Three connected topics for University of Tokyo mathematics practice

What this guide covers

The original article was written in January 2019, looking back at entrance papers and mock examinations available then. Its forecasts concerned the 2019 examination; they are not current predictions. The three topics remain useful examples of how a problem can combine several mathematical skills:

1. Probability, especially probability recurrences.
2. Complex-plane geometry.
3. Areas and volumes of solids obtained from cross-sections.

The worked examples below come from the mock-exam problems discussed in that article. Use them to practice setting up a model, rather than assuming that a future paper must contain these topics.

1. Probability recurrences

A probability problem that involves repeated steps often also tests your understanding of sequences. Instead of enumerating every successful history, identify what information about the latest step determines which transitions remain possible. See the probability-recurrence guide for the basic method.

Problem: adjacent products divisible by three

Roll a fair six-sided die $n$ times, where $n\ge2$, obtaining $a_1,\ldots,a_n$. Find the probability $p_n$ that $a_ka_{k+1}$ is divisible by $3$ for every $1\le k<n$.

A product is divisible by $3$ exactly when at least one factor is. Thus the forbidden event is a pair of consecutive outcomes neither of which is a multiple of $3$.

Let $q_n$ be the probability of a valid sequence ending in a multiple of $3$, and $r_n$ the probability of a valid sequence ending in a nonmultiple. A multiple occurs with probability $1/3$.

Appending one outcome gives

$$q_{n+1}=\frac13(q_n+r_n),\qquad r_{n+1}=\frac23q_n, \qquad q_1=\frac13,\quad r_1=\frac23.$$

We can solve the system by looking for linear combinations that form geometric sequences:

$$q_{n+1}+\frac12r_{n+1}=\frac23\left(q_n+\frac12r_n\right),$$ $$q_{n+1}-r_{n+1}=-\frac13(q_n-r_n).$$

The initial values therefore give

$$q_n+\frac12r_n=\left(\frac23\right)^n, \qquad q_n-r_n=\left(-\frac13\right)^n.$$

Solving,

$$q_n=\left(\frac23\right)^{n+1}-\left(-\frac13\right)^{n+1},$$ $$r_n=\left(\frac23\right)^{n+1}+2\left(-\frac13\right)^{n+1}.$$

Consequently,

$$\boxed{p_n=2\left(\frac23\right)^{n+1}+\left(-\frac13\right)^{n+1}.}$$

For instance, $p_2=5/9$: the only failure is that both outcomes are nonmultiples, with probability $4/9$. This provides a quick check on the formula.

2. Complex numbers and geometric transformations

The historical guide highlighted the way complex numbers connect rotations, similarity, and loci. The following problem does not explicitly tell you to introduce a complex plane. Recognizing that option is the main strategic step.

Problem: similar isosceles triangles

Triangles $OAB$ and $OCD$ are similar, noncongruent isosceles triangles with $OA=OB$ and $OC=OD$. Their corresponding vertices are $O,A,B$ and $O,C,D$, respectively, and each triple is ordered counterclockwise.

Choose $E$ so that $BCE$ is a right isosceles triangle with $EB=EC$, and $B,C,E$ are ordered counterclockwise. Let $M$ be the midpoint of $AD$. If $O,E,M$ are collinear, find $\angle AOB$.

Normalize the configuration

A translation, rotation, and positive scaling preserve the required angle. Put $O=0$, $A=1$, and write $B=b$, $C=c$, where

$$|b|=1,\qquad \operatorname{Im}b>0,\qquad |c|\ne1.$$

Similarity then gives $D=bc$ and $M=(1+bc)/2$. Because $|bc|\ne1$, the denominator $1+bc$ cannot vanish.

The right-angle and orientation conditions at $E$ give

$$\frac{c-e}{b-e}=i, \qquad e=\frac{c-ib}{1-i}.$$

Collinearity means that $e/(1+bc)$ is real. A complex number is real exactly when it equals its conjugate, so

$$\frac{e}{1+bc}=\frac{\bar e}{1+\bar b\bar c}.$$

Substitute the expression for $e$ and clear denominators:

$$(1+i)(c-ib)(1+\bar b\bar c) =(1-i)(\bar c+i\bar b)(1+bc).$$

Using $b\bar b=1$, expansion and cancellation reduce this to

$$(|c|^2-1)(b-i\bar b)=0.$$

The triangles are not congruent, so $|c|\ne1$. Hence $b=i\bar b$; multiplying by $b$ gives $b^2=i$. Of the two roots, only one lies above the real axis:

$$b=\frac{1+i}{\sqrt2}.$$

Its argument is $\pi/4$, so

$$\boxed{\angle AOB=45^\circ.}$$

The useful translations were: rotation becomes multiplication, a midpoint becomes an average, and collinearity becomes a real ratio. Fewer equations than unknown coordinates need not be a problem when the requested angle is invariant across a family of configurations.

