University of Tokyo science mathematics 2018: problems and solutions

These solutions are this site's independent explanations, not official University of Tokyo answers. The original author judged the first four problems comparatively accessible, with more geometric work in the last two. Difficulty ratings and score suggestions reflect that historical assessment. See the past-paper review.

Problem 1: a trigonometric function's variation and limits

For $0<x<\pi$, study the increase and decrease of

$$f(x)=\frac x{\sin x}+\cos x,$$

and find its limits as $x\to0+$ and $x\to\pi-$.

Solution

Differentiate and factor:

$$f'(x)=\frac{\cos x(\sin x\cos x-x)}{\sin^2x}.$$

Put $G(x)=\sin x\cos x-x$. Then $G(0)=0$ and $G'(x)=\cos2x-1<0$ on $(0,\pi)$, so $G(x)<0$. Thus $f'$ is negative before $\pi/2$, zero there, and positive afterward. The minimum is $f(\pi/2)=\pi/2$.

The sine limit gives

$$\boxed{\lim_{x\to0+}f(x)=2.}$$

For the other endpoint, put $t=\pi-x\to0+$:

$$f(\pi-t)=\frac{\pi-t}{\sin t}-\cos t =\frac\pi{\sin t}-\frac t{\sin t}-\cos t\longrightarrow+\infty.$$

Hence the second limit is $+\infty$. The divergent numerator term is $\pi$, which must not be replaced by $t$. The useful calculus habit is to differentiate a sign-controlling factor when its sign is not immediately clear; see standard limits.

Problem 2: binomial coefficients divided by factorials

Define

$$a_n=\frac1{n!}\binom{2n+1}{n}\quad(n\ge1).$$

1. For $n\ge2$, write $a_n/a_{n-1}=q_n/p_n$ in lowest terms, with positive integer denominator $p_n$, and find $p_n,q_n$.
2. Find all $n\ge1$ for which $a_n$ is an integer.

Part 1

Factorial cancellation gives

$$\frac{a_n}{a_{n-1}}=\frac{2(2n+1)}{n(n+1)}.$$

Euclid's algorithm gives $\gcd(2n+1,n)=\gcd(2n+1,n+1)=1$. Since $n(n+1)$ is even and $2n+1$ is odd, the reduced form is

$$\boxed{p_n=\frac{n(n+1)}2,\qquad q_n=2n+1.}$$

Part 2

We have $a_1=3$ and $a_2=5$. For $n\ge3$,

$$a_n=a_1\prod_{j=2}^{n}\frac{q_j}{p_j}.$$

Every factor in the product numerator is odd, while the denominator includes $p_3=6$. No factor of two in that denominator can cancel with the odd numerator. Thus the result cannot be an integer. Therefore

$$\boxed{n=1,2.}$$

A nonintegral last ratio alone would not prove a nonintegral product; the uncancelled factor of two is the decisive argument.

Problem 3: the area swept by two moving points

Let C be the part of $y=x^2$ with $-1\le x\le1$. Let $O=(0,0)$, $A=(1,0)$, and $k>0$. As P moves on C and Q moves on segment OA, define R by

$$\overrightarrow{OR}=\frac1k\overrightarrow{OP}+k\overrightarrow{OQ}.$$

Find the area $S(k)$ of R's region, and its limits as $k\to0+$ and $k\to\infty$.

Fix the height

Write $P=(t,t^2)$, $-1\le t\le1$, and $Q=(u,0)$, $0\le u\le1$. Then

$$R=\left(ku+\frac tk,\frac{t^2}{k}\right).$$

Fix $y=p\in[0,1/k]$. The possible $t$ are $\pm\sqrt{kp}$, so the possible x-values are the union of the intervals

$$\left[-\sqrt{p/k},\ k-\sqrt{p/k}\right] \quad\text{and}\quad \left[\sqrt{p/k},\ k+\sqrt{p/k}\right].$$

They overlap exactly when $p\le k^3/4$. Thus their union has length

$$L_k(p)=\begin{cases}k+2\sqrt{p/k}&0\le p\le k^3/4,\\2k&p\ge k^3/4,\end{cases}$$

restricted in both cases to $p\le1/k$. Every point of each interval occurs because $u$ ranges over a full interval; hence this is the actual cross-section, not just an outer bound.

Integrate the union length

The switching height lies within the vertical range exactly when $k\le\sqrt2$. Therefore

$$\begin{aligned} S(k)&=\int_0^{k^3/4}(k+2\sqrt{p/k})\,dp+\int_{k^3/4}^{1/k}2k\,dp\\ &=2-\frac{k^4}{12}\qquad(0<k\le\sqrt2). \end{aligned}$$

For $k\ge\sqrt2$, the intervals overlap at every height, so

$$S(k)=\int_0^{1/k}(k+2\sqrt{p/k})\,dp =1+\frac4{3k^2}.$$

Consequently,

$$\boxed{S(k)=\begin{cases}2-k^4/12&0<k\le\sqrt2,\\1+4/(3k^2)&k\ge\sqrt2. \end{cases}}$$

The plus sign in the second expression is essential; the source's minus sign would make the formulas disagree at $k=\sqrt2$. Both correct expressions give $5/3$ there.

