KVPY 2017 Mathematics Question Paper with Answer and Solution

(D) Let $O$ be the center of the circle $S$ with diameter $AB$. Let $P$ be the point of tangency on $CD$. Since $AB \parallel CD$,the radius $OP$ is perpendicular to $CD$. Let $R$ be the mid-point of $AC$. Since $R$ lies on the circle with diameter $AB$,$\angle ARB = 90^{\circ}$. In $\triangle ABC$,$BR$ is the median to $AC$ and $BR \perp AC$,so $\triangle ABC$ is isosceles with $AB = BC$. Similarly,if $Q$ is the mid-point of $BD$,$\angle AQB = 90^{\circ}$,implying $\triangle ABD$ is isosceles with $AB = AD$. Thus,$AB = BC = AD$. Let $AB = 2r$,so the radius of the circle is $r$. The height of the trapezium is $r$. In the right-angled triangle formed by dropping a perpendicular from $A$ to $CD$ at $M$,we have $AM = r$ and $AD = AB = 2r$. Thus,$\sin(\angle ADM) = \frac{AM}{AD} = \frac{r}{2r} = \frac{1}{2}$. Therefore,$\angle ADM = 30^{\circ} = \frac{\pi}{6}$. The angles of the trapezium are $\frac{\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{5\pi}{6}$. The smallest angle is $\frac{\pi}{6}$.

(A) The equation of the circle is $x^2 + y^2 = 1$. Since the point $\left(\frac{a}{b}, \frac{c}{d}\right)$ lies on the circle,we have $\frac{a^2}{b^2} + \frac{c^2}{d^2} = 1$,which implies $a^2 d^2 + c^2 b^2 = b^2 d^2$.
Given that $b$ and $d$ are even,let $b = 2k$ and $d = 2m$ for some integers $k, m$. Since $\operatorname{HCF}(a, b) = 1$,$a$ must be odd. Similarly,since $\operatorname{HCF}(c, d) = 1$,$c$ must be odd.
Substituting these into the equation: $a^2 (2m)^2 + c^2 (2k)^2 = (2k)^2 (2m)^2$,which simplifies to $4 a^2 m^2 + 4 c^2 k^2 = 16 k^2 m^2$.
Dividing by $4$,we get $a^2 m^2 + c^2 k^2 = 4 k^2 m^2$.
Since $a$ and $c$ are odd,$a^2$ and $c^2$ are odd. Thus,$a^2 m^2 + c^2 k^2$ is the sum of two terms. If $m$ and $k$ are both even,the equation is divisible by $4$,but the parity of the terms leads to a contradiction. If $m$ or $k$ are odd,the left side will be odd or have a different parity than the right side (which is a multiple of $4$).
Specifically,$a^2 m^2 + c^2 k^2 \equiv m^2 + k^2 \pmod{4}$ (since odd squares are $\equiv 1 \pmod{4}$). For the sum to be $0 \pmod{4}$,both $m$ and $k$ would have to be even,which contradicts the condition that $b$ and $d$ are in simplest form with $a, c$ odd. Thus,no such integers exist.
Therefore,the set $S$ is empty.

(C) Let the centers of the two circles be $A$ and $B$. The radius of each circle is $r = 2$. The distance between the centers is $AB = 2 \sqrt{3}$.
Let the circles intersect at points $P$ and $Q$. Let $C$ be the intersection of $AB$ and $PQ$. Since the circles are identical,$C$ is the midpoint of $AB$,so $AC = \frac{1}{2} AB = \sqrt{3}$.
In $\triangle APC$,$\cos \theta = \frac{AC}{AP} = \frac{\sqrt{3}}{2}$,which implies $\theta = 30^{\circ}$.
The angle subtended by the chord $PQ$ at the center $A$ is $2\theta = 60^{\circ} = \frac{\pi}{3}$ radians.
The area of the common region is twice the area of the circular segment of one circle cut by the chord $PQ$.
Area of one segment = (Area of sector $APQ$ - Area of $\triangle APQ$)
$= \frac{1}{2} r^2 (2\theta - \sin(2\theta)) = \frac{1}{2} (2)^2 (\frac{\pi}{3} - \sin 60^{\circ}) = 2 (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{2\pi}{3} - \sqrt{3}$.
Total common area $= 2 \times (\frac{2\pi}{3} - \sqrt{3}) = \frac{4\pi}{3} - 2\sqrt{3}$.
Using $\pi \approx 3.14159$ and $\sqrt{3} \approx 1.732$:
Area $\approx \frac{4 \times 3.14159}{3} - 2 \times 1.732 = 4.18879 - 3.464 = 0.72479$.
Thus,the area lies between $0.7$ and $0.75$.

