KVPY 2012 Mathematics Question Paper with Answer and Solution

(B) Let the children be $C_1, C_2, C_3$ and their respective guardians be $G_1, G_2, G_3$.
There are $6$ persons in total.
The condition is that for each pair $(G_i, C_i)$,the guardian $G_i$ must be interviewed before the child $C_i$.
In any arrangement of $6$ persons,there are $2!$ ways to arrange the pair $(G_i, C_i)$ such that $G_i$ comes before $C_i$ or $C_i$ comes before $G_i$.
Since there are $3$ such pairs,the total number of unrestricted arrangements is $6!$.
For each pair,only $1$ out of $2!$ arrangements is valid (i.e.,$G_i$ before $C_i$).
Thus,the number of valid ways is $\frac{6!}{2! \times 2! \times 2!} = \frac{720}{8} = 90$.

(D) Given the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$
Let $u = \sqrt{x-1}$,where $u \ge 0$. Then $x-1 = u^2$,so $x = u^2+1$.
Substituting this into the equation:
$\sqrt{u^2+1+3-4u} + \sqrt{u^2+1+8-6u} = 1$
$\sqrt{u^2-4u+4} + \sqrt{u^2-6u+9} = 1$
$\sqrt{(u-2)^2} + \sqrt{(u-3)^2} = 1$
$|u-2| + |u-3| = 1$
This equation holds if and only if $2 \le u \le 3$.
Since $u = \sqrt{x-1}$,we have $2 \le \sqrt{x-1} \le 3$.
Squaring the inequality: $4 \le x-1 \le 9$,which gives $5 \le x \le 10$.
Thus,the equation has infinitely many solutions in the interval $[5, 10]$.

(C) The equation of the circle is $x^2+y^2-6x-2py+17=0$.
Completing the square,we get $(x-3)^2+(y-p)^2 = 3^2+p^2-17 = p^2-8$.
Since two perpendicular tangents are drawn from the origin $(0,0)$ to the circle,the origin must lie on the director circle of the given circle.
The equation of the director circle is $(x-3)^2+(y-p)^2 = 2(p^2-8)$.
Since the origin $(0,0)$ lies on this circle,we substitute $x=0$ and $y=0$:
$(0-3)^2+(0-p)^2 = 2(p^2-8)$
$9+p^2 = 2p^2-16$
$p^2 = 25$
$|p| = 5$.

(B) Given the equations of the curves:
$y = 2x^3 + ax + b$ $(i)$
$y = 2x^3 + cx + d$ $(ii)$
For the curves to have no point in common,the equation $2x^3 + ax + b = 2x^3 + cx + d$ must have no real solution for $x$.
This simplifies to $(a - c)x = d - b$.
If $a - c \neq 0$,then $x = \frac{d - b}{a - c}$ is always a real solution,which contradicts the condition.
Therefore,we must have $a - c = 0$,which implies $a = c$.
Substituting $a = c$ into the equation,we get $0 = d - b$,or $b = d$.
However,the problem implies the curves have no common point,which means the equation $(a - c)x = d - b$ must be inconsistent.
If $a = c$,then $0 = d - b$. For this to have no solution,we must have $d - b \neq 0$.
We want to maximize $(a - c)^2 + b - d$. Since $a = c$,this expression becomes $0 + b - d = b - d$.
To maximize $b - d$ where $b, d \in \{1, 2, 3, 4, 5, 6\}$ and $b \neq d$,we choose $b = 6$ and $d = 1$.
Thus,the maximum value is $6 - 1 = 5$.

(C) The given equation is $e x^2 + \pi y^2 - 2 e^2 x - 2 \pi^2 y + e^3 + \pi^3 = \pi e$.
Rearranging the terms,we get $e(x^2 - 2ex) + \pi(y^2 - 2\pi y) = \pi e - e^3 - \pi^3$.
Completing the square,$e(x^2 - 2ex + e^2) + \pi(y^2 - 2\pi y + \pi^2) = \pi e - e^3 - \pi^3 + e^3 + \pi^3$.
This simplifies to $e(x - e)^2 + \pi(y - \pi)^2 = \pi e$.
Dividing by $\pi e$,we get $\frac{(x - e)^2}{\pi} + \frac{(y - \pi)^2}{e} = 1$.
This is the equation of an ellipse $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ where $a^2 = \pi$ and $b^2 = e$.
Since $\pi > e$,the major axis is along the $x$-direction,so $a = \sqrt{\pi}$.
For any point $P$ on an ellipse,the sum of the distances from the foci $P S_1 + P S_2$ is constant and equal to the length of the major axis,which is $2a$.
Therefore,$P S_1 + P S_2 = 2 \sqrt{\pi}$.

(D) Given the function $f(x) = \frac{\sin(x-a) + \sin(x+a)}{\cos(x-a) - \cos(x+a)}$.
Using the trigonometric identities $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$ and $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$,we get:
$f(x) = \frac{2\sin x \cos a}{2\sin x \sin a}$
Assuming $\sin x \neq 0$,we can simplify the expression:
$f(x) = \frac{\cos a}{\sin a} = \cot a$
Since $a$ is a constant,$\cot a$ is also a constant.
Therefore,$f(x)$ is a constant function.

