University of Tokyo science mathematics 2019: problems and solutions

The following are this site's independent solutions to the 2019 science-track paper, not official University of Tokyo answers. The original article also includes a related mock-exam problem; it is translated and solved below. Difficulty and timing judgments are the author's retrospective opinions. See the past-paper review.

Problem 1: a definite integral

Evaluate

$$I=\int_0^1\left(x^2+\frac{x}{\sqrt{1+x^2}}\right) \left(1+\frac{x}{(1+x^2)\sqrt{1+x^2}}\right)\,dx.$$

Solution by expansion

Expanding gives four manageable integrals:

$$I=\int_0^1\left\{x^2+\frac{x}{\sqrt{1+x^2}} +\frac{x^3}{(1+x^2)^{3/2}}+\frac{x^2}{(1+x^2)^2}\right\}\,dx.$$

The first three are

$$\frac13,\qquad \sqrt2-1,\qquad \left[\sqrt{1+x^2}+\frac1{\sqrt{1+x^2}}\right]_0^1 =\frac{3\sqrt2}{2}-2.$$

For the last, substitute $x=\tan\theta$:

$$\int_0^1\frac{x^2}{(1+x^2)^2}\,dx =\int_0^{\pi/4}\sin^2\theta\,d\theta=\frac\pi8-\frac14.$$

Adding,

$$\boxed{I=-\frac{35}{12}+\frac{5\sqrt2}{2}+\frac\pi8.}$$

An integration-by-parts alternative

Put $F=x+1/\sqrt{1+x^2}$. Since $F'=1-x/(1+x^2)^{3/2}$, the original integrand is $-xFF'+2xF$. Hence

$$I=\left[-\frac x2F^2\right]_0^1+\frac12\int_0^1F^2\,dx+2\int_0^1xF\,dx.$$

Expanding $F^2=x^2+2x/\sqrt{1+x^2}+1/(1+x^2)$ gives the same answer. The source accidentally squares this expanded sum again in one intermediate line; it should appear only to the first power. The author regarded the problem as accessible with direct integration; a clever rearrangement is optional.

Problem 2: two equal triangle areas inside a square

Let ABCD be a unit square. Points P, Q, R lie on AB, AD, CD, respectively, and both triangles APQ and PQR have area $1/3$. Find the maximum and minimum of $DR/AQ$.

Solution

Set $AQ=x$, $DR=y$. The first area condition gives $AP=2/(3x)$. The side constraints therefore require

$$\frac23\le x\le1,\qquad0\le y\le1.$$

Choose coordinates $A=(0,0)$, $B=(0,1)$, $C=(1,1)$, $D=(1,0)$. Then $P=(0,2/(3x))$, $Q=(x,0)$, and $R=(1,y)$. The determinant for PQR gives

$$\frac12\left\{xy+\frac2{3x}(1-x)\right\}=\frac13.$$

Thus

$$y=\frac{4x-2}{3x^2},\qquad \frac yx=f(x)=\frac4{3x^2}-\frac2{3x^3}.$$

For every $x\in[2/3,1]$, the resulting $y$ is in $[1/2,2/3]$, so no additional side restriction removes part of this interval.

Since $f'(x)=(6-8x)/(3x^4)$, the function increases up to $x=3/4$ and decreases afterward. Its relevant values are

$$f(2/3)=\frac34,\qquad f(3/4)=\frac{64}{81},\qquad f(1)=\frac23.$$

Therefore

$$\boxed{\max\frac{DR}{AQ}=\frac{64}{81},\qquad \min\frac{DR}{AQ}=\frac23.}$$

The equality constraints reduce the geometry to one variable; checking that every parameter value gives actual points on the sides makes the extrema attainable.

Problem 3: an octagonal section of an octahedron

In space, put

$$A=(2,0,0),\ B=(0,2,0),\ C=(-2,0,0),\ D=(0,-2,0),\ E=(0,0,-2),$$

and $P=(p,0,2)$, where $2<p<4$. Let M and N be the midpoints of AB and AD. Let $\alpha$ be the plane through M,N parallel to AE.