The limits are

$$\boxed{\lim_{k\to0+}S(k)=2,\qquad\lim_{k\to\infty}S(k)=1.}$$

Geometrically, when $k$ is small, most horizontal sections are two separate intervals of length $k$, spread over height $1/k$; their overlap correction tends to zero, giving area two. When $k$ is large, the sections overlap throughout and have base length $k$, height $1/k$, and a vanishing width correction, giving area one. These are checks on the calculation rather than substitutes for its interval analysis. The swept-region guide develops the same coordinate-fixing method.

Problem 4: three cubic roots with the middle root above one

Let $a>0$ and $f(x)=x^3-3a^2x$. Find and sketch all $(a,b)$ such that $f(x)=b$ has three distinct real roots $\alpha<\beta<\gamma$, with $\beta>1$.

Solution

Since $f'(x)=3(x-a)(x+a)$, the local maximum is $f(-a)=2a^3$ and the local minimum is $f(a)=-2a^3$. Three distinct roots exist exactly when

$$-2a^3<b<2a^3.$$

The middle root lies on the strictly decreasing branch $(-a,a)$. It is greater than one exactly when $a>1$ and $f(1)>b$, or $b<1-3a^2$.

For $a>1$, the latter upper bound is already below $2a^3$. Also $1-3a^2>-2a^3$, because their difference is $(a-1)^2(2a+1)>0$. Hence

$$\boxed{a>1,\qquad -2a^3<b<1-3a^2.}$$

All boundaries are excluded.

Separating the horizontal level $b$ is the key simplification. See root placement and parameter separation.

Problem 5: reflect a point in a circle tangent

Let C be the unit circle centered at zero in the complex plane. Take $P=z\in C$, with $z\ne1$, and reflect $A=1$ in the tangent to C at P to obtain $Q=u$. Put $w=1/(1-u)$.

1. Express $u$ and $\bar w/w$ as polynomials in $z$, and calculate $|w+\bar w-1|/|w|$.
2. Find the locus of $w$ when $|z|=1$ and $\operatorname{Re}z\le1/2$.

Part 1: translate reflection into two conditions

The vector from A to Q is normal to the tangent, hence parallel to OP. Thus $u-1=\lambda z$ for a nonzero real $\lambda$. Also P is equidistant from A and Q:

$$|z-1|=|z-u|=|(1-\lambda)z-1|.$$

Squaring and using $z\bar z=1$ gives

$$\lambda^2-2\lambda+\lambda(z+\bar z)=0.$$

Therefore $\lambda=2-z-\bar z$, and

$$\boxed{u=-z^2+2z.}$$

This is the actual reflection: the tangent's normal direction gives the perpendicular relation, and equal distances place its midpoint on the tangent. Since $z\ne1$, the reflection is nontrivial.

Now $1-u=(z-1)^2$, so $w=1/(z-1)^2\ne0$. As $\bar z=1/z$,

$$\boxed{\frac{\bar w}{w}=\frac{(z-1)^2}{(\bar z-1)^2}=z^2.}$$

Finally,

$$\frac{|w+\bar w-1|}{|w|} =\left|1+\frac{\bar w}{w}-\frac1w\right| =|1+z^2-(z-1)^2|=\boxed{2}.$$

Part 2: obtain the parabola and its exact interval

Write $w=x+iy$. The last identity becomes

$$|2x-1|=2\sqrt{x^2+y^2}, \qquad x=\frac14-y^2.$$

To determine which part is attained, write $z=e^{i\theta}$ with $\pi/3\le\theta\le5\pi/3$. Using $z-1=2i\sin(\theta/2)e^{i\theta/2}$,

$$w=-\frac{e^{-i\theta}}{4\sin^2(\theta/2)} =\frac14-\frac14\cot^2\frac\theta2 +\frac i2\cot\frac\theta2.$$

As $\theta/2$ increases from $\pi/6$ to $5\pi/6$, its cotangent decreases continuously from $\sqrt3$ to $-\sqrt3$. Therefore every value of $y$ in $[-\sqrt3/2,\sqrt3/2]$ occurs, giving

$$\boxed{x=\frac14-y^2,\qquad-\frac{\sqrt3}{2}\le y\le\frac{\sqrt3}{2}.}$$

The equality of distances also identifies a parabola with focus zero and directrix $x=1/2$, but the parameter restriction is still needed to select this finite arc.