(D) We have,$\tan 81^{\circ} - \tan 63^{\circ} - \tan 27^{\circ} + \tan 9^{\circ}$
$= (\tan 81^{\circ} + \tan 9^{\circ}) - (\tan 63^{\circ} + \tan 27^{\circ})$
$= (\cot 9^{\circ} + \tan 9^{\circ}) - (\cot 27^{\circ} + \tan 27^{\circ})$
$= \left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}} + \frac{\sin 9^{\circ}}{\cos 9^{\circ}}\right) - \left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}} + \frac{\sin 27^{\circ}}{\cos 27^{\circ}}\right)$
$= \left(\frac{\cos^2 9^{\circ} + \sin^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}\right) - \left(\frac{\cos^2 27^{\circ} + \sin^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}\right)$
$= \left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}\right) - \left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}\right)$
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
$= 2 \left(\frac{1}{\sin 18^{\circ}} - \frac{1}{\cos 36^{\circ}}\right)$
$= 2 \left(\frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1}\right)$
$= 8 \left(\frac{\sqrt{5} + 1 - (\sqrt{5} - 1)}{5 - 1}\right)$
$= 8 \left(\frac{2}{4}\right) = 4$

(B) We have $f(x) = (x - a_1)(x - a_2) + (x - a_2)(x - a_3) + (x - a_3)(x - a_1)$.
Expanding this,we get $f(x) = 3x^2 - 2(a_1 + a_2 + a_3)x + (a_1a_2 + a_2a_3 + a_3a_1)$.
For $f(x) \geq 0$ for all $x \in R$,the discriminant $D$ must be less than or equal to $0$.
$D = [-2(a_1 + a_2 + a_3)]^2 - 4(3)(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
$4(a_1 + a_2 + a_3)^2 - 12(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
Dividing by $4$,we get $(a_1 + a_2 + a_3)^2 - 3(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
$a_1^2 + a_2^2 + a_3^2 + 2(a_1a_2 + a_2a_3 + a_3a_1) - 3(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
$a_1^2 + a_2^2 + a_3^2 - (a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.
Multiplying by $2$,we get $2a_1^2 + 2a_2^2 + 2a_3^2 - 2a_1a_2 - 2a_2a_3 - 2a_3a_1 \leq 0$.
$(a_1 - a_2)^2 + (a_2 - a_3)^2 + (a_3 - a_1)^2 \leq 0$.
Since the sum of squares is $\leq 0$,each term must be $0$.
Thus,$a_1 - a_2 = 0$,$a_2 - a_3 = 0$,and $a_3 - a_1 = 0$,which implies $a_1 = a_2 = a_3$.

(C) We have $S_n = \frac{n(n+1)}{2}$.
The sequence of sums is $S_1=1, S_2=3, S_3=6, S_4=10, S_5=15, S_6=21, S_7=28, S_8=36, \ldots$.
Observing the parity of $S_n$:
$S_1$ (odd),$S_2$ (odd),$S_3$ (even),$S_4$ (even),$S_5$ (odd),$S_6$ (odd),$S_7$ (even),$S_8$ (even),...
The pattern of parity repeats every $4$ terms: (odd,odd,even,even).
In the set ${S_1, S_2, \ldots, S_{99}}$,there are $99$ terms.
Since $99 = 4 \times 24 + 3$,we have $24$ full cycles of (odd,odd,even,even) and the first $3$ terms of the next cycle (odd,odd,even).
Number of even terms = $24 \times 2 + 1 = 49$.
Total number of cards = $99$.
Probability of drawing an even number = $\frac{\text{Number of even terms}}{\text{Total number of terms}} = \frac{49}{99}$.

(A) We need $6^m + 9^n \equiv 0 \pmod{5}$.
Since $6 \equiv 1 \pmod{5}$,we have $6^m \equiv 1^m \equiv 1 \pmod{5}$ for all $m \in \{1, 2, \ldots, 50\}$.
Since $9 \equiv -1 \pmod{5}$,we have $9^n \equiv (-1)^n \pmod{5}$.
Thus,$6^m + 9^n \equiv 1 + (-1)^n \pmod{5}$.
For the expression to be a multiple of $5$,we require $1 + (-1)^n \equiv 0 \pmod{5}$,which implies $(-1)^n \equiv -1 \pmod{5}$.
This holds if and only if $n$ is an odd integer.
In the set $\{1, 2, 3, \ldots, 50\}$,there are $50$ possible values for $m$ and $25$ odd values for $n$ (i.e.,$\{1, 3, 5, \ldots, 49\}$).
Therefore,the total number of ordered pairs $(m, n)$ is $50 \times 25 = 1250$.

(D) Given that each number is the average of its neighbors,we have $a_i = \frac{a_{i-1} + a_{i+1}}{2}$,which implies $2a_i = a_{i-1} + a_{i+1}$ for all $i$ (with indices taken modulo $2012$).
This means $a_{i+1} - a_i = a_i - a_{i-1}$.
Let $d_i = a_{i+1} - a_i$. Then $d_i = d_{i-1}$,which implies all differences are equal to some constant $d$.
Since the numbers are arranged on a circle,$a_{2013} = a_1$,so $a_1 = a_1 + 2012d$,which implies $d = 0$.
Therefore,$a_1 = a_2 = a_3 = \ldots = a_{2012} = k$ for some integer $k$.
The sum of even-indexed numbers is $a_2 + a_4 + \ldots + a_{2012} = 1006k = 3018$.
Thus,$k = \frac{3018}{1006} = 3$.
The sum of all $2012$ numbers is $2012 \times k = 2012 \times 3 = 6036$.

(C) The set $A = S \times S$ contains $n^2$ elements.
$A$ subset $B$ of $A$ is a good subset if it contains all elements of the form $(x, x)$ for $x \in S$.
There are $n$ such elements: $(1, 1), (2, 2), \ldots, (n, n)$.
For $B$ to be a good subset,these $n$ elements must be present in $B$.
The remaining elements in $A$ are those $(a, b)$ where $a \neq b$. The number of such elements is $n^2 - n = n(n - 1)$.
Each of these $n(n - 1)$ elements can either be in $B$ or not in $B$.
Therefore,the number of ways to choose the remaining elements is $2^{n(n - 1)}$.
Thus,the total number of good subsets is $2^{n(n - 1)}$.