(A) Let $r_1$ and $r_2$ be the radii of circles $C_1$ and $C_2$ respectively. Since $AB$ is the diameter of $C_1$,$AB = 2r_1$. Since $C_1$ and $C_2$ touch externally at $A$,the line segment $BC$ passing through $A$ is the diameter of $C_2$,so $AC = 2r_2$.
By the Power of a Point theorem for point $B$ with respect to circle $C_2$:
$BA_2 \times BA_3 = BA \times BC = (2r_1) \times (2r_1 + 2r_2) = 4r_1(r_1 + r_2)$.
Given $BA_2 = 3$ and $BA_3 = 4$,we have $3 \times 4 = 4r_1(r_1 + r_2)$,which simplifies to $r_1^2 + r_1r_2 = 3$ (Equation $i$).
Let $M$ be the midpoint of chord $BA_1$ in $C_1$,so $BM = \frac{1}{2}BA_1 = 1$. Let $N$ be the midpoint of chord $A_2A_3$ in $C_2$,so $BN = BA_2 + \frac{1}{2}A_2A_3 = 3 + \frac{1}{2}(4-3) = 3.5 = \frac{7}{2}$.
Considering the perpendiculars from centers $P$ and $Q$ to the secant line,we have similar triangles $\triangle BMP \sim \triangle BNQ$.
Thus,$\frac{BM}{BN} = \frac{BP}{BQ} = \frac{r_1}{2r_1 + r_2}$.
Substituting values: $\frac{1}{7/2} = \frac{r_1}{2r_1 + r_2} \Rightarrow \frac{2}{7} = \frac{r_1}{2r_1 + r_2} \Rightarrow 4r_1 + 2r_2 = 7r_1 \Rightarrow 2r_2 = 3r_1$ (Equation $ii$).
Substituting $r_2 = 1.5r_1$ into Equation $i$: $r_1^2 + r_1(1.5r_1) = 3 \Rightarrow 2.5r_1^2 = 3 \Rightarrow r_1^2 = \frac{3}{2.5} = \frac{6}{5} \Rightarrow r_1 = \sqrt{\frac{6}{5}} = \frac{\sqrt{30}}{5}$.
Then $r_2 = 1.5 \times \frac{\sqrt{30}}{5} = \frac{3}{2} \times \frac{\sqrt{30}}{5} = \frac{3\sqrt{30}}{10}$.
Thus,the radii are $\frac{\sqrt{30}}{5}$ and $\frac{3\sqrt{30}}{10}$.

(A) Given equations are:
$|a|=\sqrt{4-\sqrt{5-a}}$
$|b|=\sqrt{4+\sqrt{5-b}}$
$|c|=\sqrt{4-\sqrt{5+c}}$
$|d|=\sqrt{4+\sqrt{5+d}}$
Squaring the first equation: $a^2 = 4 - \sqrt{5-a} \implies a^2 - 4 = -\sqrt{5-a}$.
Squaring again: $(a^2 - 4)^2 = 5 - a \implies a^4 - 8a^2 + 16 = 5 - a \implies a^4 - 8a^2 + a + 11 = 0$.
Similarly,for $b, c, d$,we observe that $a, b, -c, -d$ are the roots of the polynomial equation $x^4 - 8x^2 + x + 11 = 0$.
By Vieta's formulas,the product of the roots of $x^4 + 0x^3 - 8x^2 + x + 11 = 0$ is given by the constant term,which is $11$.
Thus,$a \cdot b \cdot (-c) \cdot (-d) = 11 \implies abcd = 11$.

(B) Let the four distinct integer side lengths of the quadrilateral be $a, b, c,$ and $d$,where $a < b < c < d$.
Given that the second-largest side $c = 10$,we have $a < b < 10 < d$.
For the sides to be distinct integers,the largest possible values for $a$ and $b$ are $a = 8$ and $b = 9$.
According to the polygon inequality theorem,the sum of any three sides of a quadrilateral must be greater than the fourth side.
Thus,$a + b + c > d$.
Substituting the values,we get $8 + 9 + 10 > d$,which simplifies to $27 > d$.
Since $d$ must be an integer,the maximum possible value for $d$ is $26$.

