Answer: C

Solution(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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(s):

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.