such, $\angle MDC = \angle DMC ,$ making $MDC$ isoceles and $MC = DC = 6.$

As we can see, $\sin (CMB) = \frac 12,$ making $\angle CMB = 30^\circ .$

Therefore, $\angle AMC = 150^\circ .$ Since $\angle AMD$ is half of that, $\angle AMD = 75^\circ .$

Thus, the correct answer is E.

Answer: A

Solution:

The equation is equal to $$\sqrt{5 | x | + 8} = \sqrt{|x|^2 - 16}.$$ Solving, we get that: $$\begin{align*} 5|x|+8 &= |x|^2-16 \\ |x|^2-5|x|-24&=0 \\ (|x|-8)(|x|+3)&=0 \end{align*}$$ This makes $|x|=-3,8,$ making $|x|=8$ the only possible value. Thus, $x =8,-8$ with a product of $-64.$

Thus, the correct answer is A.

Answer: C

Solution:

To find the points closest to $B,$ we must find the points closer to $B$ when taking the perpendicular bisector between $B$ and any other point.

If we take perpendicular bisector of $B$ and $D,$ the amount on the side closer to $B$ is equal to the area of $ABC$ which is equal to $$\dfrac{2\cdot 2\cdot \sin(120^\circ)}2 = \sqrt 3.$$

If we take perpendicular bisector of $B$ and $A,$ the amount on the side closer to $C$ is equal to the area of $CEF$ which has $CE = 1$ and $EF = \tan(30^\circ) = \dfrac{1}{\sqrt 3} .$ Therefore, the amount subtracted here is the area of $CEF$ which is equal to $$\dfrac{1\cdot \frac 1{\sqrt 3}}2 = \dfrac{\sqrt 3}6.$$

Doing the same with the perpendicular bisector of $B$ and $C$ has the same area subtracted as above, so the area is equal to $$\begin{align*} \sqrt 3 - 2\cdot \dfrac{\sqrt 3}6 &= \sqrt 3 - \dfrac{\sqrt 3}3 \\&=\dfrac{2 \sqrt 3}3.\end{align*}$$

Thus, the correct answer is C.

Answer: B

Solution:

The largest difference must be $9$ so $w-z=9.$ Then, $(w-x)+(x-z) = 9,$ so they must both be two different pariwise differences that add to $9.$ Thus, $$w-x=6,x-z=3$$ or $$w-x=5,x-z=4.$$ or the other way around.

Similarly, $$w-y=6,y-z=3$$ or $$w-y=5,y-z=4$$ or the other way around.

Then, we must have $x-y=1$ since it is the only place for it to be. Thus, we could have $$y-z=3,x-z=4$$ or $$y-z=5,x-z=6.$$

Thus, the cases are $$w-z=9,$$$$w-y=6,$$$$w-y=5$$ or $$w-z=9,$$$$w-y=5,$$$$w-y=3.$$

Then, for the first case, we have: $$w+x+y+z=44$$$$w+w-5+w-6+w-9=44$$ $$4w=64$$$$w=16.$$

Also, for the second case, we have: $$w+x+y+z=44$$ $$w+w-3+w-4+w-9=44$$$$4w=60$$ $$w=15.$$

The sum over all cases is then $31.$

Thus, the correct answer is B.

Answer: A

Solution:

Let the side length of the cube be $s.$ Then, we take the diagonal cross section of the cube.

This would have a $1-1\sqrt 2$ right triangle. Then, the base has $s \sqrt 2$ and two legs of right isoceles triangles. The legs of the isoceles triangle is $s,$ so the side equal to $\sqrt 2$ is also equal to $(\sqrt 2 +1)s.$

Therefore, $$(\sqrt 2+2)s = \sqrt 2$$$$s = \dfrac{\sqrt 2}{\sqrt 2+2}$$$$s = \dfrac{(\sqrt 2)(2-\sqrt 2 )}{(\sqrt 2+2)(2-\sqrt 2)}$$$$s = \sqrt 2-1.$$

Then the volume is $$s^3= (\sqrt 2-1)^3 = 5 \sqrt 2-7.$$

Thus, the correct answer is A.