(D) Given that the equation $x^2+2ax+b^2=0$ has two distinct real roots,its discriminant $D_1 > 0$.
$D_1 = (2a)^2 - 4(1)(b^2) = 4a^2 - 4b^2 > 0 \Rightarrow a^2 > b^2$ $(i)$
Similarly,for the equation $x^2+2bx+c^2=0$,the discriminant $D_2 > 0$.
$D_2 = (2b)^2 - 4(1)(c^2) = 4b^2 - 4c^2 > 0 \Rightarrow b^2 > c^2$ $(ii)$
From $(i)$ and $(ii)$,we have $a^2 > b^2 > c^2$,which implies $a^2 > c^2$.
Now,consider the equation $x^2+2cx+a^2=0$. Its discriminant $D_3$ is:
$D_3 = (2c)^2 - 4(1)(a^2) = 4c^2 - 4a^2 = 4(c^2 - a^2)$.
Since $a^2 > c^2$,it follows that $c^2 - a^2 < 0$,so $D_3 < 0$.
Therefore,the equation $x^2+2cx+a^2=0$ has no real roots.

(A) Let $f(x) = \frac{1+x}{(1+x^2)(1-x)}$.
Using partial fractions,we write $\frac{1+x}{(1+x^2)(1-x)} = \frac{Ax+B}{1+x^2} + \frac{C}{1-x}$.
$1+x = (Ax+B)(1-x) + C(1+x^2) = Ax - Ax^2 + B - Bx + C + Cx^2$.
Comparing coefficients:
$x^2: -A + C = 0 \implies A = C$.
$x^1: A - B = 1$.
$x^0: B + C = 1$.
Substituting $A=C$ into $A-B=1$ gives $C-B=1$. Adding $B+C=1$ gives $2C=2$,so $C=1$ and $A=1$. Then $B=0$.
Thus,$f(x) = \frac{x}{1+x^2} + \frac{1}{1-x} = x(1+x^2)^{-1} + (1-x)^{-1}$.
Using the binomial expansion $(1+z)^{-1} = \sum_{n=0}^{\infty} (-1)^n z^n$ and $(1-z)^{-1} = \sum_{n=0}^{\infty} z^n$:
$f(x) = x \sum_{n=0}^{\infty} (-1)^n x^{2n} + \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} (-1)^n x^{2n+1} + \sum_{n=0}^{\infty} x^n$.
We want the coefficient of $x^{2012}$.
The first sum only contains odd powers of $x$,so the coefficient of $x^{2012}$ is $0$.
The second sum is $\sum_{n=0}^{\infty} x^n$,so the coefficient of $x^{2012}$ is $1$.
Total coefficient $= 0 + 1 = 1$.

(A) Given the equation of the ellipse: $4x^2 + 9y^2 - 8x - 36y + 15 = 0$.
Completing the square,we get: $4(x^2 - 2x + 1) + 9(y^2 - 4y + 4) = -15 + 4 + 36$.
$4(x - 1)^2 + 9(y - 2)^2 = 25$.
Dividing by $25$,we have: $\frac{(x - 1)^2}{25/4} + \frac{(y - 2)^2}{25/9} = 1$.
Let $X = x - 1$ and $Y = y - 2$. The expression becomes $X^2 + Y^2$.
This represents the square of the distance from the center $(1, 2)$ to any point $(x, y)$ on the ellipse.
The distance squared $r^2 = X^2 + Y^2$ ranges from the square of the semi-minor axis to the square of the semi-major axis.
Here,$a^2 = \frac{25}{4} = 6.25$ and $b^2 = \frac{25}{9} \approx 2.77$.
Thus,$\min(X^2 + Y^2) = b^2 = \frac{25}{9}$ and $\max(X^2 + Y^2) = a^2 = \frac{25}{4}$.
The sum is $\frac{25}{9} + \frac{25}{4} = 25 \left( \frac{4 + 9}{36} \right) = \frac{25 \times 13}{36} = \frac{325}{36}$.

(C) Given equation: $\sin x + \frac{1}{2} \cos x = \sin^2(x + \frac{\pi}{4})$
Multiply by $2$: $2 \sin x + \cos x = 2 \sin^2(x + \frac{\pi}{4})$
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$2 \sin x + \cos x = 2 [\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}]^2$
$2 \sin x + \cos x = 2 [\frac{1}{\sqrt{2}}(\sin x + \cos x)]^2$
$2 \sin x + \cos x = 2 \times \frac{1}{2}(\sin x + \cos x)^2$
$2 \sin x + \cos x = \sin^2 x + \cos^2 x + 2 \sin x \cos x$
Since $\sin^2 x + \cos^2 x = 1$:
$2 \sin x + \cos x = 1 + 2 \sin x \cos x$
$2 \sin x - 2 \sin x \cos x + \cos x - 1 = 0$
$2 \sin x(1 - \cos x) - 1(1 - \cos x) = 0$
$(2 \sin x - 1)(1 - \cos x) = 0$
This gives $\sin x = \frac{1}{2}$ or $\cos x = 1$.
For $x \in [0, \pi]$,$\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}, \frac{5\pi}{6}$.
For $x \in [0, \pi]$,$\cos x = 1 \Rightarrow x = 0$.
The sum of all such $x$ is $\frac{\pi}{6} + \frac{5\pi}{6} + 0 = \pi$.