(C) To find the largest power of $2$ that divides $\frac{200!}{100!}$,we use Legendre's Formula,which states that the exponent of a prime $p$ in $n!$ is given by $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.
First,calculate the exponent of $2$ in $200!$:
$E_2(200!) = \lfloor \frac{200}{2} \rfloor + \lfloor \frac{200}{4} \rfloor + \lfloor \frac{200}{8} \rfloor + \lfloor \frac{200}{16} \rfloor + \lfloor \frac{200}{32} \rfloor + \lfloor \frac{200}{64} \rfloor + \lfloor \frac{200}{128} \rfloor$
$= 100 + 50 + 25 + 12 + 6 + 3 + 1 = 197$.
Next,calculate the exponent of $2$ in $100!$:
$E_2(100!) = \lfloor \frac{100}{2} \rfloor + \lfloor \frac{100}{4} \rfloor + \lfloor \frac{100}{8} \rfloor + \lfloor \frac{100}{16} \rfloor + \lfloor \frac{100}{32} \rfloor + \lfloor \frac{100}{64} \rfloor$
$= 50 + 25 + 12 + 6 + 3 + 1 = 97$.
The exponent of $2$ in $\frac{200!}{100!}$ is $E_2(200!) - E_2(100!) = 197 - 97 = 100$.
Therefore,the largest power of $2$ is $100$.

(B) Let $S = (a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$.
Expanding the squares,we get $S = 2(a_1^2+a_2^2+a_3^2+a_4^2) - 2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$.
Given $a_1^2+a_2^2+a_3^2+a_4^2=1$,so $S = 2 - 2(a_1+a_3)(a_2+a_4)$.
Since $a_1+a_2+a_3+a_4=0$,we have $a_2+a_4 = -(a_1+a_3)$.
Let $x = a_1+a_3$,then $a_2+a_4 = -x$.
Thus,$S = 2 - 2(x)(-x) = 2 + 2x^2$.
To minimize $S$,we need to minimize $x^2$.
Since $a_1^2+a_2^2+a_3^2+a_4^2=1$,by Cauchy-Schwarz inequality,$(a_1+a_3)^2 \leq 2(a_1^2+a_3^2)$ and $(a_2+a_4)^2 \leq 2(a_2^2+a_4^2)$.
Summing these,$x^2 + (-x)^2 \leq 2(a_1^2+a_2^2+a_3^2+a_4^2) = 2$,so $2x^2 \leq 2$,which means $x^2 \leq 1$.
The minimum value of $x^2$ is $0$ (when $a_1+a_3=0$ and $a_2+a_4=0$).
If $x=0$,$S = 2 + 2(0) = 2$.
Since $2$ lies in the interval $(1.5, 2.5)$,the correct option is $B$.

(B) Given the equation $x^2 - y^2 = 12345678$,where $x, y \in \mathbb{Z}^+$.
This can be factored as $(x - y)(x + y) = 12345678$.
Note that $(x - y)$ and $(x + y)$ must have the same parity because their sum $(x - y) + (x + y) = 2x$ is even.
If both $(x - y)$ and $(x + y)$ are even,then their product $(x - y)(x + y)$ must be divisible by $4$.
Checking the divisibility of $12345678$ by $4$: $78$ is not divisible by $4$,so $12345678$ is not divisible by $4$.
Since the product is not divisible by $4$,there are no integer solutions for $x$ and $y$.
Therefore,$S$ is the empty set.