Answer: D

Solution:

We must find $2011^{2011} \mod 1000.$

This is equivalent to $11^{2011} \mod 1000.$

By the binomial theorem, we get that this is equal to $$(10+1)^{2011} = \sum_{i=0}^{2011} 10^i \binom {2011}i.$$

Then, if $i \geq 3,$ it is a multiple of $1000,$ so $$\begin{align*}&\sum_{i=0}^{2} 10^i \binom {2011}i\\ &\equiv 100\cdot \dfrac{2011\cdot 2010}2\\&+100\cdot 2011\cdot 1005\\&+2011\cdot 10+1 \\ &\equiv 100\cdot 11\cdot 5 + 11\cdot 10+1 \\ &\equiv 5651\\ &\equiv 651 \end{align*}$$ This has a hundreds digit of $6.$

Thus, the correct answer is D.

Answer: B

Solution:

The lattice point $x,y$ is a coordingate intesected by $y=mx+2,$ if and only if $y=mx$ intersects the lattice point $(x,y-2),$ so it suffices to look at $y=mx$ instead.

Thus, we must find the smallest $m>$ such that it intersects a lattice point. We will inspect each $x$ and find the smallest $m$ that intersects that lattice point and take the maximum.

If $x$ is even, then the number would be $$\dfrac{x+2}{2x} = \dfrac 12 + \dfrac 1x.$$ The minimum of this would be $x=100$ which is $$\dfrac 12 + \dfrac 1{100} = \dfrac {51}{100}.$$

If $x$ is even, then the number would be $$\dfrac{x+1}{2x} = \dfrac 12 + \dfrac 1{2x}.$$ The minimum of this would be $x=99$ which is $$\dfrac 12 + \dfrac 1{2\cdot 99} = \dfrac {50}{99}.$$

The minimum of the possible $m$ is then $\dfrac{50}{99}$ since it is less than $\dfrac{51}{100}.$

Thus, the correct answer is B.

Answer: D

Solution:

Suppose we have the side lengths of $a=BC,b=AC,c=AB$ and the side lengths of the next triangle is $x,y,z.$

Then, we know that $x=AD = AF$ by the congruence of $ADI$ and $AFI$ since they are right triangles with an equal hypotenuse and an equal length.

Similarly, $y=BD = BE$ and $z=CF=CE.$

Thus, $$x+y = AB = c,$$ $$x+z = AC = b,$$ and $$y+z = BC = a.$$

Then, we have $$x+y+z = \dfrac{a+b+c}2$$ so $$z = \dfrac{a+b-c}2,$$$$y = \dfrac{a-b+c}2,$$ $$x= \dfrac{-a+b+c}2.$$

Suppose $a=b-1,c=b+1.$ Then, $$z = \dfrac{b-1+b-(b-1)}2$$$$= \dfrac b2+1,$$$$y =\dfrac{b-1-b+b+1}2$$$$= \dfrac b2,$$ $$x= \dfrac{-(b+1)+b+b+1}2$$$$= \dfrac b2-1.$$ Thus, $y = \frac 12 y,$ $x=y+1,$ and $z=y-1.$ This means that the triangle would always be in the form $s-1,s,s+1$ for all $T_i.$

This would satisfy the triangle inequality if $s-1+s > s+1$ making $s > 2.$ Also, the perimeter is equal to $3s.$

Thus, the middle term of $T_i$ is equal to $\dfrac{2012}{2^{i-1}}$ since it always halves, so the first time it is less than $2$ is if $i=11.$ This makes the last $i=10,$ making the middle term equal to $$\dfrac{2012}{2^{10}} = \dfrac{2012}{2^9} = \dfrac{503}{128}.$$ Then, the perimeter equals $$3\cdot \dfrac{503}{128} = \dfrac{1509}{128}.$$

Thus, the correct answer is D.