Irish Intervarsity Mathematics Competition 1992

Trinity College Dublin 9.30–12.30 15^{th} February

Answer all questions. Calculators permitted.

1. Solve the equation

(x − 2)(x − 3)(x + 4)(x + 5) = 44.

Answer: Let x − 2 = y − 3, ie y = x + 1. Then the equation becomes

(y − 3)(y − 4)(y + 3)(y + 4) = 44.

In other words,

(y^{2} − 9)(y^{2} − 16) = 44.

Let z = y^{2}. We have a quadratic

(z − 9)(z − 16) = 44.

This gives

z^{2} − 25z + 100 = 0.

Let z = 5u. Then

25u^{2} − 125u + 100 = 0 ⟹ u^{2} − 5u + 4 = 0
⟹ (u − 4)(u − 1) = 0
⟹ u = 1 or 4
⟹ z = 5 or 20
⟹ y = ±\sqrt{5} or ± 2\sqrt{5}
⟹ x = ±\sqrt{5} − 1 or ± 2\sqrt{5} − 1.

2. Find the greatest value of

\frac{x + 2}{2x^{2} + 3x + 6}.

Answer: To maximize

f(x) = \frac{x + 2}{2x^{2} + 3x + 6},

it is sufficient to minimize

g(x) = \frac{2x^{2} + 3x + 6}{x + 2}.

Setting x + 2 = y,

g(x) = \frac{2(y − 2)^{2} + 3(y − 2) + 6}{y}
= \frac{2y^{2} − 5y + 8}{y}
= 2y − 5 + \frac{8}{y}
= 2 (\sqrt{y} − \frac{2}{\sqrt{y}} )^{2} + 3.

Thus

g_{min} = 3,

the minimum being attained when y = 2. Hence

f_{max} = \frac{1}{3},

the minimum being attained when x = 0.

3. Writing numbers to the base 8, show that there are infinitely many numbers which are doubled by reversing their ‘digits’. Answer: Consider 2 ‘digit’ numbers x = ab, ie

x = 8a + b (0 ≤ a, b ≤ 7).

This is doubled on reversal if

8b + a = 2(8a + b) ⟹ 15a = 6b
⟹ 5a = 2b.

This has the solution a = 2, b = 5, ie

2 · 25_{8} = 52_{8}.

Now we see that any number of the form

2525 ⋯ 25_{8}

has the required property.

4. Show that for any positive real numbers a, b with a > b,

a^{n} − b^{n} > n(a − b)(ab)^{(n−1)/2}.

Answer: We have

\frac{a^{n} − b^{n}}{a − b} = a^{n−1} + ba^{n−2} + ⋯ + b^{n−1}.

Thus

\frac{a^{n} − b^{n}}{n(a − b)} = \frac{1}{n} (a^{n−1} + ba^{n−2} + ⋯ + b^{n−1}) .

The quantity on the right is the arithmetic mean of a^{n−1}, ba^{n−2}, …, b^{n−1}. Using the result that Arithmetic Mean ≥ Geometric Mean, with equality only when all terms are equal, we deduce that

\frac{a^{n} − b^{n}}{n(a − b)} > (a^{n−1} · ba^{n−2} · · · · · b^{n−1})^{\frac{1}{n}}
= (ab)^{\frac{n−1}{2}} .

5. Describe geometrically the points P, Q in an arbitrary triangle ABC that minimize

(a) AP + BP + CP

(b) AQ^{2} + BQ^{2} + CQ^{2}.

Answer: (a) The point P minimizing AP + BP + CP is the unique point P such that

AP̂B = BP̂C = CP̂A = \frac{2π}{3} (= 120^{∘}).

(To construct this point, draw a circular arc on AB such that AP̂B = 2π/3 for all points P on the arc; and a similar circular arc on BC such that BP̂C = 2π/3 for all point on the arc. Then the required point P is the 2nd point—apart from the point B—where these 2 arcs meet.)

Suppose that P minimizes AP + BP + CP. Keep CP fixed, allowing P to vary on a circle centred on C. Draw the tangent t to this circle at P.

A small variation dP of P along the tangent t must leave AP + BP unchanged, to the first order in dP. It follows that P must be the point on t minimizing AP +BP.

But by the Law of Reflection, this will occur when AP and BP make the same angle θ with t. (To see this, let B′ be the reflection of B in t. Then

AP + BP = AP + P B′

will be minimized when AP B′ is a straight line.)

But now

AP̂C = θ + \frac{π}{2} = BP̂C.

Thus at the minimizing point,

AP̂C = BP̂C.

By the same argument with B in place of C,

AP̂B = CP̂B.

Hence all 3 angles AP̂B, BP̂C, CP̂A are equal; and since they add up to 2π (360^{∘}), each must be equal to 2π/3 (120^{∘}).

