In the spirit of the Olympic Games going on right now, I went and found an entrance exam
from the University of Tokyo. I thought it would be interesting to see how things are
done in Japan. But to be honest, it turned out to be a frightening and stressful experience.
For some reason these questions I'm about to show you scared me more than any other
exam I've looked at recently, and they're only at the undergraduate level to gain entrance
to the university. In Japan, students applying to public universities will first take the
National Center test. And with these scores, they can then apply to particularly universities,
where they might be invited to take a department specific test. The University of Tokyo is
one of Japan's top ranking universities and this is their 2021 mathematics test. It's
notorious enough that it's inspired a couple of animes, and researchers are even working
on a robot that can pass this exam. Professor Noriko Arai said that the robot is already
scoring in the top 1%. But perhaps what this means is that our exams often focus too much
on things that can be memorised - even by a computer, and instead should be focusing
on understanding, which is where we still have an advantage over the AI. For this exam,
students have 150 minutes. And with six questions, that's 25 minutes per question. I've translated
them into English so let's have a read through them. For question one, a and b are real numbers
and we have this parabola C, which has two shared points with another parabola, y is
equal to minus x squared. The x coordinate of one shared point is between minus one and
zero. And the other shared points x coordinate is between zero and one. Part one asks to
draw the possible ranges of the points a and b on the coordinate plane. And Part two is
to draw the possible paths of the parabola C on the coordinate plane. And the solutions
for this question look a bit like this for part one, and part two. In question two, it's
all about complex numbers, we have this equation f of z is equal to a z squared plus b z plus
c. a, b and c are complex numbers, and i is the imaginary unit. For part one, let alpha,
beta and gamma be complex numbers. If f of zero is alpha, f of one is beta, and f of
i is gamma, then denote a, b and c by alpha, beta and gamma respectively. Part Two asks
if f of zero, f of one and f of i are all real numbers between one and two, graphically
represent the possible range of f of two on the complex number plane. Now I'm going to
come back in a moment and show my full solution to this one. But in the meantime, maybe have
a think about it. I have only used online tools for the translation. So that might make
it a little harder than it needs to be to understand but these still seem really difficult
to me, at least, it seems hard to form a picture in my mind of how this solution is going to
play out. Here's question three, we have a function called C, a tangent to that function
at a point A. And we're asked to show that there's only one other shared point between
C and the tangent that is different from A, and to find the coordinate of that point.
Then for part two, we're using that x coordinate and having to calculate an integral. An online
solution I found for this one shows pages of integrals. And I just thought, excuse me,
what on earth? With 25 minutes per question, they really can ask some quite involved calculations.
This fourth question was an absolute mind bender. It uses this notation here, which
I believe refers to a binomial coefficient. And I think I would need a lot more practice
with those to really understand how to do this question. Question five here is kind
of about examining the increase and decrease of a function, and in doing that, finding
derivatives. And then there's question six here, which I found the hardest to understand.
Maybe that's the translation or maybe I'm just not used to doing problems that are set
out like this. Either way, I think I can be confident that it's pretty difficult. But
let's jump back to question two, and I'll show you my solution to that one. So here's
question two, we have this equation f of z. We are told that alpha, beta and gamma are
these particular three values of f of z. And we're asked to denote a, b and c using alpha,
beta and gamma. So we can take what they've told us about f zero, and set z to zero in
the equation. And that would give us c is equal to alpha, which is the first of the
three done and definitely the easiest one. Next, we can set z to be one, and that would
be equal to beta. But we don't get any further than knowing that a plus b is equal to beta
minus alpha. So we take a look at the third piece of information we're given and set z
to i. That will give us this equation and rearranging it, we get minus a plus bi is
gamma minus alpha. So we've ended up with two simultaneous equations to solve for a
and b. We can solve them by elimination to find out what our value of b is in terms of
alpha, beta and gamma. Now that we know b, we can sub it back and find a, which has a
similar looking form. I will say the process up to here involves quite a bit of algebra
and manipulation of i and negative numbers all over the place, which leads plenty of
room to make a mistake, what you're not seeing on this page is all the times I screwed it
up myself. But this is the answer for part one, we've represented a, b and c as they
wanted us to. Then for part two, they're saying that alpha, beta and gamma are all real numbers
between one and two. And if we set z to be equal to two, they want us to graphically
represent the range of that function. So if we set z to two, we get four a plus two b
plus c, then if we sub in our previous answers for a, b, and c, we get this pretty long equation
here. But we can simplify it all the way down to this line. And what's useful to notice
is that whilst alpha, beta and gamma are just real numbers between one and two, these coefficients
in front of them are three complex numbers. So the function is the addition of three complex
numbers. And we can even call them z one, z two, and z three, we want to plot the range
of this function. And we can actually just find the vertices directly by setting alpha,
beta and gamma to either one or two. Here's how I've done it here. All the possible values
of alpha, beta and gamma, would give us a combination that falls between these points.
And I've plotted them here on the complex plane, that gives us this kind of hexagon
shape. And the range of f two, which is what they asked us for is the boundary and interior
of the shape. You can notice that when alpha, beta and gamma are all equal to one, or all
equal to two, that point falls within the shape, and so isn't actually necessary to
be able to draw the boundary. But you might ask, why does the shape look like this? Well,
we can go back and consider the function with just alpha and beta the first two terms, when
we add these terms together with alpha and beta between one and two, we get something
like this, it's actually a parallelogram of possible solutions. Here is the point where
alpha and beta are one, and here is the point where they're both two, then adding on that
third term, z three gamma, that will transform each point of the parallelogram in the direction
of z three, which is minus one plus i. And that would be a transformation out in this
angle, essentially, to the left one and up one. So here's what the parallelogram looks
like when it's transformed like that. If this is where it was originally marked by P, then
it will first move up to here, when gamma is equal to one, and its other limit will
be out here, when gamma is equal to two, and so will exist in all the possible places between
these two points. It is just the same hexagon shape that we had before. This kind of smeared
out parallelogram shows the range of our solution. I would say this exam deserves a gold medal
for difficulty. Of course, with a lot of practice, it might start to look easier if you notice
similarities between a lot of past questions. But getting into the University of Tokyo certainly
seems very competitive, and you at least need to be very on top of your mathematics. Thanks
for watching this video. And thank you to my Patreon supporters for making these videos
possible. And a special shout out to today's Patron cat of the day Doorknob2.