1. Describe the sections of octahedron PABCDE and plane $\alpha$ by $y=0$, drawing them in the same plane.
2. Find the values of p for which the section by $\alpha$ is an octagon.
3. Under that condition, take the part of the section with $y\ge0,z\ge0$. Find the area of its orthogonal projection onto the yz-plane.

Part 1: work in an axial plane

The octahedron's section by $y=0$ is quadrilateral PCEA in the xz-plane, with vertices

$$(p,2),\quad(-2,0),\quad(0,-2),\quad(2,0).$$

The plane $\alpha$ is $z=x-1$: it contains $M=(1,1,0)$ and $N=(1,-1,0)$, and its direction along AE has equal changes in x and z. Its trace on $y=0$ is therefore the line through $(1,0)$ and $(0,-1)$.

These coordinates specify the three sketches in the source. If $2<p<3$, P lies above that line; at $p=3$, P lies on it; if $3<p<4$, P lies below it. In every sketch, the line cuts the lower edge CE at $(x,z)=(-1/2,-3/2)$. Its other intersection lies on PA in the first case, at P in the second, and on PC in the third.

Part 2: count crossed edges

Below or at $z=0$, the plane cuts the five edges AB, AD, BE, CE, and DE. Above that plane, it cuts PB, PC, and PD precisely when P lies below $z=x-1$, namely $2<p-1$. This supplies eight distinct section vertices exactly when

$$\boxed{3<p<4.}$$

At the transition $p=3$, intersections merge at P, so it is not an octagon. One can verify the count by evaluating $z-x+1$ at the six vertices and checking which edges connect opposite signs.

Part 3: only three upper-quadrant vertices are needed

Let $F=PB\cap\alpha$ and $G=PC\cap\alpha$. In the region $y,z\ge0$, the section's relevant vertices are M,F,G, and its intersection with $y=z=0$ projects to the origin.

On PB, write $P$-weight $\lambda$ from B. Then $(x,y,z)=(p\lambda,2(1-\lambda),2\lambda)$. The plane condition gives $\lambda=1/(p-2)$, so

$$(y_F,z_F)=\left(\frac{2p-6}{p-2},\frac2{p-2}\right).$$

On PC, write $(x,z)=(-2+(p+2)\mu,2\mu)$. The plane condition gives $p\mu=3$, hence

$$(y_G,z_G)=\left(0,\frac6p\right),\qquad(y_M,z_M)=(1,0).$$

Split the quadrilateral into two triangles based at the origin. Its area is

$$\frac12\frac2{p-2}+\frac12\frac{2p-6}{p-2}\frac6p =\boxed{\frac{7p-18}{p(p-2)}}.$$

Identifying the requested projection avoids calculating every vertex of the octagon.

Problem 4: a product that is never a square

For integers $n\ge1$,

1. find $d_n=\gcd(n^2+1,5n^2+9)$;
2. prove that $(n^2+1)(5n^2+9)$ is never an integer square.

Part 1

Since $5n^2+9=5(n^2+1)+4$,

$$d_n=\gcd(n^2+1,4)= \boxed{\begin{cases}1&n\text{ even},\\2&n\text{ odd}.\end{cases}}$$

An even n has $n^2\equiv0\pmod4$, while an odd n has $n^2\equiv1\pmod4$.

Part 2

After division by the gcd, the two factors are coprime. If their product were a square, each reduced factor would be a square, since every prime exponent in their product is even and no prime is shared.

For even n, $d_n=1$, so $n^2+1$ would be a square. But

$$n^2<n^2+1<(n+1)^2\quad(n\ge1),$$

which is impossible.

For odd n, we would have

$$n^2+1=2u^2,\qquad5n^2+9=2v^2$$

for coprime positive integers u,v. Every odd square is one modulo eight, so the second equation would give $2v^2\equiv6\pmod8$. Squares modulo eight are $0,1,4$, and twice a square is therefore $0$ or $2$, never six. Again this is impossible.