Problem 6: a union of three rounded tubes

Let

$$O=(0,0,0),\quad A=(1,0,0),\quad B=(1,1,0),\quad C=(1,1,1), \qquad\frac12<r<1.$$

For a moving point on OA, AB, or BC, take the closed ball of radius $r$ centered at that point. The three swept solids are $V_1,V_2,V_3$, respectively.

1. Find the heights $t$ for which the plane $y=t$ meets both $V_1$ and $V_3$, and describe their sections in that plane.
2. Find the condition on $r$ for $V_1\cap V_3\subseteq V_2$.
3. Under that condition, let $S=\operatorname{Vol}(V_1)$ and $T=\operatorname{Vol}(V_1\cap V_2)$. Express the volume of $V_1\cup V_2\cup V_3$ using S and T.
4. Calculate S, T, and the union volume.

Part 1: two capsule sections

The planes meeting $V_1$ have $-r\le t\le r$, while those meeting $V_3$ have $1-r\le t\le1+r$. Their common range is

$$\boxed{1-r\le t\le r.}$$

Put

$$a_t=\sqrt{r^2-t^2},\qquad b_t=\sqrt{r^2-(1-t)^2}.$$

In the xz-plane at that height, the section of $V_1$ is the closed capsule of radius $a_t$ about the horizontal segment from $(0,0)$ to $(1,0)$. The section of $V_3$ is the closed capsule of radius $b_t$ about the vertical segment from $(1,0)$ to $(1,1)$.

For $t\le1/2$, $a_t\ge b_t$; for $t\ge1/2$ the inequality reverses. The illustrations show the corresponding two shapes.

Part 2: include the intersection in the middle tube

Throughout the relevant height range, $0<t<1$, so the section of $V_2$ is the disk $(x-1)^2+z^2\le r^2$.

The farthest points of the two-capsule intersection from $(1,0)$ include $(1-b_t,a_t)$, at squared distance $a_t^2+b_t^2$. This is also the maximum. Indeed, in the upper-left quadrant relative to $(1,0)$, the capsule constraints give $1-x\le b_t$ and $z\le a_t$; in the remaining quadrants one of the radius-$a_t$ or radius-$b_t$ end disks bounds the distance by a smaller value. The corner lies in both rectangular portions because $a_t,b_t<1$.

Thus inclusion of the whole section is equivalent to

$$a_t^2+b_t^2\le r^2 \quad\Longleftrightarrow\quad r^2\le t^2+(1-t)^2 =2(t-1/2)^2+1/2.$$

The allowed interval contains $t=1/2$, where the right side has minimum $1/2$. Therefore

$$\boxed{\frac12<r\le\frac1{\sqrt2}.}$$

Part 3: inclusion–exclusion

The three individual volumes equal S, and the two adjacent intersections have volume T by congruence. The nonadjacent intersection $V_1\cap V_3$ equals the triple intersection under the inclusion condition. Those two terms cancel in inclusion–exclusion:

$$\boxed{\operatorname{Vol}(V_1\cup V_2\cup V_3)=3S-2T.}$$

The source has S and T reversed in one displayed answer; the correct expression agrees with its final calculation.

Part 4: calculate the two building blocks

A radius-r ball swept along a unit segment forms a cylinder of radius r and length one, plus two hemispherical caps. Hence

$$\boxed{S=\pi r^2+\frac{4\pi}{3}r^3.}$$

For the adjacent intersection $V_1\cap V_2$, cut at $z=p$, $-r\le p\le r$. Put $\rho=\sqrt{r^2-p^2}$. The intersection consists of a square of side $\rho$ in the upper-left quadrant about A and three quarter-disks of radius $\rho$ in the other quadrants. The segment lengths are one and $\rho\le r<1$, so these pieces lie within the relevant tube sections.

Consequently,

$$\begin{aligned} T&=\int_{-r}^{r}\left(1+\frac{3\pi}{4}\right)(r^2-p^2)\,dp\\ &=\boxed{\left(\pi+\frac43\right)r^3.} \end{aligned}$$

The final volume is

$$\boxed{3\pi r^2+\left(2\pi-\frac83\right)r^3, \qquad\frac12<r\le\frac1{\sqrt2}.}$$

Using unit-radius cap volumes would be incorrect: all radius powers must remain in S and T.

Reviewing this paper

The original author's historical practice targets were approximately 90–100 points for the strongest medical-track candidates and 70–80 for other science-track candidates. These are personal strategy suggestions, not official cutoffs. The main review themes are controlling derivative signs, finding uncancelled prime factors, integrating actual unions of intervals, locating a middle root on a monotone branch, proving a locus is fully attained, and applying inclusion–exclusion to solids.