University of Tokyo science mathematics 2017: problems and solutions

These explanations are independently written by this site and are not official University of Tokyo solutions. The original author considered the 2017 science-track paper comparatively accessible. Difficulty and score suggestions below are personal assessments, not official admissions criteria. See the past-paper review for study suggestions.

Problem 1: a trigonometric quotient becomes a quadratic

For real $a,b$, define

$$f(\theta)=\cos3\theta+a\cos2\theta+b\cos\theta, \qquad g(\theta)=\frac{f(\theta)-f(0)}{\cos\theta-1},\quad0<\theta<\pi.$$

1. Express $f$ and $g$ as polynomials in $x=\cos\theta$.
2. Find and sketch the parameters $(a,b)$ for which $g$ has attained minimum zero on this interval.

The source screenshot places the minus one outside the denominator, which is inconsistent with the polynomial question and its solution. The denominator used here is $\cos\theta-1$.

Solution

Using $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\cos2\theta=2\cos^2\theta-1$,

$$\boxed{f=4x^3+2ax^2+(b-3)x-a.}$$

The value $f(0)$ is at $\theta=0$, corresponding to $x=1$. Therefore

$$\begin{aligned} g&=\frac{4(x^3-1)+2a(x^2-1)+(b-3)(x-1)}{x-1}\\ &=\boxed{4x^2+2(a+2)x+2a+b+1}. \end{aligned}$$

Complete the square:

$$g=4\left(x+\frac{a+2}{4}\right)^2+b+1-\frac{(a-2)^2}{4}.$$

Because $x\in(-1,1)$, an attained minimum requires the vertex to lie strictly inside the interval. Its height must be zero, giving

$$\boxed{-6<a<2,\qquad b=\frac{(a-2)^2}{4}-1.}$$

The parameter set is the indicated open arc of a parabola. The endpoints $(-6,15)$ and $(2,-1)$ are excluded.

The author rated this as a straightforward use of triple angles and quadratic optimization. Remembering the open interval prevents confusing a minimum with an unattained infimum.

Problem 2: a six-step lattice walk

A point starts at the origin of the integer lattice. Each second it independently moves one unit right, up, left, or down, each with probability $1/4$.

1. Find the probability that after six seconds it lies on $y=x$.
2. Find the probability that it is back at the origin.

A short solution using diagonal coordinates

Let $U=x+y$ and $V=y-x$. Each step changes each coordinate by either plus or minus one. The four pairs of changes $(+1,+1),(+1,-1),(-1,+1),(-1,-1)$ occur equally often among the four possible moves. Thus the two step increments are independent fair signs. Across independent steps, the entire U-increment sequence is independent of the V-increment sequence.

The condition $y=x$ means $V=0$. After six steps this requires three positive and three negative V-increments, so

$$\boxed{\Pr(y=x)=\binom63\left(\frac12\right)^6=\frac5{16}.}$$

The origin requires both $U=0$ and $V=0$. Their independence, justified by the increment pairs above rather than by symmetry alone, gives

$$\boxed{\Pr((x,y)=(0,0))=\left(\frac5{16}\right)^2=\frac{25}{256}.}$$

An endpoint-counting alternative

All $4^6=4096$ paths are equally likely. The diagonal endpoints and numbers of paths to each are:

| Endpoint | Paths to each endpoint | | --- | --- | | $(0,0)$ | 400 | | $(1,1)$ and $(-1,-1)$ | 300 each | | $(2,2)$ and $(-2,-2)$ | 120 each | | $(3,3)$ and $(-3,-3)$ | 20 each |

For example, the origin count sums $6!/(r!r!u!u!)$ over nonnegative $r+u=3$. More generally, at $(j,j)$ the right and up counts exceed their opposite counts by $j$, and summing the corresponding multinomial counts gives the table. It yields $1280/4096=5/16$ on the diagonal and $400/4096=25/256$ at the origin. The shorter diagonal-coordinate method saves time while the table provides a direct check.

Problem 3: the complex reciprocal of a line segment

For nonzero complex $z$, put $w=1/z$.

