The Oral Exam
I wrote how the written entrance exam was used to keep Jewish students from studying at Moscow State University, but the real brutality happened at the oral exam. Undesirable students were given very difficult problems. Here is a sample “Jewish” problem:
Solve the following equation for real y:
Here is how my compatriots who studied algebra in Soviet high schools would have approached this problem. First, cube it and get a 9th degree equation. Then, try to use the Rational Root Theorem and find that y = 1 is a root. Factoring out y − 1 gives an 8th degree equation too messy to deal with.
The most advanced students would have checked if the polynomial in question had multiple roots by GCDing it with its derivative, but in vain.
We didn’t study any other methods. So the students given that problem would have failed it and the exam.
Unfortunately, this problem is impossible to appeal, because it has an elementary solution that any applicant could have understood. It goes like this:
Let us introduce a new variable: x = (y3 + 1)/2. Now we need to solve a system of equations:
This system has a symmetry which we can exploit. The graphs of the functions x = (y3 + 1)/2 and y = (x3 + 1)/2 are reflections of each other across the line x = y. As both functions are increasing, the solution to the system of equations should lie on the line x = y. Hence, we need to solve the cubic y = (y3 + 1)/2, one of whose roots we already know.
Now I offer you another problem without telling you the solution:
Share:Four points on a plane used to belong to four different sides of a square. Reconstruct the square by compass and straightedge.
PL:
Can the the points lie on the extensions of the sides of the square we construct?
17 May 2011, 10:40 pmIlya:
Hi Tanya,
18 May 2011, 1:56 am(at least) for some positions of the points there may exist several squares. Are you supposed to reconstruct one of them?
I. J. Kennedy:
Yes, the square is not unique given four points. Maybe we are supposed to assume the square is upright? (Sides parallel or perpendicular to axes.)
18 May 2011, 10:06 amTanya Khovanova:
As often with such problems, the geometry problem is not well-defined. Let’s assume a general case.
18 May 2011, 3:44 pmIlya:
In fact, for general position of points the problem is not very difficult. 1. one joins the points with intervals and gets two intervals that intersect (these join the points on the opposite sides). 2. One finds the centers of the intervals and constructs two intervals parallel to these two of the same length but intersecting in the middle. 3. Then one constructs the cosine of the angle (beta) between one of the intervals and the corresponding side (one expresses it as a rational function of the cotangent of the angle between the intervals (alpha) and their lengths. If the angle between the intervals is different from 90 (which is the case if the points are in general position) then there will be at most two possibilities). 4. one constructs the sides of the square for the translated intervals. 5. Finally, one makes a parallel transport to reconstruct the original square. I think this solution is straightforward and is based on standard constructions. Olympiad-trained kids can find it, but, of course, this is unsolvable for a good candidate that knows only high-school math.
19 May 2011, 1:54 amMV:
Ilya, could you elucidate the step 3 you mentioned? What cotangent relationship are you exploiting?
19 May 2011, 4:29 amIlya:
MV, Observe that if the original points belonged to the sides of a square then the translated points (the ends of the translated intervals) belong to the sides of a parallel transport of the original square. Furthermore, the center of the new square is precisely the intersection point of the new intervals. Denote it by A. Pick two of the new points B and C such that the angle alpha=BAC is less than 90. There are two possibilities: either B and C belong to a quadrant or AB and AC belong to different quadrants. The computations in the two cases are different, and I’ll explain only the first case: Denote the corner of the quadrant containing A and B by D, and the angle ACD by beta. Thus, the law of sines for the triangles ABD and ACD implies that ACsin(beta)=ADsin(45)=ABsin(270-alpha-beta)=-ABcos(alpha+beta) AB(sin(alpha)sin(beta)-cos(alpha)cos(beta)). Hence cot(beta)=(ABsin(alpha)-AC)/ABcos(alpha). The RH side can be constructed by standard procedures. Thus, we constructed cot(beta), and we can reconstruct the line DC. The rest is easy. Note that this doesn’t work if alpha=90. In this case there are infinitely many squares.
19 May 2011, 12:14 pmGS:
(Please remove my previous posts, they’re all messed up)
Here’s my (actual) construction. https://i56.tinypic.com/2q1zcys.jpg
A B C and D are the given points.
1. construct the circle with AB as a diameter, call this C1, and the circles with BC as a diameter, call this C2.
2. Select a point E on C1, and construct the line joining E and A, call this L1.
3. Construct the line perpendicular from D to L1, intersecting at point F.
4. Construct the circle with center E and radius EF, call this C3.
5. Where C3 and L2 intersect, label this point G.
6. Repeat steps 3 4 and 5 to get point K.
7. Construct the circle inscribing the triangle BGK. Call this C5.
8. C2 and C5 intersect at point B, and at another point W. W is one vertex of the square
9. On the line joining W and C, the perpendicular from D is the next vertex X. The remaining vertices Y and Z are found in the same manner.
19 May 2011, 4:46 pmPhilip Petrov:
Hi GS,
I did your construction on Geonext:
https://www.cphpvb.net/wp-content/uploads/2011/05/construct.zip
Not perfect, but a good start 🙂
21 May 2011, 5:14 pmPeter Gerdes:
The tragic side of P != NP.
What was the point of all the pretending. It’s trivial to observe that if P != NP then it’s easy to generate solutions with `easy’ solutions which are extremely hard to find. Even non-mathematicians understand that problems easy in retrospect can be quite hard prospectively.