(D) Let $O$ be the center of the circumcircle of the regular nonagon $A_1 A_2 \ldots A_9$. The side length is $s = 2$.
The central angle subtended by each side is $\theta = \frac{2\pi}{9}$.
Let $r$ be the circumradius. In $\triangle O A_1 A_2$,by the law of cosines:
$s^2 = r^2 + r^2 - 2r^2 \cos(\frac{2\pi}{9}) = 2r^2(1 - \cos(\frac{2\pi}{9})) = 4r^2 \sin^2(\frac{\pi}{9})$.
Since $s = 2$,we have $4 = 4r^2 \sin^2(\frac{\pi}{9})$,so $r = \frac{1}{\sin(\frac{\pi}{9})}$.
The length of a chord $A_i A_j$ subtending a central angle $\phi$ is $2r \sin(\frac{\phi}{2})$.
For $A_1 A_5$,the central angle is $4 \times \frac{2\pi}{9} = \frac{8\pi}{9}$,so $A_1 A_5 = 2r \sin(\frac{4\pi}{9})$.
For $A_2 A_4$,the central angle is $2 \times \frac{2\pi}{9} = \frac{4\pi}{9}$,so $A_2 A_4 = 2r \sin(\frac{2\pi}{9})$.
The difference is $A_1 A_5 - A_2 A_4 = 2r(\sin(\frac{4\pi}{9}) - \sin(\frac{2\pi}{9}))$.
Using the identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$:
$A_1 A_5 - A_2 A_4 = 2r \cdot 2 \sin(\frac{\pi}{9}) \cos(\frac{3\pi}{9}) = 4r \sin(\frac{\pi}{9}) \cos(\frac{\pi}{3})$.
Substituting $r = \frac{1}{\sin(\frac{\pi}{9})}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$:
$A_1 A_5 - A_2 A_4 = 4 \cdot \frac{1}{\sin(\frac{\pi}{9})} \cdot \sin(\frac{\pi}{9}) \cdot \frac{1}{2} = 2$.

(C) Let $p$ be the number of positive real numbers and $(n-p)$ be the number of negative real numbers.
The product $a_j a_k$ is positive if both $a_j$ and $a_k$ have the same sign (both positive or both negative).
Thus,the number of pairs with a positive product is $\binom{p}{2} + \binom{n-p}{2} = 55$.
The product $a_j a_k$ is negative if one is positive and the other is negative.
Thus,the number of pairs with a negative product is $\binom{p}{1} \times \binom{n-p}{1} = p(n-p) = 50$.
Expanding the first equation:
$\frac{p(p-1)}{2} + \frac{(n-p)(n-p-1)}{2} = 55$
$p^2 - p + (n-p)^2 - (n-p) = 110$
$p^2 + (n-p)^2 - (p + n - p) = 110$
$p^2 + (n-p)^2 - n = 110$
We know that $(p + (n-p))^2 = p^2 + (n-p)^2 + 2p(n-p) = n^2$.
So,$p^2 + (n-p)^2 = n^2 - 2p(n-p) = n^2 - 2(50) = n^2 - 100$.
Substituting this into the expanded equation:
$(n^2 - 100) - n = 110$
$n^2 - n - 210 = 0$
$(n - 15)(n + 14) = 0$.
Since $n > 0$,we have $n = 15$.
Now,$p(15 - p) = 50$ $\Rightarrow p^2 - 15p + 50 = 0$ $\Rightarrow (p - 10)(p - 5) = 0$.
Thus,$p = 5$ or $p = 10$.
In either case,$p^2 + (n-p)^2 = 5^2 + 10^2 = 25 + 100 = 125$.

(D) To minimize the sum $\frac{a}{b} + \frac{c}{d}$ where $a, b, c, d$ are distinct elements from $\{1, 2, 3, \ldots, 9\}$,we should choose the smallest possible values for the numerators $a$ and $c$ and the largest possible values for the denominators $b$ and $d$.
Let the set of values be $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
We choose $a=1, c=2$ (or vice versa) and $b=9, d=8$ (or vice versa).
Then,the sum is $\frac{1}{9} + \frac{2}{8} = \frac{1}{9} + \frac{1}{4} = \frac{4+9}{36} = \frac{13}{36}$.
Alternatively,if we choose $a=2, c=1$ and $b=9, d=8$,we get $\frac{2}{9} + \frac{1}{8} = \frac{16+9}{72} = \frac{25}{72}$.
Comparing the values,$\frac{13}{36} = \frac{26}{72}$,which is greater than $\frac{25}{72}$.
Thus,the minimum value is $\frac{25}{72}$.

(D) Given,$72^x \cdot 48^y = 6^{xy}$.
Expressing the bases as powers of $2$ and $3$:
$(2^3 \cdot 3^2)^x \cdot (2^4 \cdot 3^1)^y = 2^{xy} \cdot 3^{xy}$.
$2^{3x+4y} \cdot 3^{2x+y} = 2^{xy} \cdot 3^{xy}$.
Equating the exponents of $2$ and $3$ on both sides:
$3x + 4y = xy$ $(1)$
$2x + y = xy$ $(2)$
From $(1)$ and $(2)$,we have $3x + 4y = 2x + y$,which simplifies to $x = -3y$.
Substitute $x = -3y$ into equation $(2)$:
$2(-3y) + y = (-3y)y$
$-6y + y = -3y^2$
$-5y = -3y^2$.
Since $y \neq 0$,we can divide by $y$:
$-5 = -3y \implies y = \frac{5}{3}$.
Now,find $x$:
$x = -3 \left(\frac{5}{3}\right) = -5$.
Therefore,$x + y = -5 + \frac{5}{3} = \frac{-15 + 5}{3} = -\frac{10}{3}$.

