Consider the quadratic function in $x$
$$y = - \frac { 1 } { 8 } x ^ { 2 } + a x + b .$$
When we denote the coordinates of the vertex of the graph of (1) by $( p , q )$, we have
$$p = \mathbf { A } a , \quad q = \mathbf { B } a ^ { 2 } + b .$$
(1) When the vertex ( $p , q$ ) is on the straight line $x + y = 1 , a$ and $b$ satisfy
$$b = \mathbf { C D } a ^ { 2 } - \mathbf { E E } a + \mathbf { F } ,$$
and so $8 a + b$ is maximized at $a = \mathbf { G }$, and the maximum value is $\mathbf { H }$.
(2) When the graph of (1) is tangent to the $x$-axis, the range of values of $a + b$ is
$$a + b \leqq \frac { \mathbf { I } } { \mathbf { J } }$$

On a coordinate plane, 12 points are arranged as shown in the figure to the right. If we are to select three points as vertices of a triangle, how many triangles are possible in total?
First, there are $\mathbf { K L M }$ ways to select three points from the 12 points.
Next, let us find how many ways it is possible to select three or more points in a straight line.
Let us look at the two cases.
(i) There are $\mathbf { N }$ straight lines that pass through four points.
(ii) There are $\mathbf { O }$ straight lines that pass through three points.
Hence, among all combinations of three points that are in a straight line and so cannot be the vertices of a triangle, $\mathbf { P Q }$ combinations belong to case (i), and $\mathbf { Q }$ combinations belong to case (ii).
Thus, the total number of possible triangles is $\mathbf{STU}$.
In particular, if we set $( 1,1 )$ as point A and $( 4,1 )$ as point B , then $\mathbf { V W }$ triangles have two vertices on segment AB.

We are to find the value of $a$ such that $15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a$ can be factorized as the product of linear expressions in $x$ and $y$.
First of all, the first three terms of the above expression form a quadratic expression in $x$ and $y$ that can be factorized as
$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } = ( \mathbf { A } x - \mathbf { B } y ) ( \mathbf { C } x + \mathbf { D } y ) .$$
Hence, when we set
$$\begin{aligned} & 15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } - 11 x + 22 y + a \\ & \quad = ( \mathbf { A } x - \mathbf { B } y + b ) ( \mathbf { C } x + \mathbf { D } y + c ) , \end{aligned}$$
the right-hand side of equation (1) can be expanded into
$$15 x ^ { 2 } - 2 x y - 8 y ^ { 2 } + ( \mathbf { E } b + \mathbf { F } c ) x + ( \mathbf { G } b - \mathbf { H } c ) y + b c .$$
When we compare the coefficients of this expression with the coefficients of the left-hand side of equation (1), we have
$$b = \mathbf { I } , \quad c = - \mathbf { J } ,$$
and hence $a = - \mathbf { K L }$.

We are to find a two-digit natural number $a$ such that $a + 9$ is a multiple of 7 and $a + 8$ is a multiple of 13.
First of all, $a + 9$ and $a + 8$ can be represented as
$$a + 9 = \mathbf { M } m , \quad a + 8 = \mathbf { N O } n ,$$
where $m$ and $n$ are natural numbers. From these two equalities, we have
$$\mathbf { M } m - \mathbf { N O } n = \mathbf { P } .$$
Since the pair of $m = \mathbf { Q }$ and $n = \mathbf { R }$ is an integral solution of (1), we have
$$\mathbf { M } ( m - \mathbf { Q } ) = \mathbf { NO } ( n - \mathbf { R } ) .$$
From (2), a natural number $n$ satisfying (1) can be represented as
$$n = \mathbf { S }$$
where $k$ is an integer. Thus
$$a = \mathbf { U V } k + \mathbf { W } ,$$
and since $a$ is a two-digit natural number, $a = \mathbf { X Y }$.