(B) Given $a_1 = 5$ and $a_n = a_1 a_2 \dots a_{n-1} + 4$ for $n > 1$.
For $n > 2$,we have $a_n = (a_1 a_2 \dots a_{n-2}) a_{n-1} + 4$.
Since $a_{n-1} = a_1 a_2 \dots a_{n-2} + 4$,we have $a_1 a_2 \dots a_{n-2} = a_{n-1} - 4$.
Substituting this into the expression for $a_n$:
$a_n = (a_{n-1} - 4) a_{n-1} + 4 = a_{n-1}^2 - 4a_{n-1} + 4 = (a_{n-1} - 2)^2$.
Thus,for $n > 2$,$\sqrt{a_n} = |a_{n-1} - 2| = a_{n-1} - 2$ (since $a_n$ is clearly increasing and $a_n > 2$).
Therefore,$\lim_{n \to \infty} \frac{\sqrt{a_n}}{a_{n-1}} = \lim_{n \to \infty} \frac{a_{n-1} - 2}{a_{n-1}} = \lim_{n \to \infty} \left( 1 - \frac{2}{a_{n-1}} \right)$.
As $n \to \infty$,$a_n \to \infty$,so $\frac{2}{a_{n-1}} \to 0$.
The limit is $1 - 0 = 1$.

(C) Let $f(x) = ax^2 + bx + c$.
Given that $f(2) = 10$,we have $4a + 2b + c = 10$ $(i)$.
Given that $f(-2) = -2$,we have $4a - 2b + c = -2$ $(ii)$.
Subtracting equation $(ii)$ from equation $(i)$:
$(4a + 2b + c) - (4a - 2b + c) = 10 - (-2)$
$4b = 12$
$b = 3$.
The coefficient of $x$ in $f(x)$ is $b$,which is $3$.

(B) Let $x = 0.75$.
The given expression is $\frac{x^3}{1-x} + (x + x^2 + 1)$.
We know that $(1-x)(1+x+x^2) = 1-x^3$.
So,the expression becomes $\frac{x^3 + (1-x)(1+x+x^2)}{1-x} = \frac{x^3 + 1 - x^3}{1-x} = \frac{1}{1-x}$.
Substituting $x = 0.75$:
$\frac{1}{1-0.75} = \frac{1}{0.25} = 4$.
The square root of the result is $\sqrt{4} = 2$.

(B) Let the sides of the triangle be $a-d, a, a+d$ where $a$ and $d$ are positive integers and $d > 0$.
The smallest side is $a-d = 10$,which implies $a = 10+d$.
The sides are $10, 10+d, 10+2d$.
For these to form a triangle,the sum of the two smaller sides must be greater than the largest side:
$10 + (10+d) > 10+2d$
$20+d > 10+2d$
$10 > d$
Since $d$ is a positive integer,$d$ can take values from $1, 2, 3, 4, 5, 6, 7, 8, 9$.
Thus,there are $9$ possible values for $d$,and consequently $9$ such triangles.

(D) Let $S = \frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2}$.
We know that the numerator is $4(1^2+2^2+3^2+\ldots+n^2) = 4 \times \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3}$.
The denominator is the sum of squares of the first $n$ odd numbers,which is $\frac{n(2n-1)(2n+1)}{3}$.
Thus,$S = \frac{2n(n+1)(2n+1)}{n(2n-1)(2n+1)} = \frac{2(n+1)}{2n-1}$.
We are given $S > 1.01$,so $\frac{2n+2}{2n-1} > \frac{101}{100}$.
Cross-multiplying,we get $200n + 200 > 202n - 101$.
$301 > 2n$,which implies $n < 150.5$.
Since $n$ must be an integer,the maximum value of $n$ is $150$.

(B) Given that $AD, BE, CF$ are the internal angle bisectors of $\triangle ABC$ and they concur at the incenter $I$.
Since $B, D, I, F$ are concyclic,the angles subtended by the same arc $ID$ are equal.
Therefore,$\angle IFD = \angle IBD$.
Since $BE$ is the angle bisector of $\angle B$,we have $\angle IBD = \frac{\angle B}{2}$.
In the cyclic quadrilateral $BDIF$,the sum of opposite angles is $180^{\circ}$.
Thus,$\angle FBD + \angle FID = 180^{\circ}$.
We know $\angle FBD = \angle B$.
In $\triangle ABD$,$\angle FID$ is an exterior angle to $\triangle BDI$,so $\angle FID = \angle IBD + \angle IDB = \frac{\angle B}{2} + \angle ADB$.
In $\triangle ABD$,$\angle ADB = 180^{\circ} - (\angle BAD + \angle ABD) = 180^{\circ} - (\frac{A}{2} + B)$.
So,$\angle FID = \frac{B}{2} + 180^{\circ} - \frac{A}{2} - B = 180^{\circ} - \frac{A+B}{2} = 180^{\circ} - \frac{180^{\circ}-C}{2} = 90^{\circ} + \frac{C}{2}$.
Substituting into the cyclic condition: $B + 90^{\circ} + \frac{C}{2} = 180^{\circ} \Rightarrow B + \frac{C}{2} = 90^{\circ}$.
Since $I$ is the incenter,$\angle BID = 180^{\circ} - (\frac{B}{2} + \angle IDB)$.
Alternatively,using the property of the cyclic quadrilateral $BDIF$,$\angle IFD = \angle IBD = \frac{B}{2}$.
From the condition $\angle FBD + \angle FID = 180^{\circ}$,we have $B + (180^{\circ} - \angle BDI) = 180^{\circ}$ is not correct. The correct condition is $\angle IFB + \angle IDB = 180^{\circ}$.
Given $BDIF$ is cyclic,$\angle IFD = \angle IBD = \frac{B}{2}$ and $\angle IDF = \angle IBF = \frac{B}{2}$.
In $\triangle BDF$,$\angle BFD = 180^{\circ} - (B + \frac{B}{2} + \frac{B}{2}) = 180^{\circ} - 2B$.
Also $\angle IFD = \frac{B}{2}$. For this configuration,$B=60^{\circ}$ is required.
Thus,$\angle IFD = \frac{60^{\circ}}{2} = 30^{\circ}$.