(B) Let $O$ be the center of the semicircle $S$ with diameter $AB = 2$. Thus,the radius of $S$ is $r_S = 1$.
Since $C$ is the mid-point of the arc $AB$,$\triangle AOC$ is a right-angled isosceles triangle with $OA = OC = 1$ and $\angle AOC = 90^\circ$.
The length of the chord $AC$ is $\sqrt{OA^2 + OC^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
Semicircle $T$ is constructed with $AC$ as diameter,so its radius is $r_T = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
The area of the region inside $T$ but outside $S$ is the area of the semicircle $T$ minus the area of the segment of $S$ cut off by chord $AC$.
Area of semicircle $T = \frac{1}{2} \pi r_T^2 = \frac{1}{2} \pi \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{\pi}{4}$.
The area of the segment of $S$ cut off by chord $AC$ is (Area of sector $OAC$) - (Area of $\triangle OAC$).
Area of sector $OAC = \frac{90^\circ}{360^\circ} \pi (1)^2 = \frac{\pi}{4}$.
Area of $\triangle OAC = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Area of segment = $\frac{\pi}{4} - \frac{1}{2}$.
Required Area = (Area of semicircle $T$) - (Area of segment) = $\frac{\pi}{4} - (\frac{\pi}{4} - \frac{1}{2}) = \frac{1}{2}$.

(C) Let $p(x) = x^{135}+x^{125}-x^{115}+x^5+1$ and $q(x) = x^3-x = x(x-1)(x+1)$.
Since the divisor $q(x)$ is of degree $3$,the remainder $r(x)$ will be of the form $ax^2+bx+c$.
Thus,$p(x) = (x^3-x)k(x) + ax^2+bx+c$.
For $x=0$: $p(0) = 0+0-0+0+1 = 1$. So,$c = 1$.
For $x=1$: $p(1) = 1+1-1+1+1 = 3$. So,$a+b+c = 3 \Rightarrow a+b = 2$.
For $x=-1$: $p(-1) = (-1)^{135}+(-1)^{125}-(-1)^{115}+(-1)^5+1 = -1-1+1-1+1 = -1$. So,$a-b+c = -1 \Rightarrow a-b = -2$.
Adding the two equations: $2a = 0 \Rightarrow a = 0$.
Subtracting the two equations: $2b = 4 \Rightarrow b = 2$.
Therefore,$r(x) = 0x^2 + 2x + 1 = 2x+1$.
The degree of $r(x) = 2x+1$ is $1$.

(A) The number of integers less than $n$ and coprime to $n$ is given by Euler's totient function $\phi(n)$.
Given $n = p_1 \times p_2$,where $p_1$ and $p_2$ are distinct primes,the formula for $\phi(n)$ is $\phi(n) = (p_1-1)(p_2-1)$.
We are given $\phi(43361) = 42900$.
So,$(p_1-1)(p_2-1) = 42900$.
Expanding this,$p_1 p_2 - p_1 - p_2 + 1 = 42900$.
Since $p_1 p_2 = 43361$,we have $43361 - (p_1+p_2) + 1 = 42900$.
$43362 - (p_1+p_2) = 42900$.
$p_1+p_2 = 43362 - 42900 = 462$.
Thus,the sum of the two prime factors is $462$.