Consider the two functions
$$\begin{aligned} & f ( x ) = x ^ { 2 } + 2 a x + 4 a - 3 \\ & g ( x ) = 2 x + 1 \end{aligned}$$
We are to find the condition on $a$ for which $f ( x ) \geqq g ( x )$ for all $x$ and also find the range of values of the minimum of $f ( x )$ under this condition.
We must find the condition under which
$$x ^ { 2 } + \mathbf { A } ( a - \mathbf { A } ) x + \mathbf { A C } a - \mathbf { A D } \geq 0$$
for all $x$.
For each of $\mathbf { E } \sim \mathbf{ H }$ in the following questions, choose the correct answer from among (0) $\sim$ (7) below each question.
(1) The required condition is that $a$ satisfy the quadratic inequality $\mathbf{E}$. Hence $a$ is in the range $\mathbf { F }$. (0) $a ^ { 2 } - 5 a + 4 \geqq 0$
(1) $a ^ { 2 } - 6 a + 5 \geqq 0$
(2) $a ^ { 2 } - 5 a + 4 \leqq 0$
(3) $a ^ { 2 } - 6 a + 5 \leqq 0$
(4) $a \leqq 1$ or $5 \leqq a$
(5) $1 \leqq a \leqq 5$ (6) $1 \leqq a \leqq 4$ (7) $a \leqq 1$ or $4 \leqq a$
(2) Let $m$ be the minimum value of $f ( x )$. Then, since $m = \mathbf { G }$, the range of values which $m$ can take under the condition in (1) is $\mathbf { H }$. (0) $a ^ { 2 } + 4 a - 3$
(1) $4 a ^ { 2 } + 4 a - 3$
(2) $- a ^ { 2 } + 4 a - 3$
(3) $2 a ^ { 2 } - 4 a + 3$
(4) $- 5 \leqq m \leqq 1$
(5) $- 8 \leqq m \leqq 1$ (6) $- 8 \leqq m \leqq - 1$ (7) $- 5 \leqq m \leqq - 1$

The figure to the right is a net for the tetrahedron OABC. This tetrahedron satisfies
$$\begin{gathered} \mathrm { BC } = 10 , \quad \mathrm { AC } = 8 , \quad \sin \angle \mathrm { ACB } = \frac { 3 } { 4 } , \\ \mathrm { OA } = 4 , \quad \triangle \mathrm { ABC } \equiv \triangle \mathrm { OBC } . \end{gathered}$$
(1) The area of the triangle ABC is $\mathbf { A B }$.
(2) Let AH denote the perpendicular line drawn from point A to side BC. The length of AH is $\mathbf { C }$.
(3) Let $\theta$ denote the angle formed by the plane ABC and the plane OBC. Then we have
$$\cos \theta = \frac { \mathbf { D } } { \mathbf { E } } , \quad \sin \theta = \frac { \mathbf { F } \sqrt { \mathbf { G } } } { \mathbf { H } } .$$
(4) The volume of the tetrahedron OABC is $\frac { \mathbf { IJ } \sqrt { \mathbf { K } } } { \mathbf { L } }$.

Consider the quadratic function in $x$
$$y = - \frac { 1 } { 8 } x ^ { 2 } + a x + b .$$
When we denote the coordinates of the vertex of the graph of (1) by $( p , q )$, we have
$$p = \mathbf { A } a , \quad q = \mathbf { B } a ^ { 2 } + b .$$
(1) When the vertex ( $p , q$ ) is on the straight line $x + y = 1 , a$ and $b$ satisfy
$$b = \mathbf { C D } a ^ { 2 } - \mathbf { E E } a + \mathbf { F } ,$$
and so $8 a + b$ is maximized at $a = \mathbf { G }$, and the maximum value is $\mathbf { H }$.
(2) When the graph of (1) is tangent to the $x$-axis, the range of values of $a + b$ is
$$a + b \leqq \frac { \mathbf { I } } { \mathbf { J } } .$$

On a coordinate plane, 12 points are arranged as shown in the figure to the right. If we are to select three points as vertices of a triangle, how many triangles are possible in total?
First, there are $\mathbf { K L M }$ ways to select three points from the 12 points.
Next, let us find how many ways it is possible to select three or more points in a straight line.
Let us look at the two cases.
(i) There are $\mathbf { N }$ straight lines that pass through four points.
(ii) There are $\mathbf { O }$ straight lines that pass through three points.
Hence, among all combinations of three points that are in a straight line and so cannot be the vertices of a triangle, $\mathbf { P Q }$ combinations belong to case (i), and $\mathbf { Q }$ combinations belong to case (ii).
Thus, the total number of possible triangles is $\mathbf{STU}$.
In particular, if we set $( 1,1 )$ as point A and $( 4,1 )$ as point B , then $\mathbf { V W }$ triangles have two vertices on segment AB.

The triangle $ABC$ satisfies
$$\mathrm { AB } = 2 , \quad \mathrm { BC } = 3 , \quad \mathrm { CA } = 4 .$$
(1) When we set $\angle \mathrm { ABC } = \theta$, the inner product $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } }$ of the vectors $\overrightarrow { \mathrm { AB } }$ and $\overrightarrow { \mathrm { BC } }$ is
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \mathbf { AB } \cos \theta .$$
Finding the value of $\cos \theta$ from the law of cosines, we obtain
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \frac { \mathbf { C } } { \mathbf { D } } .$$
(2) We divide the side BC into $n$ equal parts by the points $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \cdots , \mathrm { P } _ { n - 1 }$ which are arranged in ascending order of the distance from B, and set $\mathrm { B } = \mathrm { P } _ { 0 } , \mathrm { C } = \mathrm { P } _ { n }$. We are to find the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } }$.
When we calculate the inner product of $\overrightarrow { \mathrm { AP } _ { k - 1 } }$ and $\overrightarrow { \mathrm { AP } _ { k } }$ using (1), we have
$$\overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \mathbf { E } + \frac { \mathbf { F } } { 2 n } + \mathbf { G }$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \frac { \mathbf { IJ } } { \mathbf { K } }$$

