---
title: "東京大学大学院　情報理工入試過去問（数学）解答・２０２２年　第一問"
tags: u-tokyo, exam
---

# 東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">―解答・２０２２年　第一問</span>

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+ こちらの解答は非公式で個人記録用に作成されており、大学院側に認められたものではありません。正確性についての保証は致しかねます。
+ 問題本文は研究科のウェブサイトから見ることができます。
+ 満足出来た解答でしたらコメントしていただけると幸いです。

## ２０２２年　第一問
For some triplets of $x, y, z \in \mathbb{R}$, $(x, y, z) \in S$ if:
$$
\begin{cases}
0 & < & z - xy & < & 1 \\
0 & < & z - (x + y)^2 & < & -xy
\end{cases}
$$

And the set of points $\Omega = \{ (x, y) : \displaystyle\exists z : (x, y, z) \in S\}$.

### (1) Find inequalities on $x, y$ representing $\Omega$.
From $0 < z-(x + y)^2 < -xy$, we get:
$$
x^2 + xy + y^2 < z - xy < x^2 + y^2
$$

Then combining the two inequality and eliminate the $z-xy$ term, we get:
$$\Omega = \left\{ (x, y) :
\begin{cases}
x^2 + xy + y^2 < 1 \\[1ex]
0 < x^2 + y^2 \\[1ex]
x^2 + xy + y^2 < x^2 + y^2
\end{cases} \right\}
$$

From equation three, we know that $xy < 0$, inducing that equation two is always true.
Thus we can reduce the result into:
$$
\Omega = \left\{ (x, y) :
\begin{cases}
x^2 + xy + y^2 < 1 \\[1ex]
xy < 0
\end{cases} \right\}
$$

### (2) Plot the region representing $\Omega$ on the $xy$-plane.
+ $xy < 0$ tells that no points are in the I, III quadrants.
+ We use the general formula of conic sections: $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to plot the region $x^2 + xy + y^2 < 1$.

The border line of the region of $1x^2 + 1xy + 1y^2 - 1 = 0$, so for the general formula,
$$
A = B = C = 1, D = E = 0, F = -1.
$$

+ Since $B \neq 0$, the plot is a *rotated* conic section.
+ Since $B^2 - 4AC > 0$, the plot is an ellipse.
+ Using the formula $\cot 2\theta = {A - C \over B} = 0$, the rotation of the plot is $45^{\circ}$, or $\pi \over 4$.
+ Use the formulas to calculate the shape of the elliplse:
$$
\begin{align}
A' &= A\cos^2\theta + B\cos\sin + C\sin^2\theta \\
B' &= 0 \\
C' &= A\sin ^2\theta - B\cos\sin + C\cos^2\theta \\
D' &= D\cos\theta + E\sin\theta \\
E' &= -D\sin\theta + E\cos\theta \\
F' &= F
\end{align}
$$
So the shape of the ellipse is as the ellipse
$$
{3 \over 2}x^2 + {1 \over 2} y^2 = 1 \Rightarrow \frac{x^2}{\sqrt{2 \over 3}^2} + \frac{y^2}{\sqrt{2}^2} = 1
$$
which the main axis is the $y$-axis.

Also, from the formula, the boundary intersecting the axises are:
$$
(0, 1), (0, -1), (1, 0), (-1, 0)
$$

So we can know the plot is:

### (3) Find the linear transformation for the border of the region of $\Omega$ from a unit circle, where $(0, 1)$ is transformed to the point having the max curve.
From (2), we know the ellipse is one rotated $45^\circ$ counter-clockwise, which before the rotation (As the blue ellipse in the plot):
+ Major axis is the $y$-axis, intersecting the axis at $(0, \sqrt{2}), (0, -\sqrt{2})$
+ minor axis is the $x$-axis, intersecting the axis at $(\sqrt{\frac{2}{3}}, 0), (-\sqrt{\frac{2}{3}}, 0)$

However, since the point $(1, 0)$ must be transformed onto the point with max curve, which is on the major curve; So we use the following roration matrix to rotate the point $(1, 0)$ the the $y$-axis(a.k.a., rotate point $B$ to point $A$):
$$
\bmat{
\cos{\pi \over 2} & -\sin{\pi \over 2} \\
\sin{\pi \over 2} & \color{white}{-}\cos{\pi \over 2}
} = \bmat{
0 & -1 \\
1 & 0
}
$$

Then, to strech the unit circle into the unrotated ellipse, the linear transformation matrix is:
$$
\bmat{
\sqrt{\frac{2}{3}} & 0 \\ 0 & \sqrt{2}
}
$$

Lastly, rotate $\pi \over 4$ more to represent the ellipse, use one more rotation matrix:
$$
\bmat{
\cos{\pi \over 4} & -\sin{\pi \over 4} \\
\sin{\pi \over 4} & \color{white}{-}\cos{\pi \over 4}
} = \bmat{
\sqrt{1 \over 2} & -\sqrt{1 \over 2} \\
\sqrt{1 \over 2} & \sqrt{1 \over 2}
}
$$

Resulting in the linear transformation matrix:
$$
T = \bmat{
\cos{\pi \over 4} & -\sin{\pi \over 4} \\
\sin{\pi \over 4} & \color{white}{-}\cos{\pi \over 4}
}\bmat{
\sqrt{\frac{2}{3}} & 0 \\ 0 & \sqrt{2}
}\bmat{
\cos{\pi \over 2} & -\sin{\pi \over 2} \\
\sin{\pi \over 2} & \color{white}{-}\cos{\pi \over 2}
} = \bmat{
-1 & -\sqrt{1 \over 3} \\
1 & -\sqrt{1 \over 3}
}
$$
Where the point $(1, 0)$ is transformed to $(-1, 1)$.

### (4) Calculate the determinant from result of (3).
$$
\det (T) = \vmat{
-1 & -\sqrt{1 \over 3} \\
1 & -\sqrt{1 \over 3}
} = \sqrt{1 \over 3} - \left( -\sqrt{1 \over 3} \right) = {2 \over 3}\sqrt{3}
$$

### (5) Find the area of $\Omega$.
Within the whole ellipse, the parts within the I, II quadrant is not encluded.

To find the area of it, we instead find the area representing it within the original unit circle.

The border of the segments of the unit circle which got transformed into the I quadrant is calculated as:
$$
T\bmat{
a & b \\
c & d
} = \bmat{
1 & 0 \\
0 & 1
}
$$
Where $\trps{\bmat{a & c}}$ and $\trps{\bmat{b & d}}$ is the vector of the border of the sector which will be transformed into the first quadrant.
$$
\bmat{
-1 & -\sqrt{1 \over 3} \\
1 & -\sqrt{1 \over 3}
}\bmat{
a & b \\
c & d
} = \bmat{
1 & 0 \\
0 & 1
} \Rightarrow
\bmat{
a & b \\
c & d
} = \bmat{
-{1 \over 2} & {1 \over 2} \\
-{\sqrt{3} \over 2} &-{\sqrt{3} \over 2}
} = \bmat{
\cos {4 \over 3}\pi & \cos {5 \over 3}\pi \\
\sin {4 \over 3}\pi & \sin {5 \over 3}\pi
}
$$
Which represents a sector of $1 \over 6$ of the unit circles area; the area transforms into the III quadrant is the same size.

Since the area sacle factor after the transformation is simply the determinant of the transformation matrix, the result is:
$$
1^2\pi \cdot (1 - 2\cdot {1 \over 6}) \cdot \det(T) \\
= {4 \over 6} \cdot \pi \cdot {2 \over 3}\sqrt{3} \\
= \frac{4\sqrt{3}}{9}\pi
$$