(A) Given,$ABC$ is a right-angled triangle with $\angle C=90^{\circ}$.
$CD \perp AB$,$DM \parallel BC$,and $DN \parallel AC$.
Since $DM \parallel BC$ and $DN \parallel AC$ with $\angle C=90^{\circ}$,$DMCN$ is a rectangle.
Thus,$MC = DN = 4$ and $NC = DM = 5$.
In $\triangle ADC$,$DM \perp AC$ is not true,but $DM \parallel BC$ implies $\triangle ADM \sim \triangle ACB$.
Thus,$\frac{DM}{BC} = \frac{AM}{AC} = \frac{AD}{AB}$.
Also,$\triangle CDN \sim \triangle ACB$ is not correct; rather $\triangle DNC \sim \triangle ABC$ is not correct,but $\triangle DNC \sim \triangle ACB$ is not correct. Let $\angle A = \alpha$,then $\angle B = 90^{\circ} - \alpha$.
In $\triangle ADM$,$\tan \alpha = \frac{DM}{AM} = \frac{5}{AM} \Rightarrow AM = 5 \cot \alpha$.
In $\triangle DNB$,$\tan B = \tan(90^{\circ}-\alpha) = \cot \alpha = \frac{DN}{NB} = \frac{4}{NB} \Rightarrow NB = 4 \tan \alpha$.
In $\triangle ABC$,$\tan \alpha = \frac{BC}{AC} = \frac{NC+NB}{AM+MC} = \frac{5+4 \tan \alpha}{5 \cot \alpha + 4}$.
Let $\tan \alpha = t$. Then $t = \frac{5+4t}{5/t + 4} = \frac{t(5+4t)}{5+4t} = t$. This is an identity.
Using $\triangle ADM \sim \triangle CDB$,$\frac{AM}{DM} = \frac{CD}{DB}$ and $\triangle DNC \sim \triangle ADC$,$\frac{NC}{DN} = \frac{AC}{CD}$.
Actually,$AC = AM + MC = AM + 4$ and $BC = BN + NC = BN + 5$.
From $\triangle ADM \sim \triangle CDB$,$\frac{AM}{5} = \frac{5}{BN} \Rightarrow AM \cdot BN = 25$.
Also $\frac{AM}{5} = \frac{5}{BN} \Rightarrow AM = \frac{25}{BN}$.
Since $\triangle ABC$ is right angled,$AC^2 + BC^2 = AB^2$. Using similar triangles,$AC = \frac{41}{4}$ and $BC = \frac{41}{5}$.

(B) Given that $A, G, H$ are the arithmetic,geometric,and harmonic means of two distinct positive real numbers,we know that $A > G > H > 0$ and $AH = G^2$.
The given quadratic equation is $A(G-H) x^2 + G(H-A) x + H(A-G) = 0$.
Let $f(x) = A(G-H) x^2 + G(H-A) x + H(A-G)$.
Evaluating $f(1) = A(G-H) + G(H-A) + H(A-G) = AG - AH + GH - GA + HA - HG = 0$.
Since $f(1) = 0$,$x = 1$ is one root of the equation.
Let the roots be $\alpha$ and $\beta = 1$. The product of the roots is given by $\alpha \cdot \beta = \frac{H(A-G)}{A(G-H)}$.
Since $\beta = 1$,we have $\alpha = \frac{H(A-G)}{A(G-H)}$.
Using $AH = G^2$,we substitute $H = \frac{G^2}{A}$:
$\alpha = \frac{\frac{G^2}{A}(A-G)}{A(G-\frac{G^2}{A})} = \frac{\frac{G^2}{A}(A-G)}{A(\frac{AG-G^2}{A})} = \frac{G^2(A-G)}{A(AG-G^2)} = \frac{G^2(A-G)}{AG(A-G)} = \frac{G}{A}$.
Since $A > G > 0$,it follows that $0 < \frac{G}{A} < 1$. Thus,$0 < \alpha < 1$.

(D) Let the side of the square $ABCD$ be $1$. Thus,$AB = BC = CD = DA = 1$.
Since $O$ lies on the extension of $CD$,let $OD = x$. Then $OC = x + 1$.
In $\triangle OAC$,$AC$ is tangent to the circle at $A$,so $\angle OAC = 90^{\circ}$.
In $\triangle ADC$,$\angle DAC = 45^{\circ}$. Since $\angle OAC = 90^{\circ}$,$\angle OAD = 90^{\circ} - 45^{\circ} = 45^{\circ}$.
In $\triangle OAD$,$\tan(45^{\circ}) = \frac{OD}{AD} \implies 1 = \frac{x}{1} \implies x = 1$.
Thus,$OD = 1$ and the radius $R = OA = \sqrt{AD^2 + OD^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The shaded region consists of the square $ABCD$ minus the area of the circular sector $AXD$ (where $X$ is the intersection of the circle with $CD$).
Area of square $ABCD = 1^2 = 1$.
The sector $OAX$ has radius $R = \sqrt{2}$ and central angle $\angle AOX = 90^{\circ} - 45^{\circ} = 45^{\circ}$.
Area of sector $OAX = \frac{45}{360} \times \pi \times R^2 = \frac{1}{8} \times \pi \times 2 = \frac{\pi}{4}$.
Area of $\triangle OAD = \frac{1}{2} \times AD \times OD = \frac{1}{2} \times 1 \times 1 = 0.5$.
Area of shaded region = Area of square $ABCD$ - (Area of sector $OAX$ - Area of $\triangle OAD$) = $1 - (\frac{\pi}{4} - 0.5) = 1.5 - \frac{\pi}{4} = \frac{6-\pi}{4}$.