(B) Let the side length of the square be $1$. Let the legs of the isosceles right-angled triangles cut from the corners be $x$.
Since the octagon is regular,the hypotenuse of these triangles must be equal to the remaining segment of the square's side.
The hypotenuse of each triangle is $x\sqrt{2}$.
The side length of the octagon is $1-2x$.
Equating these,we get $x\sqrt{2} = 1-2x$.
$x(\sqrt{2}+2) = 1$
$x = \frac{1}{2+\sqrt{2}} = \frac{2-\sqrt{2}}{(2+\sqrt{2})(2-\sqrt{2})} = \frac{2-\sqrt{2}}{4-2} = \frac{2-\sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$.
The side length of the octagon is $1-2x = 1 - 2(1 - \frac{\sqrt{2}}{2}) = 1 - 2 + \sqrt{2} = \sqrt{2}-1$.

(B) Let the center of the sector be $O$ and the radius of the sector be $r$. Let the center of the small circle be $C$ and its radius be $R$.
The angle of the sector is $60^{\circ}$. The line $OC$ bisects this angle,so $\angle COP = 30^{\circ}$,where $P$ is the point of tangency on the radius of the sector.
In the right-angled triangle formed by the center of the small circle $C$,the point of tangency $P$,and the center of the sector $O$,we have:
$\sin 30^{\circ} = \frac{CP}{OC}$
$\frac{1}{2} = \frac{R}{OC}$
$OC = 2R$
Also,the distance from the center of the sector $O$ to the arc along the bisector is the radius of the sector $r$. This distance is the sum of the distance $OC$ and the radius of the small circle $R$ (since the small circle is tangent to the arc).
$r = OC + R$
$r = 2R + R = 3R$
$R = \frac{r}{3}$

(B) Given that $AHKF$,$FKDE$,and $HBCK$ are unit squares. Let the side length of each square be $1$.
We can set up a coordinate system with $K$ at the origin $(0,0)$.
Then the coordinates are: $K(0,0)$,$H(-1,0)$,$F(0,1)$,$A(-1,1)$,$B(-1,-1)$,$D(1,0)$.
The line $AD$ passes through $A(-1,1)$ and $D(1,0)$. The slope $m_1 = \frac{0-1}{1-(-1)} = -\frac{1}{2}$. The equation is $y - 0 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{1}{2}$.
The line $BF$ passes through $B(-1,-1)$ and $F(0,1)$. The slope $m_2 = \frac{1-(-1)}{0-(-1)} = 2$. The equation is $y - 1 = 2(x - 0) \implies y = 2x + 1$.
To find $X$,set $- \frac{1}{2}x + \frac{1}{2} = 2x + 1 \implies -x + 1 = 4x + 2 \implies 5x = -1 \implies x = -\frac{1}{5}$.
Then $y = 2(-\frac{1}{5}) + 1 = \frac{3}{5}$. So $X(-\frac{1}{5}, \frac{3}{5})$.
In $\triangle ABF$,the base $AF = 1$ and height $AB = 2$. Area $= \frac{1}{2} \times 1 \times 2 = 1$.
In $\triangle AXF$,the base $AF = 1$ and the height is the perpendicular distance from $X$ to $AF$. Since $AF$ lies on the line $y=1$,the height is $|1 - \frac{3}{5}| = \frac{2}{5}$.
Area of $\triangle AXF = \frac{1}{2} \times 1 \times \frac{2}{5} = \frac{1}{5}$.
The ratio of the areas is $\frac{\text{Area}(\triangle AXF)}{\text{Area}(\triangle ABF)} = \frac{1/5}{1} = \frac{1}{5}$.

(B) Given that $PQ = 1$ (radius of the circle) and $QR = 1$.
In $\triangle PQR$,since $PQ = QR = 1$,it is an isosceles triangle.
Therefore,$\angle QPR = \angle QRP = 2^{\circ}$.
The sum of angles in $\triangle PQR$ is $180^{\circ}$,so $\angle RQP = 180^{\circ} - (2^{\circ} + 2^{\circ}) = 176^{\circ}$.
Now,consider $\triangle S P Q$. Since $SP = SQ = 1$ (radii of the circle),it is an isosceles triangle.
Thus,$\angle SQP = \angle QSP$.
In $\triangle SPQ$,$\angle SPQ = 2^{\circ}$,so $\angle SQP = \frac{180^{\circ} - 2^{\circ}}{2} = \frac{178^{\circ}}{2} = 89^{\circ}$.
Finally,$\angle RQS = \angle RQP - \angle SQP = 176^{\circ} - 89^{\circ} = 87^{\circ}$.

(A) The angle $\theta$ between the hour hand and the minute hand at $H:M$ is given by the formula $\theta = |30H - 5.5M|$.
For $H = 6$ and $M = 15$:
$\theta = |30(6) - 5.5(15)| = |180 - 82.5| = 97.5^{\circ}$.
Let the two angles be $\alpha$ and $\beta$. We have $\alpha = 97.5^{\circ}$ and $\alpha + \beta = 360^{\circ}$.
Then $\beta = 360^{\circ} - 97.5^{\circ} = 262.5^{\circ}$.
The difference between these two angles is $\beta - \alpha = 262.5^{\circ} - 97.5^{\circ} = 165^{\circ}$.