Consider complex numbers $z$ such that
$$z \bar { z } - ( 1 - 2 i ) z - ( 1 + 2 i ) \bar { z } \leqq 15 .$$
(1) On a complex number plane, the figure represented by inequality (1) is the interior and circumference of the circle having the center $\mathbf{L} + \mathbf{M} i$ and the radius $\mathbf{NO}$.
(2) Let us consider all complex numbers $z$ which are on the straight line
$$( 1 - i ) z - ( 1 + i ) \bar { z } = 2 i$$
and satisfy the inequality (1). Of those, let $z _ { 1 }$ be the $z$ such that $| z |$ is maximized and $z _ { 2 }$ be the $z$ such that $| z |$ is minimized. Then we have
$$z _ { 1 } = \sqrt { \mathbf { P Q } } + \mathbf{Q} + ( \sqrt { \mathbf { S T } } + \mathbf { U } ) i ,$$ $$z _ { 2 } = - \frac { \mathbf { U } } { \mathbf { V } } + \frac{\mathbf{W}}{\mathbf{P}} i .$$

Let us consider the real numbers $x , y , t$ and $u$ satisfying the following four conditions:
$$\begin{aligned} & y \geqq | x | \\ & x + y = t \\ & x ^ { 2 } + y ^ { 2 } = 12 \\ & x ^ { 3 } + y ^ { 3 } = u \end{aligned}$$
We are to find the ranges of values which $t$ and $u$ can take.
(1) From (1) and (3), we see that the point $( x , y )$ is located on the arc which is a quadrant of the circle having its center at the origin and the radius $\mathbf { A }$. Moreover, the coordinates of the end points of this arc are
$$( \sqrt { \mathbf { C } } , \sqrt { \mathbf { CD } } ) \text { and } ( - \sqrt { \mathbf { C } } , \sqrt { \mathbf { D } } ) .$$
From this and (2), we also see that the range of values which $t$ can take is
$$\mathbf { E } \leqq t \leqq \mathbf { F } . \mathbf { G } .$$
(2) Next, from (2) and (3), we have
$$x y = \frac { \mathbf { H } } { \mathbf { I } } \left( t ^ { 2 } - \mathbf { J K } \right)$$
and further, using (4) we also have
$$u = \frac { \mathbf { L } } { \mathbf{L}} \left( \mathbf{NO} \, t - t ^ { 3 } \right)$$
Hence, since
$$\frac { d u } { d t } = \frac { \mathbf { P } } { \mathbf { Q } } \left( \mathbf { RS } - t ^ { 2 } \right)$$
the range of values which $u$ can take under the condition (5) is
$$\mathbf { TL } \leqq u \leqq \mathbf { UV } \sqrt { \mathbf{ W } } .$$

Let $a > 1$. We divide the region defined by the two inequalities
$$0 \leqq x \leqq \frac { \pi } { 6 } , \quad 0 \leqq y \leqq a \cos 3 x$$
into two sections by the straight line $y = 1$. Let us denote the area of the section where $y \geq 1$ by $S$ and the area of the section where $y \leq 1$ by $T$. We are to find the value of $a$ such that $T - S$ is maximized, and also find the maximum value of $T - S$.
Let $t$ denote the value of $x \left( 0 \leqq x \leqq \frac { \pi } { 6 } \right)$ satisfying the equation $a \cos 3 x = 1$. Then we have
$$\begin{aligned} S & = \frac { \sin 3 t } { \mathbf { A } \cos 3 t } - t \\ S + T & = \frac { 1 } { \mathbf { B } \cos 3 t } . \end{aligned}$$
When we set $f ( t ) = T - S$, we see that
$$f ^ { \prime } ( t ) = \frac { ( \mathbf { C } - \mathbf { D } \sin 3 t ) \sin 3 t } { \cos ^ { \mathbf { E } } 3 t } .$$
Hence $T - S$ is maximized at $t = \frac { \pi } { \mathbf { F G } }$. Thus, $T - S$ is maximized at $a = \frac { \mathbf { H } \sqrt { \mathbf { I } } } { \mathbf{J} }$, and the maximum value is $\frac { \pi } { \mathbf { K } }$.