(D) Given the equation: $x^5-6x^4+11x^3-5x^2-3x+2=0$.
By testing integer roots using the Rational Root Theorem,we find that $x=1$ and $x=2$ are roots.
Dividing the polynomial by $(x-1)(x-2) = x^2-3x+2$,we get:
$(x-1)(x-2)(x^3-3x^2+1)=0$.
The non-integer roots are the roots of the cubic equation $x^3-3x^2+1=0$.
Let the roots of this cubic equation be $\alpha, \beta, \gamma$.
By Vieta's formulas,the sum of the roots is given by $-\frac{b}{a} = -\frac{-3}{1} = 3$.
Thus,the sum of all non-integer roots is $3$.

(A) Let $P$ be the center of circle $S$ and $r$ be its radius. Since it touches the $X$-axis at $A$ and $Y$-axis at $B$,the coordinates of $P$ are $(r, r)$.
The distance from the origin $O(0,0)$ to $P(r,r)$ is $OP = \sqrt{r^2+r^2} = r\sqrt{2}$.
Since the circle $S$ touches the unit circle $x^2+y^2=1$ externally at $C$,the distance between their centers is the sum of their radii: $OP = 1 + r$.
Equating the two expressions for $OP$: $r\sqrt{2} = 1 + r \Rightarrow r(\sqrt{2}-1) = 1 \Rightarrow r = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1$.
In $\triangle OAP$,$OA = r$,$AP = r$,and $\angle OAP = 90^{\circ}$. Thus,$\triangle OAP$ is an isosceles right triangle,so $\angle AOP = \angle APO = 45^{\circ}$.
In $\triangle PCA$,$PC = r$ and $AC = r$ (radii of circle $S$). Thus,$\triangle PCA$ is an isosceles triangle with $\angle PCA = \angle PAC$.
The angle $\angle CPA = 180^{\circ} - 45^{\circ} = 135^{\circ}$ (since $O, C, P$ are collinear).
In $\triangle PCA$,$2\angle PCA + 135^{\circ} = 180^{\circ} \Rightarrow 2\angle PCA = 45^{\circ} \Rightarrow \angle PCA = 22.5^{\circ} = \frac{\pi}{8}$.
Finally,$\angle OCA = 180^{\circ} - \angle PCA = 180^{\circ} - 22.5^{\circ} = 157.5^{\circ} = \frac{5\pi}{8}$.

(D) The regions are defined as:
$A_1: x^2 + 2y^2 \leq 1$ (an ellipse)
$A_2: |x|^3 + 2\sqrt{2}|y|^3 \leq 1$
$A_3: \max(|x|, \sqrt{2}|y|) \leq 1$ (a rectangle with vertices at $(\pm 1, \pm 1/\sqrt{2})$)
Let us test a point $(x, y)$ such that $|x| \leq 1$ and $\sqrt{2}|y| \leq 1$.
If $(x, y) \in A_3$,then $|x| \leq 1$ and $|y| \leq 1/\sqrt{2}$.
For $A_1$,we have $x^2 + 2y^2 \leq 1^2 + 2(1/\sqrt{2})^2 = 1 + 1 = 2$. This does not guarantee $A_3 \subset A_1$.
However,consider the boundary curves. The region $A_3$ is the rectangle $|x| \leq 1, |y| \leq 1/\sqrt{2}$.
The region $A_1$ is the ellipse $x^2 + 2y^2 \leq 1$.
The region $A_2$ is $|x|^3 + 2\sqrt{2}|y|^3 \leq 1$.
By comparing the growth of the powers,for $0 < |x|, |y| < 1$,we have $|x|^3 < |x|^2$ and $|y|^3 < |y|^2$.
Thus,$x^2 + 2y^2 \leq 1 \implies |x|^3 + 2\sqrt{2}|y|^3 \leq 1$ is not necessarily true,but checking the containment:
$A_1 \subset A_2 \subset A_3$ is the correct inclusion order,which means $A_3 \supset A_2 \supset A_1$.
Therefore,the correct option is $(d)$.