Alternatively, the first equation forces u odd; the second would force v odd as well. Adding and dividing by two yields $3n^2+5=u^2+v^2$, whose sides are zero and two modulo four, respectively. Either contradiction proves the result. See integer-problem strategies.

Problem 5: unique roots and three related limits

For $n\ge1$, consider

$$x^{2n-1}=\cos x.$$

1. Prove there is exactly one real solution $a_n$.
2. Prove $\cos a_n>\cos1$.
3. Find

$$a=\lim a_n,\qquad b=\lim a_n^n,\qquad c=\lim\frac{a_n^n-b}{a_n-a}.$$

Parts 1 and 2: locate the root before differentiating

For $|x|>1$, $|x^{2n-1}|>1\ge|\cos x|$, so there is no root. For $-1\le x<0$, the odd power is negative and cosine is positive, again excluding roots.

On $[0,1]$, $F_n(x)=x^{2n-1}-\cos x$ has $F_n(0)=-1$ and $F_n(1)=1-\cos1>0$. It strictly increases: its derivative $(2n-1)x^{2n-2}+\sin x$ is positive for $0<x\le1$. Therefore it has exactly one root, with

$$0<a_n<1.$$

Since cosine strictly decreases there, $\cos a_n>\cos1$.

Part 3: use the equation defining the root

From $a_n^{2n-1}=\cos a_n>\cos1$,

$$(\cos1)^{1/(2n-1)}<a_n<1.$$

The lower bound tends to one, so $a=1$. Also

$$a_n^n=(\cos a_n)^{n/(2n-1)}\longrightarrow\sqrt{\cos1},$$

giving $b=\sqrt{\cos1}$.

For the last limit, the exact identity

$$a_n^n=\sqrt{a_n\cos a_n}$$

follows from the defining equation and positivity. Put $H(x)=\sqrt{x\cos x}$ in a neighborhood of one. Then the requested quotient is a derivative along $a_n\to1$:

$$\boxed{c=H'(1)=\frac{\cos1-\sin1}{2\sqrt{\cos1}}.}$$

Equivalently, split the derivative into that of $\sqrt{\cos x}$ and that of $\sqrt x$. Merely raising a loose bound to a growing power would not justify this limit.

Related mock-exam example included in the source

For $n\ge1$, let $F_n(x)=x^{2n+1}-x$, and let its tangent at $(-1,0)$ be $\ell$. Show there is exactly one other common point, of x-coordinate $t_n$, with $t_n<1+\sqrt{2/n}$. If $S_n$ is the enclosed area, find $\lim S_n/n$.

The tangent is $y=2nx+2n$. Intersections satisfy

$$G_n(x)=x^{2n+1}-(2n+1)x-2n=0.$$

Its derivative is $(2n+1)(x^{2n}-1)$: positive outside $[-1,1]$ and negative inside. We have $G_n(-1)=0$ and $G_n(1)=-4n<0$. Thus minus one is the tangential root, and there is exactly one other root $t_n>1$.

For $d=\sqrt{2/n}$, the first three terms of the binomial expansion give

$$G_n(1+d)\ge\binom{2n+1}{2}d^2-4n=2>0.$$

Hence $1<t_n<1+\sqrt{2/n}$, and $t_n\to1$.

The curve lies below the tangent on $(-1,t_n)$, so

$$S_n=\frac{1-t_n^{2n+2}}{2n+2} +\frac{2n+1}{2}(t_n^2-1)+2n(t_n+1).$$

Do not replace a power of a quantity tending to one by one. In particular, $(1+\sqrt{2/n})^{2n+2}$ diverges; after division by $n^2$ it still diverges, so it does not give the lower squeeze bound claimed by a tempting incorrect argument.