(b) The point minimizing AP^{2}+BP^{2}+CP^{2} is the centroid G of ABC. For suppose Q is the minimal point. Let A′ be mid-point of BC. Then

AQ^{2} + BQ^{2} = 2A′Q^{2} + 2A′B^{2}.

So we have to minimize

2A′Q^{2} + QA^{2}.

Evidently this will be minimal when AQA′ lie in a straight line. Thus Q must lie on the median AA′. Similarly it must lie on the other 2 medians. Hence it lies at the centroid G of ABC.

6. For m a positive integer, let k(m) denote the largest integer k such that 2^{k} divides m!. Let c(m) denote the number of 1’s in the binary representation of m. Show that k(m) = m − c(m).

Answer: To compute k(m), consider how many of the numbers 1, 2, …, m are divisible by 2^{i}.

[\frac{m}{2}] numbers are divisible by 2. [\frac{m}{4}] numbers are divisible by 4. …

We deduce that

k(m) = [\frac{m}{2}] + [\frac{m}{4}] + · · · .

Note that

m = \frac{m}{2} + \frac{m}{4} + · · · .

Consider repeated division of m by 2. At the first step we obtain [\frac{m}{2}], with remainder 1 if m is odd, ie if the last bit of m is 1. At the second step we obtain [\frac{m}{4}], with remainder 1 if the second last bit of m is 1. And so on.

We conclude that the sum of the remainders is equal to c(m), the number of bits of m equal to 1. Thus

k(m) = m − c(m).

7. If x_{1}, x_{2}, …, x_{n} are positive numbers and s is their sum, prove that

(1 + x_{1})(1 + x_{2}) ⋯ (1 + x_{n}) ≤ 1 + s + \frac{s^{2}}{2!} + ⋯ + \frac{s^{n}}{n!} .

Answer: We have

e^{x} = 1 + x + \frac{x^{2}}{2!} + · · · .

Thus

(1 + x_{1}) ⋯ (1 + x_{n}) ≤ e^{x_{1}} ⋯ x^{x_{n}} = e^{s} = 1 + s + \frac{s^{2}}{2!} + · · · .

But we can go further. The left hand side only contains terms of degree ≤ n in x_{1}, …, x_{n}. Therefore we need only include terms on the right of degree ≤ n. But s^{n+1}, s^{n+2}, … only contain terms of degree > n, and can therefore be omitted. Thus

(1 + x_{1})(1 + x_{2}) ⋯ (1 + x_{n}) ≤ 1 + s + \frac{s^{2}}{2!} + ⋯ + \frac{s^{n}}{n!} .

8. A table tennis club with 20 members organises 14 singles games (2-player games) one Saturday morning in such a way that each member plays at least once. Show that there must be 6 games involving 12 different players.

Answer: Call the players 1, …, 20. Let player i play 1 + n_{i} games. Then the total number of games is

\frac{1}{2} ∑(1 + n_{i}) = 10 + \frac{1}{2} ∑ n_{i}.

But we know that there are 14 games. Hence

∑ n_{i} = 8.

Now let us go through the 20 players. For each player with n_{i} > 0, delete n_{i} of his games. This leaves at least

\frac{1}{2}(20 − 8) = 6 games.

By construction, no player plays in more than 1 of these 6 games.

9. Let a_{1}, a_{2}, … be the sequence of all positive integers with no 9’s in their decimal representation. Show that the series

\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + ⋯

converges.

Answer: Consider numbers with exactly k digits. Each such number is

≥ 10 ⋯ 0 = 10^{k−1}.

There are 10^{k} − 10^{k−1} k-digit numbers, of which 9^{k} − 9^{k−1} do not contain the digit 9. Thus the k-digit numbers not containing 9 contribute

≤ (9^{k} − 9^{k−1}) 10^{−(k−1)}
≤ 10 (\frac{9}{10})^{k}

to the sum. Hence

∑ \frac{1}{a_{i}} ≤ 10 ∑ (\frac{9}{10})^{k} = 100.

In particular the series converges.

10. In a convex quadrilateral ABCD, let E and F be the midpoints of the sides BC and DA (respectively). Show that the sum of the areas of the triangles EDA and F BC is equal to the area of the quadrilateral.

Answer: Writing |XY Z| for the area of the triangle XY Z,

|EDA| = \frac{1}{2}|DA| × p,

where p is the perpendicular distance from E to DA. But

p = \frac{1}{2}(p_{1} + p_{2}).

where p_{1}, p_{2} are the perpendicular distances from B, C to DA, respectively. Hence

|EDA| = \frac{1}{4}|DA| × p_{1} + \frac{1}{4}|DA| × p_{2}
= \frac{1}{2} (|BDA| + |CDA|) .

Similarly

|F BC| = \frac{1}{2} (|ABC| + |DBC|) .

Hence

|EDA| + |F BC| = \frac{1}{2} (|BDA| + |DBC| + |CDA| + |ABC|)
= \frac{1}{2} (|ABCD| + |ABCD|)
= |ABCD|