1. Let $\alpha\ne0$ and let $L$ be the perpendicular bisector of the segment from zero to $\alpha$. Show its image is a circle minus one point, and find the center and radius.
2. Let $\beta$ be the cube root of one with positive imaginary part. Find the locus of $w$ as $z$ ranges over the closed segment joining $\beta$ to $\beta^2$.

Part 1

On $L$, $|z|=|z-\alpha|$. Substituting $z=1/w$, with $w\ne0$, gives

$$1=|1-\alpha w|, \qquad\left|w-\frac1\alpha\right|=\left|\frac1\alpha\right|.$$

Thus the center is $1/\alpha$ and radius $|1/\alpha|$, with $w=0$ excluded. Conversely, any nonzero point on this circle gives a point $z=1/w$ on the bisector, so no other point is missing.

The map $1/z$ combines inversion with reflection in the real axis. It is not exactly the geometric inversion $1/\bar z$; see the inversion guide.

Part 2

We have $\beta=(-1+i\sqrt3)/2$ and $\beta^2=(-1-i\sqrt3)/2$. Their segment is exactly

$$\operatorname{Re}z=-\frac12,\qquad |z|\le1.$$

The line is the perpendicular bisector of zero and minus one. Part 1 therefore gives $|w+1|=1$, while $|w|=1/|z|\ge1$. The complete answer is

$$\boxed{|w+1|=1,\qquad |w|\ge1.}$$

Every point satisfying these conditions has a nonzero reciprocal in the original segment. The two endpoints are included.

Problem 4: conjugate powers and a gcd

Let $p=2+\sqrt5$ and, for $n\ge1$, define

$$a_n=p^n+\left(-\frac1p\right)^n.$$

Find $a_1,a_2$; for $n\ge2$, express $a_1a_n$ using $a_{n+1},a_{n-1}$; prove every $a_n$ is a positive integer; and find $\gcd(a_{n+1},a_n)$.

Initial values and recurrence

Since $-1/p=2-\sqrt5$, the two bases have sum four and product minus one. Thus

$$\boxed{a_1=4,\qquad a_2=18.}$$

For two bases $u,v$,

$$(u+v)(u^n+v^n)=u^{n+1}+v^{n+1}+uv(u^{n-1}+v^{n-1}).$$

Here $uv=-1$, so

$$\boxed{a_1a_n=a_{n+1}-a_{n-1},\qquad a_{n+1}=4a_n+a_{n-1}.}$$

The coefficient four multiplies $a_n$, not $a_{n-1}$. The two positive integer initial values and this recurrence prove positivity and integrality by two-step induction.

Alternatively, expanding $(2+\sqrt5)^n$ and $(2-\sqrt5)^n$ makes the odd radical terms cancel. Only even radical powers remain, giving an even integer. The conjugate expansion must reverse the signs of the odd radical terms.

Descend the gcd

Euclid's algorithm and the recurrence give

$$\gcd(a_{n+1},a_n)=\gcd(a_n,a_{n-1})=\cdots=\gcd(18,4)=2.$$

Hence

$$\boxed{\gcd(a_{n+1},a_n)=2\quad(n\ge1).}$$

The conjugate-power guide explains this technique and its relation to integer and fractional parts.

Problem 5: common tangents to two reflected parabolas

For real $k$, consider

$$C:x=y^2+k,\qquad D:y=x^2+k.$$

1. If $y=ax+b$ is a common tangent with $a\ne-1$, express $k,b$ in terms of $a$.
2. Determine $k$ when a common tangent has slope two. Show that there are three common tangents and find their slopes and intercepts.

Part 1: impose two repeated-root conditions

A common tangent cannot be horizontal: substituting $y=b$ into C gives a single intersection with a nonhorizontal tangent. It cannot be vertical either, because D is a polynomial graph with finite derivative. Thus write a nonvertical line with $a\ne0$.

For C, substitution gives

$$a^2x^2+(2ab-1)x+b^2+k=0.$$

The discriminant condition is $1-4ab-4a^2k=0$. For D, the corresponding equation is $x^2-ax+k-b=0$, giving $a^2-4(k-b)=0$. Substitute $b=k-a^2/4$ from the second equation into the first:

$$1+a^3-4ak(1+a)=0.$$

With $a\ne0,-1$, use $1+a^3=(a+1)(a^2-a+1)$ and divide by $4a(a+1)$. The resulting formulas are

$$\boxed{k=\frac{a^2-a+1}{4a},\qquad b=\frac{-a^3+a^2-a+1}{4a}.}$$

Keeping the sign of the cubic correct is essential to this factorization.