Since presumably they refused to release useful statistical facts jury rigging these problems only seems to make the discrimination more apparent by the sucpiscious reluctance to release figures refuting the discrimination accusations after seeming to care enough to change the rules in apparent response to these discrimination worries.
I mean if a college is accused of discriminating against women they can just deny it and insist that admission records and the considerations each year are private and any statistical difference was an artifact. Or after the accusation you could respond by saying, “Here let us show you we don’t discriminate by baseing all our admissions on exams which will be graded blindly but then refusing to respond to the allegation girls get different tests than boys.”
I’m just curious about the psychology of this. I know this kind of thing happens but if your insincere efforts to stop discrimination will only increase the evidence that you discriminate why take them?
22 May 2011, 4:41 amPhilip Petrov:
@Peter – the things are not that simple. The fact that USA did a great job in the politics to make a “peaceful living” multicultural society is only because you are a young nation and you was able to afford it – there is no tradition to keep. And yes, the tradition usually holds the roots of the system from collapsing. The nation and the country are a very complicated systems and sometimes they depend on the needs of discrimination (I know – it sounds disgusting, but kill me – it’s true!). Sometimes you must choose to have a discriminated minority than a non-discriminated chaos and anarchy. Oh, yes – sometimes you must if you are “in charge”…
23 May 2011, 1:19 amPratik Poddar:
Incomplete but a different approach to the original problem:
Cube both the LHS and the RHS.
16y-8 = (y^3+1)^3
Treat both LHS and RHS as two seperate functions.
Note that both functions have equal value 8 at y=1.
Also note that derivative of LHS is 16
derivative of RHS is 3*((y^3+1)^2)*3*y^2 = 9y^2*(1+y^3)^2
9y^2*(1+y^3)^2 > 16 for y>1
So, the equation can have no roots for y>1 and one root as y=1
24 May 2011, 2:19 pmCan anyone complete this argument? Thanks
Peter Gerdes:
@Phillip
As I said I wasn’t doubting that such a situation occurred. I’ve heard enough stories about the Soviet Union from my Russian friends not even to be all that surprised that things happened the way they did but there is always a reason (perhaps silly) why they bothered to go thru an elaborate ruse that seems to do little in the way of hiding the discrimination.
I mean was this for the purpose of international relations, e.g., to make it a less in your face and thus more deniable by would be apoligists? Was this the result of tension between genuine idealism on the part of some and the desire to discriminate by others? I was just looking for the context, however, bizarre which lead to this bizarre outcome.
I mean if someone asked this about “seperate but equal” in the American South I would explain that it arose as a way to discriminate that circumvented most federal laws/efforts to curtail discrimination. I was wondering what the equivalent explanation was in this case.
28 May 2011, 12:36 amPhilip Petrov:
From the point of an American – having no real enemy and rich enough – it is easy to not discriminate anyone.
If you are phd assistant-professor in university and your salary is $416 gross salary ($291 net) the philosophy for the things in the world slightly change for you. Guess what – the cars, TV’s, hardware for your PC, clothes, food… all are not cheaper than what you drive, watch, play with, dress and eat in USA. At the same time you see some “ethnic-right”, “non discriminative”, “democratic” countries that buy your country factories, shut them down, and make you buy production from their factories.
Finally you start to be a little bit “not tolerant” to some people. You start to protect yourself.
Be happy to live on the “strong side of the planet”. And be “tolerant” as much as you can… since you can afford it.
30 May 2011, 12:34 pmThe Dude Minds… › Reconstruire le carré:
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21 September 2011, 12:26 pmTanya Khovanova’s Math Blog » Blog Archive » This is Not a Consultation:
[…] my essays The Oral Exam and A Math Exam’s Hidden Agenda, I gave some examples of math problems that were used during […]
9 October 2011, 7:11 amTanya Khovanova’s Math Blog » Blog Archive » Jewish Problems:
[…] already gave an example of problems of the kinds of problems that were given to Jewish people at the oral entrance exam to the math […]
10 October 2011, 11:41 amVictor Polinger:
Today, to an American math instructor or just a person who likes challenging problems, some of these “Jewish” problems look easy enough because they can be solved by, say, a graphing calculator. It is important to note that graphing calculators did not exist at that time or/and were not allowed to be used during exams. Without a calculator, these problems are true “coffins”. Another comment is that this anti-Semitic policy was not practiced in just Moscow State University. Actually, being the all-state policy, similar drastic measures applied to every Jewish applicant all over the USSR including some top-level institutes, like, say, MIFI, MFTI, etc. Well, there is nothing new in these facts. However, to an American, they are difficult to comprehend. As a rule, Americans have just a fuzzy knowledge of why and what happened in the former USSR… They try to find an explanation of such a stupid policy, but there is none. To my understanding, as a rule, any kind of ethnic and racial discrimination is meaningless and has no rational roots or reasons. What stands behind is just an enormous complex of inferiority and rage.
The “Jewish” problems per se represent some mathematical interest … Thank you for collecting and presenting them to the public.
Victor
11 October 2011, 9:18 pmEmery Nickols:
This is one awesome blog post.Much thanks again. Fantastic.
22 April 2012, 7:51 amTom Smith:
Yet another magical and unexpected find of the golden ratio (well, really its negative)!! Also the geometry one is neat. Great post, thanks for sharing.
12 August 2014, 10:43 pmTom Smith:
Also, a, most likely hopelessly naive approach, to it: can we just back track Euclid’s construction? I.e. take adjacent sides and create semi-circles hoping for the desired property that the other two vertices will lye on the connecting line between the midpoints of them? Sorry for possibly atrocious explanation.
12 August 2014, 10:58 pm