3. Solid volumes from cross-sections

For a complicated solid, fixing one coordinate can turn a spatial problem into a plane region. Determine its area as a function of the fixed coordinate, then integrate. The main work is often identifying the values at which the shape of the section changes.

Problem: shortest axis-parallel paths in a cube

Let $C=[-1,1]^3$, let $D$ be its surface, and let $E$ be the closed square pyramid with apex $O=(0,0,0)$ and base the face $z=1$ of $C$.

For points $P,Q$, let $d(P,Q)$ be the shortest length of a polygonal path joining them using segments parallel to coordinate axes. For $P\in C$, define

$$m(P)=\min_{Q\in D}d(P,Q).$$

Let $F$ consist of the points $P\in C$ satisfying $m(P)\ge d(O,P)$.

1. Express $m(P)$ in coordinates for $P=(x,y,z)\in E$.
2. For $0\le t\le1$, find the area $S(t)$ of the section of $E\cap F$ by $z=t$, taking an empty section to have area zero.
3. Find the volume of $F$.

Part 1: translate the distance definition

Any axis-parallel path must cover at least the absolute change in each coordinate, and moving along the three axes in succession attains that bound. Therefore

$$d(P,Q)=|x_P-x_Q|+|y_P-y_Q|+|z_P-z_Q|.$$

To reach a specified face of the cube, the other two coordinates can remain unchanged. Thus, throughout $C$,

$$m(P)=\min\{1-|x|,1-|y|,1-|z|\}.$$

Inside $E$ we have $0\le z\le1$, $|x|\le z$, and $|y|\le z$, giving

$$\boxed{m(P)=1-z\quad(P\in E).}$$

Part 2: intersect a square and a diamond

At height $z=t$, the condition defining $F$ becomes

$$1-t\ge|x|+|y|+t.$$

Together with the pyramid constraints, the section is

$$|x|\le t,\qquad |y|\le t,\qquad |x|+|y|\le1-2t.$$

The square expands as $t$ increases, while the diamond shrinks.

For $0\le t\le1/4$, the square lies inside the diamond, since its corners satisfy $2t\le1-2t$.

For $1/4\le t\le1/3$, four corner triangles are cut from the square. Each triangle has perpendicular side length $4t-1$.

For $1/3\le t\le1/2$, the diamond lies inside the square, since $1-2t\le t$.

For $t>1/2$ there is no section. At $t=1/2$ it is a single point and still has area zero. Therefore

$$\boxed{S(t)=\begin{cases} 4t^2&0\le t\le\frac14,\\ 4t^2-2(4t-1)^2&\frac14\le t\le\frac13,\\ 2(1-2t)^2&\frac13\le t\le\frac12,\\ 0&\frac12\le t\le1. \end{cases}}$$

The formulas agree at every shared endpoint.

Part 3: integrate and restore symmetry

The volume inside this one pyramid is

$$\begin{aligned} \operatorname{Vol}(E\cap F) &=\int_0^{1/4}4t^2\,dt +\int_{1/4}^{1/3}\{4t^2-2(4t-1)^2\}\,dt\\ &\quad+\int_{1/3}^{1/2}2(1-2t)^2\,dt =\frac1{18}. \end{aligned}$$

The cube decomposes into six congruent pyramids with apex $O$, one for each face. Their common boundaries have volume zero, and $F$ is symmetric under the corresponding coordinate permutations and sign changes. Hence

$$\boxed{\operatorname{Vol}(F)=6\cdot\frac1{18}=\frac13.}$$

The factor six matters: integrating $S(t)$ finds only the portion within $E$.

What to carry into your practice

For repeated probability experiments, choose states that retain the information needed for the next step. For geometric similarity and rotation, try complex coordinates. For solids, work with cross-sections and locate every change of shape before integrating. These are ways of connecting familiar ideas, rather than a prediction of a particular future paper.