Instead use the exact root equation

$$t_n^{2n+1}=(2n+1)t_n+2n, \qquad t_n^{2n+2}=(2n+1)t_n^2+2nt_n.$$

Substituting into $S_n/n$, the first term tends to zero because its numerator is $O(n)$ and denominator $n(2n+2)$; the second tends to zero because $t_n^2-1\to0$; the last tends to four. Therefore

$$\boxed{\lim_{n\to\infty}\frac{S_n}{n}=4.}$$

The coefficient $(2n+1)$ must remain in the root equation; omitting its one would change the exact substitution.

Problem 6: a quartic with a purely imaginary paired sum

Distinct complex numbers $\alpha,\beta,\gamma,\delta$ are the four roots of

$$z^4-2z^3-2az+b=0,$$

where a,b are real. Suppose $\alpha\beta+\gamma\delta$ has real part zero and nonzero imaginary part.

1. Show that exactly two roots are real and the other two form a nonreal conjugate pair.
2. Express b in terms of a and determine the allowed a-values.
3. Find the locus of $\alpha+\beta$.

Part 1: use conjugate roots

Conjugating a polynomial equation with real coefficients shows that a nonreal root brings its conjugate. Thus the possibilities are four real roots, two real roots plus a conjugate pair, or two nonreal conjugate pairs.

Four real roots would make every paired product sum real. For two conjugate pairs $u,\bar u,v,\bar v$, all three possible pairings give real sums:

$$u\bar u+v\bar v,\qquad uv+\bar u\bar v, \qquad u\bar v+\bar uv.$$

These also violate the nonzero-imaginary-part condition. Therefore only the middle possibility remains.

Part 2: compare coefficients in two cases

Write the real roots as s,t and the nonreal pair as $u=x+iy$, $\bar u=x-iy$, with $y\ne0$. The pairing $st+u\bar u$ is real, so the required pairing is a cross pairing, say $su+t\bar u$. Its real part is $(s+t)x$ and its imaginary part is $(s-t)y\ne0$ because the roots are distinct. Hence

$$(s+t)x=0.$$

If $s+t=0$, factor the quartic as

$$(z^2-s^2)(z^2-2xz+x^2+y^2).$$

Coefficient comparison yields $x=1$, $s^2=-a$, $y^2=-a-1$, and $b=-a^2$. A genuine nonreal pair requires $a<-1$. The four roots are $\pm\sqrt{-a}$ and $1\pm i\sqrt{-a-1}$, which are distinct and satisfy the condition under a cross pairing.

If $x=0$, factor instead as

$$(z-s)(z-t)(z^2+y^2).$$

Comparison gives $s+t=2$, $st=-y^2$, $y^2=a$, and $b=-a^2$. Now $a>0$, with real roots $1\pm\sqrt{1+a}$ and nonreal roots $\pm i\sqrt a$. They are distinct and also admit the required cross pairing.

Therefore necessity and sufficiency give

$$\boxed{b=-a^2,\qquad a<-1\text{ or }a>0.}$$

A direct factorization checks both cases:

$$z^4-2z^3-2az-a^2=(z^2+a)(z^2-2z-a).$$

Part 3: retain the excluded points

The permitted pairing makes $\alpha+\beta$ a sum of one real and one nonreal root. For $a>0$, its possible values are

$$1\pm\sqrt{1+a}\ \pm i\sqrt a,$$

with independent signs. For $a<-1$, substituting $q=-a-1>0$ gives precisely the same family with a replaced by q.

Writing $\alpha+\beta=X+iY$, we obtain

$$\boxed{(X-1)^2-Y^2=1,\qquad Y\ne0.}$$

Conversely, every point of this hyperbola with $Y\ne0$ is realized by choosing $a=Y^2>0$ and the appropriate roots and pairing. The points $(0,0)$ and $(2,0)$ are excluded.

Reviewing this paper

The original author's historical practice targets were about 90 points for the strongest medical-track candidates and about 70 for other science-track candidates. They are strategy suggestions rather than official admissions thresholds. The latter questions reward careful use of a defining equation and explicit case analysis; partial progress can be valuable when a full complex-root classification would take too long.