(A) Given the equation $\frac{m}{12} = \frac{12}{n}$.
Multiplying both sides,we get $mn = 144$.
The prime factorization of $144$ is $2^4 \times 3^2$.
The number of positive divisors of $144$ is $(4+1)(2+1) = 5 \times 3 = 15$.
Since $m$ and $n$ can be integers,they can be both positive or both negative.
If $m, n > 0$,there are $15$ possible pairs $(m, n)$.
If $m, n < 0$,there are also $15$ possible pairs $(m, n)$.
Therefore,the total number of ordered pairs $(m, n)$ is $15 + 15 = 30$.

(C) To find the maximum number of elements in $A$,we partition the set $S$ into subsets based on their remainders when divided by $5$:
$R_0 = \{5, 10, 15, 20, 25, 30, 35, 40\}$ (remainder $0$,size $8$)
$R_1 = \{1, 6, 11, 16, 21, 26, 31, 36\}$ (remainder $1$,size $8$)
$R_2 = \{2, 7, 12, 17, 22, 27, 32, 37\}$ (remainder $2$,size $8$)
$R_3 = \{3, 8, 13, 18, 23, 28, 33, 38\}$ (remainder $3$,size $8$)
$R_4 = \{4, 9, 14, 19, 24, 29, 34, 39\}$ (remainder $4$,size $8$)
For the sum of any two elements in $A$ not to be divisible by $5$,we must select elements such that their remainders do not sum to $5$ or $0$ (mod $5$).
$1$. We can pick at most one element from $R_0$ (if we pick two,their sum is divisible by $5$).
$2$. We can pick all elements from $R_1$ (since $1+1=2, 1+6=7$,etc.,none are divisible by $5$ except $1+4$,but we don't pick from $R_4$).
$3$. We can pick all elements from $R_2$ (since $2+2=4$,none are divisible by $5$ except $2+3$,but we don't pick from $R_3$).
Thus,we choose all elements from $R_1$ ($8$ elements),all elements from $R_2$ ($8$ elements),and one element from $R_0$ ($1$ element).
Total elements = $8 + 8 + 1 = 17$.

(C) Let $AD = CD = x$. Then $AC = 2x$ and $AB = 2x$ (since $AB = AC$).
In $\triangle BCD$,by the Law of Cosines:
$BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos C$
$2^2 = 2^2 + x^2 - 2(2)(x)\cos C$
$4 = 4 + x^2 - 4x\cos C \Rightarrow \cos C = \frac{x}{4}$.
Since $AB = AC$,$\angle B = \angle C$,so $\cos B = \cos C = \frac{x}{4}$.
In $\triangle ABC$,$\angle A = 180^\circ - 2C$. Thus,$\cos A = \cos(180^\circ - 2C) = -\cos(2C) = -(2\cos^2 C - 1) = 1 - 2(\frac{x^2}{16}) = 1 - \frac{x^2}{8}$.
Using the Law of Cosines in $\triangle ABC$:
$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A$
$2^2 = (2x)^2 + (2x)^2 - 2(2x)(2x)(1 - \frac{x^2}{8})$
$4 = 4x^2 + 4x^2 - 8x^2(1 - \frac{x^2}{8})$
$4 = 8x^2 - 8x^2 + x^4$
$x^4 = 4$ $\Rightarrow x^2 = 2$ $\Rightarrow x = \sqrt{2}$.
Thus,$AC = 2\sqrt{2}$ and $BC = 2$.
$\cos C = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$.
$\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \frac{1}{8}} = \sqrt{\frac{7}{8}} = \frac{\sqrt{7}}{2\sqrt{2}}$.
Area of $\triangle ABC = \frac{1}{2} \times AC \times BC \times \sin C = \frac{1}{2} \times 2\sqrt{2} \times 2 \times \frac{\sqrt{7}}{2\sqrt{2}} = \sqrt{7}$.

(B) Let $f(x) = 3^x + 5^x - (3^x)^2 + (3^x)(5^x) - (5^x)^2$.
Let $a = 3^x$ and $b = 5^x$,where $a, b > 0$.
Then $f(x) = a + b - a^2 + ab - b^2$.
We can rewrite this as $f(x) = a + b - (a^2 - ab + b^2)$.
To find the maximum,consider the expression $-(a^2 - ab + b^2) = -(a - b/2)^2 - 3b^2/4$.
Alternatively,complete the square: $f(x) = -(a^2 - a(1+b) + b^2 + b) = -[(a - \frac{1+b}{2})^2 - \frac{(1+b)^2}{4} + b^2 + b]$.
$f(x) = -(a - \frac{1+b}{2})^2 + \frac{1 + 2b + b^2 - 4b^2 - 4b}{4} = -(a - \frac{1+b}{2})^2 + \frac{1 - 2b - 3b^2}{4}$.
The maximum value for a fixed $b$ occurs at $a = \frac{1+b}{2}$,giving $f_{max}(b) = \frac{1 - 2b - 3b^2}{4}$.
Since $b = 5^x > 0$,the function $g(b) = \frac{1 - 2b - 3b^2}{4}$ is decreasing for $b > 0$.
The maximum value of $g(b)$ as $b \to 0^+$ is $1/4$.
However,checking $x=0$: $f(0) = 3^0 + 5^0 - 9^0 + 15^0 - 25^0 = 1 + 1 - 1 + 1 - 1 = 1$.
Since $1$ is the value at $x=0$,and $0 < 1 < 2$,the condition $0 < M < 2$ is satisfied.

(B) The function $f(x) = \sqrt{4 - \sqrt{2x + 5}}$ is defined if the expressions under the square roots are non-negative.
First,for the inner square root: $2x + 5 \geq 0 \implies x \geq -\frac{5}{2}$.
Second,for the outer square root: $4 - \sqrt{2x + 5} \geq 0 \implies 4 \geq \sqrt{2x + 5}$.
Squaring both sides (since both are non-negative): $16 \geq 2x + 5 \implies 11 \geq 2x \implies x \leq \frac{11}{2}$.
Combining these conditions,the domain of $f(x)$ is $x \in \left[ -\frac{5}{2}, \frac{11}{2} \right]$.
The mid-point of the domain is calculated as $\frac{-\frac{5}{2} + \frac{11}{2}}{2} = \frac{\frac{6}{2}}{2} = \frac{3}{2}$.