Part 2: include the exceptional slope

At $a=2$, the formula gives $k=3/8$. For $a\ne-1$, solving $k=(a^2-a+1)/(4a)=3/8$ gives $a=2$ or $a=1/2$. Their intercepts are $-5/8$ and $5/16$.

The excluded case $a=-1$ must be examined separately. Substitution of $y=-x+b$ into D gives $x^2+x+3/8-b=0$, whose discriminant vanishes at $b=1/8$. Substitution into C gives the same double-root condition. Thus the three pairs are

$$\boxed{(a,b)=\left(2,-\frac58\right),\quad \left(\frac12,\frac5{16}\right),\quad \left(-1,\frac18\right).}$$

All possible slopes have been covered, and no vertical line is tangent to D, proving completeness. The main lesson is to return to a parameter value excluded during division. See the common-tangent guide.

Problem 6: rotate an equilateral triangle and sweep its edge

In three-dimensional space, O is the origin and $OPQ$ is an equilateral triangle of side one. Put $A=(1,0,0)$ and $\theta=\angle AOP\in[0^\circ,180^\circ]$.

1. When $Q=(0,0,1)$, find the ranges of the x-coordinate of P and of $\theta$.
2. Let Q move in the plane $x=0$. Find the volume of the region K swept out by the segment OP.

Part 1: the endpoint circle

The constraints $OP=PQ=OQ=1$ imply $P_z=1/2$ and $P_x^2+P_y^2=3/4$. Thus P runs over a circle of radius $\sqrt3/2$ at height $1/2$.

Since $OP=1$, $\cos\theta=P_x$. Therefore

$$\boxed{-\frac{\sqrt3}{2}\le P_x\le\frac{\sqrt3}{2},\qquad 30^\circ\le\theta\le150^\circ.}$$

Every intermediate value is attained by moving continuously around the circle.

Part 2: prove the full set of possible directions

Q must remain at unit distance from O, so it moves on the unit circle in the yz-plane. For a unit endpoint direction $u=(u_x,u_y,u_z)$, the equilateral condition is $u\cdot Q=1/2$.

As Q moves on that circle, $u\cdot Q$ takes all values from $-\sqrt{u_y^2+u_z^2}$ to $+\sqrt{u_y^2+u_z^2}$. Thus such a Q exists exactly when

$$\sqrt{1-u_x^2}\ge\frac12, \qquad |u_x|\le\frac{\sqrt3}{2}.$$

The segment OP then supplies every radius between zero and one along each admissible direction. Consequently K is the part of the closed unit ball outside the two open cones about the positive and negative x-axes with half-angle $30^\circ$.

At a fixed $x$, put $\rho=\sqrt{y^2+z^2}$. The ball requires $\rho^2\le1-x^2$, and the allowed direction condition requires $\rho^2\ge x^2/3$. Hence the section is the full annulus

$$\frac{x^2}{3}\le\rho^2\le1-x^2, \qquad |x|\le\frac{\sqrt3}{2}.$$

Its area is $\pi(1-4x^2/3)$. Integrating,

$$\begin{aligned} \operatorname{Vol}(K) &=\pi\int_{-\sqrt3/2}^{\sqrt3/2}\left(1-\frac43x^2\right)\,dx\\ &=\boxed{\frac{2\sqrt3}{3}\pi.} \end{aligned}$$

Equivalently, subtract from the corresponding central slab of the ball two cones of radius $1/2$ and height $\sqrt3/2$. It is the moving segment, not only the endpoint P, that fills the volume.

Reviewing this paper

The original author recommended ambitious historical practice targets—110–120 for the strongest medical-track candidates and 80–90 for other science-track candidates—because the set was judged comparatively accessible. These are personal strategy opinions, not official cutoffs or a guarantee of admission.

The time-saving ideas are worth retaining: reduce trigonometry to a quadratic, use independent diagonal increments, track a reciprocal map's missing point, exploit conjugates, restore an excluded tangent slope, and prove every attainable direction before integrating a swept volume.