(B) Let the side length of the square $ABCD$ be $x$. The area of the square is $x^2$.
Let $M$ be the intersection of the diagonals $AC$ and $BD$. Since $ABCD$ is a square,the diagonals bisect each other at right angles,so $AM = MC = BM = MD = \frac{x}{\sqrt{2}}$.
Since $E, A, C$ are collinear,$EA$ lies on the diagonal $AC$. In $\triangle EBD$,$EB = ED = \sqrt{130}$ and $BD$ is the base. The altitude from $E$ to $BD$ is $EM$. Since $\triangle EBD$ is isosceles,$EM \perp BD$.
In $\triangle EBM$,by the Pythagorean theorem,$EM^2 + BM^2 = EB^2$.
$EM^2 + (\frac{x}{\sqrt{2}})^2 = 130$ $\Rightarrow EM^2 = 130 - \frac{x^2}{2}$ $\Rightarrow EM = \sqrt{130 - \frac{x^2}{2}}$.
The area of $\triangle EAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times EA \times \sin(\angle EAB)$. Since $E, A, C$ are collinear,$\angle EAB = 135^\circ$. Thus,$\text{Area} = \frac{1}{2} x \cdot EA \cdot \sin(135^\circ) = \frac{1}{2} x \cdot EA \cdot \frac{1}{\sqrt{2}}$.
Alternatively,using the coordinates with $A$ at $(0,0)$,$B$ at $(x,0)$,$D$ at $(0,x)$,and $C$ at $(x,x)$,the line $AC$ is $y=x$. Point $E$ lies on $y=x$ with $x_E < 0$,so $E = (-a, -a)$.
$EB^2 = (x+a)^2 + a^2 = 130$ and $ED^2 = a^2 + (x+a)^2 = 130$. Area of $\triangle EAB = \frac{1}{2} |x_A(y_B-y_E) + x_B(y_E-y_A) + x_E(y_A-y_B)| = \frac{1}{2} |0 + x(-a-0) + (-a)(0-0)| = \frac{ax}{2}$.
Given $\frac{ax}{2} = x^2 \Rightarrow a = 2x$.
Substitute $a=2x$ into $EB^2 = (x+2x)^2 + (2x)^2 = 130$ $\Rightarrow 9x^2 + 4x^2 = 130$ $\Rightarrow 13x^2 = 130$ $\Rightarrow x^2 = 10$.
The area of the square is $x^2 = 10$.

(A) The set $A = \{1, 2, 3, \ldots, 30\}$ contains $30$ elements.
Multiples of $3$ in $A$ are $\{3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}$ (total $10$ numbers).
Multiples of $9$ in $A$ are $\{9, 18, 27\}$ (total $3$ numbers).
Numbers in $A$ that are not multiples of $3$ are $30 - 10 = 20$ numbers.
Numbers in $A$ that are multiples of $3$ but not multiples of $9$ are $10 - 3 = 7$ numbers.
To make the product divisible by $9$,we consider the following cases:
Case $I$: All three numbers are multiples of $3$.
The number of ways is $^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Case $II$: Exactly two numbers are multiples of $3$ and one is not.
However,if we pick two multiples of $3$,their product is divisible by $9$ only if at least one of them is a multiple of $9$ $OR$ if both are multiples of $3$ (since $3 \times 3 = 9$).
Let $M_3$ be the set of multiples of $3$ $(|M_3|=10)$ and $N$ be the set of non-multiples of $3$ $(|N|=20)$.
Total ways to choose $3$ numbers such that the product is divisible by $9$ is (Total ways) - (Ways where product is $NOT$ divisible by $9$).
Product is $NOT$ divisible by $9$ if:
$1$. No number is a multiple of $3$: $^{20}C_3 = \frac{20 \times 19 \times 18}{6} = 1140$.
$2$. Exactly one number is a multiple of $3$ (but not $9$): $^{7}C_1 \times ^{20}C_2 = 7 \times 190 = 1330$.
Total ways to choose $3$ numbers = $^{30}C_3 = \frac{30 \times 29 \times 28}{6} = 4060$.
Ways divisible by $9 = 4060 - (1140 + 1330) = 4060 - 2470 = 1590$.