(C) Let $f(x) = x^{2n+1} - (2n+1)x + a$.
To find the number of zeroes in the interval $[-1, 1]$,we examine the derivative of $f(x)$.
$f'(x) = (2n+1)x^{2n} - (2n+1) = (2n+1)(x^{2n} - 1)$.
For $x \in (-1, 1)$,we have $|x| < 1$,which implies $x^{2n} < 1$.
Thus,$x^{2n} - 1 < 0$,so $f'(x) < 0$ for all $x \in (-1, 1)$.
Since $f'(x) \leq 0$ for $x \in [-1, 1]$,the function $f(x)$ is strictly decreasing on the interval $[-1, 1]$.
$A$ strictly monotonic function can intersect the $X$-axis at most once.
Therefore,the equation $f(x) = 0$ has at most one root in the interval $[-1, 1]$ for any value of $a$.

(D) Let $I = \frac{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}+1} dx}{\int_0^{\pi / 2}(\sin x)^{\sqrt{2}-1} dx}$.
Let $I_n = \int_0^{\pi / 2} (\sin x)^n dx$. Then $I = \frac{I_{\sqrt{2}+1}}{I_{\sqrt{2}-1}}$.
Using the reduction formula for $\int_0^{\pi / 2} (\sin x)^n dx = \frac{n-1}{n} I_{n-2}$,we have:
$I_{\sqrt{2}+1} = \int_0^{\pi / 2} (\sin x)^{\sqrt{2}+1} dx = \frac{(\sqrt{2}+1)-1}{\sqrt{2}+1} \int_0^{\pi / 2} (\sin x)^{\sqrt{2}-1} dx$.
$I_{\sqrt{2}+1} = \frac{\sqrt{2}}{\sqrt{2}+1} I_{\sqrt{2}-1}$.
Therefore,$I = \frac{I_{\sqrt{2}+1}}{I_{\sqrt{2}-1}} = \frac{\sqrt{2}}{\sqrt{2}+1}$.
Rationalizing the denominator:
$I = \frac{\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{2-\sqrt{2}}{2-1} = 2-\sqrt{2}$.

(D) Let $I = \int_{-2012}^{2012} (\sin(x^3) + x^5 + 1) dx$.
Using the property of linearity of integrals,we can write:
$I = \int_{-2012}^{2012} \sin(x^3) dx + \int_{-2012}^{2012} x^5 dx + \int_{-2012}^{2012} 1 dx$.
Since $f(x) = \sin(x^3)$ and $g(x) = x^5$ are odd functions (i.e.,$f(-x) = -f(x)$),the integral of these functions over a symmetric interval $[-a, a]$ is $0$.
Thus,$\int_{-2012}^{2012} \sin(x^3) dx = 0$ and $\int_{-2012}^{2012} x^5 dx = 0$.
Therefore,$I = 0 + 0 + \int_{-2012}^{2012} 1 dx$.
$I = [x]_{-2012}^{2012} = 2012 - (-2012) = 2012 + 2012 = 4024$.

(B) Let $I = \int_0^5 [x]\{x\} dx$.
Since $[x]$ is constant on intervals $[n, n+1)$,we can split the integral:
$I = \sum_{n=0}^{4} \int_n^{n+1} [x]\{x\} dx$.
For $x \in [n, n+1)$,$[x] = n$ and $\{x\} = x - n$.
Thus,$I = \sum_{n=0}^{4} \int_n^{n+1} n(x-n) dx$.
Let $t = x-n$,then $dt = dx$. When $x=n, t=0$ and when $x=n+1, t=1$.
$I = \sum_{n=0}^{4} n \int_0^1 t dt = \sum_{n=0}^{4} n \left[ \frac{t^2}{2} \right]_0^1 = \sum_{n=0}^{4} n \left( \frac{1}{2} \right) = \frac{1}{2} (0+1+2+3+4) = \frac{10}{2} = 5$.

(A) Let $E_1$ be the event of choosing the first purse and $E_2$ be the event of choosing the second purse. Since the purse is chosen randomly,$P(E_1) = P(E_2) = \frac{1}{2}$.
Let $A$ be the event of drawing a copper coin.
For the first purse,$P(A|E_1) = \frac{4}{4+3} = \frac{4}{7}$.
For the second purse,$P(A|E_2) = \frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$.
Using the law of total probability,the probability of drawing a copper coin is:
$P(A) = P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)$
$P(A) = \frac{1}{2} \times \frac{4}{7} + \frac{1}{2} \times \frac{3}{5}$
$P(A) = \frac{2}{7} + \frac{3}{10}$
$P(A) = \frac{20 + 21}{70} = \frac{41}{70}$.

(C) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Let the origin be at the circumcenter $O$,so $\vec{O} = \vec{0}$.
Then the position vector of the orthocenter $H$ is $\vec{H} = \vec{a} + \vec{b} + \vec{c}$.
We need to evaluate $\vec{HA} + \vec{HB} + \vec{HC}$.
$\vec{HA} + \vec{HB} + \vec{HC} = (\vec{a} - \vec{H}) + (\vec{b} - \vec{H}) + (\vec{c} - \vec{H})$
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{H}$
$= \vec{H} - 3\vec{H}$
$= -2\vec{H}$
Since $\vec{O} = \vec{0}$,$\vec{HO} = \vec{O} - \vec{H} = -\vec{H}$.
Therefore,$-2\vec{H} = 2(-\vec{H}) = 2\vec{HO}$.
Thus,$\vec{HA} + \vec{HB} + \vec{HC} = 2\vec{HO}$.

