Mathematics Interview Questions (VI)

26. Prove Ptolemy's Theorem.

Solution. This theorem states that, for a cyclic quadrilateral, the product of the diagonals equals the sum of the products of the opposite sides.

Proof: Let ABCD be a cyclic quadrilateral. On the chord BC, the inscribed angles ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB. Construct K on AC such that ∠ABK = ∠CBD; since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.

Now, by common angles △ABK is similar to △DBC, and likewise △ABD is similar to △KBC. Thus AK/AB = CD/BD, and CK/BC = DA/BD; equivalently, AK⋅BD = AB⋅CD, and CK⋅BD = BC⋅DA. By adding two equalities we have AK⋅BD + CK⋅BD = AB⋅CD + BC⋅DA, and factorizing this gives (AK+CK)·BD = AB⋅CD + BC⋅DA. But AK+CK = AC, so AC⋅BD = AB⋅CD + BC⋅DA.

27. Find roots of the equation considering different values of .

Solution. By a sketch, we can see that the number of roots is infinity when . Firstly, consider when . We know that, when , , so if the line intersects with between and , there are intersections. That is when . Secondly, if , then there is only one intersection, namely . Finally, if , consider translating left by , which will then give the same situation as .

28. Integrate between 0 and for cases .

Solution. For , . When it equals 0, . The integral is therefore

where , so it equals

For , since ,

29. If , prove is a multiple of 60.

Solution. Since for any we have that , this means one of is even and is odd, or else all three are even.

Also, , so either all are multiples of 5 or else exactly one is.

Also, , so either all or exactly two are multiples of 3.

Hence, there are two cases:

Case 1: All three are even. Then clearly and thus .

Case 2: WLOG, suppose that is even while are odd. We then have , and the RHS is always a multiple of 8, so must be a multiple of 4, so again .

30. Two people are playing a game which involves taking it in turns to eat chillies. There are 5 mild chillies and 1 hot chilli. Assuming the game is over when the hot chilli is eaten (and that I don't like hot chillies), is it a disadvantage to go first? What is the probability that I will eat the chilli if I go first? How about if there are 6 mild and 2 hot?


So each player gets the hot chili with probability , and no advantage if you go first.

If there are 6 mild and 2 hot, then

So each player gets the hot chili with probability , and it will be a disadvantage if you go first.