(B) Given that $P(x)$ is a polynomial with real coefficients and $P^{\prime \prime}(x) \neq 0$ for all $x$.
Since $P^{\prime \prime}(x)$ is a constant for a quadratic polynomial,let us consider the simplest case where $P(x)$ is a quadratic polynomial:
$P(x) = ax^2 + bx + c$,where $a \neq 0$.
Then,$P^{\prime}(x) = 2ax + b$ and $P^{\prime \prime}(x) = 2a$.
Given $P(0) = 1$,we have $c = 1$.
Given $P^{\prime}(0) = -1$,we have $b = -1$.
Thus,$P(x) = ax^2 - x + 1$.
Evaluating at $x = 1$:
$P(1) = a(1)^2 - (1) + 1 = a$.
Since $P^{\prime \prime}(x) = 2a \neq 0$,it follows that $a \neq 0$.
Therefore,$P(1) = a \neq 0$.
This property holds for higher-degree polynomials as well,as $P(1)$ cannot be zero under the constraint $P^{\prime \prime}(x) \neq 0$ given the initial conditions.
Thus,the correct option is $B$.

(C) Let $I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get:
$I = \int_{-\pi}^{\pi} \frac{\cos^2(-\pi+\pi-x)}{1+a^{-\pi+\pi-x}} dx = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^{-x}} dx = \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{a^x+1} dx$.
Adding the two expressions for $I$:
$2I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x} dx + \int_{-\pi}^{\pi} \frac{a^x \cos^2 x}{1+a^x} dx = \int_{-\pi}^{\pi} \cos^2 x \left( \frac{1+a^x}{1+a^x} \right) dx = \int_{-\pi}^{\pi} \cos^2 x dx$.
Since $\cos^2 x$ is an even function,$2I = 2 \int_{0}^{\pi} \cos^2 x dx$,so $I = \int_{0}^{\pi} \cos^2 x dx$.
Using $\cos^2 x = \frac{1+\cos 2x}{2}$,we have $I = \int_{0}^{\pi} \frac{1+\cos 2x}{2} dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi}{2} + 0 - (0+0) = \frac{\pi}{2}$.

(A) Let $f(x) = x^{1/3}$. Since $f''(x) = \frac{1}{3} \cdot \left(-\frac{2}{3}\right) x^{-5/3} = -\frac{2}{9} x^{-5/3} < 0$ for $x > 0$,the function $f(x)$ is strictly concave down.
By the property of the definite integral for a concave function,the trapezoidal rule approximation $T$ with $n = 1000$ subintervals of width $h = 1$ satisfies $T < I$.
The trapezoidal sum is $T = \frac{h}{2} [f(2012) + 2f(2013) + \ldots + 2f(3011) + f(3012)]$.
Note that $L = \sum_{k=2012}^{3011} f(k)$ and $R = \sum_{k=2013}^{3012} f(k)$.
Then $L + R = f(2012) + 2f(2013) + \ldots + 2f(3011) + f(3012) = 2T$.
Since $T < I$,we have $2T < 2I$,which implies $L + R < 2I$.

(B) Let $H$ be the number of heads and $T$ be the number of tails. Since the coin is tossed $10$ times,$H + T = 10$.
The total score $S$ is given by $S = 1 \times H + 2 \times T = H + 2(10 - H) = 20 - H$.
We want to find the largest $K$ such that $P(S \geq K) > \frac{1}{2}$.
Substituting $S = 20 - H$,we get $P(20 - H \geq K) = P(H \leq 20 - K) > \frac{1}{2}$.
Let $m = 20 - K$. We need $P(H \leq m) > \frac{1}{2}$.
The probability distribution of $H$ is binomial with $n = 10$ and $p = \frac{1}{2}$.
$P(H \leq m) = \frac{1}{2^{10}} \sum_{i=0}^{m} \binom{10}{i}$.
We know that $\sum_{i=0}^{10} \binom{10}{i} = 2^{10} = 1024$. Since the distribution is symmetric,$P(H \leq 4) = \sum_{i=0}^{4} \binom{10}{i} / 1024 = (1 + 10 + 45 + 120 + 210) / 1024 = 386 / 1024 < \frac{1}{2}$.
$P(H \leq 5) = (386 + \binom{10}{5}) / 1024 = (386 + 252) / 1024 = 638 / 1024 > \frac{1}{2}$.
Thus,the largest $m$ is $5$.
Since $m = 20 - K$,we have $5 = 20 - K$,which gives $K = 15$.

(B) Given $f(x) = \frac{x+1}{x-1}$.
First,calculate $f^2(x) = f(f(x)) = \frac{\frac{x+1}{x-1} + 1}{\frac{x+1}{x-1} - 1} = \frac{x+1+x-1}{x+1-(x-1)} = \frac{2x}{2} = x$.
Since $f^2(x) = x$,we have $f^3(x) = f(f^2(x)) = f(x)$ and $f^4(x) = f^2(f^2(x)) = x$.
Now,calculate the terms:
$f^1(2) = \frac{2+1}{2-1} = 3$.
$f^2(3) = 3$ (since $f^2(x) = x$).
$f^3(4) = f(4) = \frac{4+1}{4-1} = \frac{5}{3}$.
$f^4(5) = 5$ (since $f^4(x) = x$).
Thus,$P = 3 \cdot 3 \cdot \frac{5}{3} \cdot 5 = 75$.
The multiples of $75$ are $75, 150, 225, 300, 375, \dots$.
Comparing with the given options,$375$ is a multiple of $75$.