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Solutions



91.
A square and a regular pentagon are inscribed in a circle. The nine vertices are all distinct and divide the circumference into nine arcs. Prove that at least one of them does not exceed 1/40 of the circumference of the circle.
Solution. Let the four points of the square be at A, B, C, D. We can partition the circumference into eight arcs as follows: (1) four closed arcs (containing endpoints), each of length 1/20 of the circumference centered at A, B, C and D; call these the shorter arcs: (2) four open arcs (not containing endpoints), each of length 1/5 of the circumference and interpolated between two of the arcs centred at the vertices of the square; call these the longer arcs.

Suppose a regular pentagon is inscribed in a circle. Since any pair of adjacent vertices terminate a closed arc whose length is 1/5 of the circumference, no two vertices of the pentagon can belong to the same longer arc. Since there are only four longer arcs, at least one vertex of the pentagon must lie inside a shorter arc and so be no more distant that 1/2 ×[1/20] = [1/40] from a vertex of the square.



92.
Consider the sequence 200125, 2000125, 20000125, , 20000125, (in which the nth number has n+1 digits equal to zero). Prove that none of these numbers is the square, cube or fifth power of an integer.
Solution. Each number has the form 2 ×10n+1 + 125 where n 4. Since both 10n and 120 are multiples of 8, each number in the sequence is congruent to 5, modulo 8, and so cannot be square.

Observe that 10n+1 = 10n-2 ×8 ×125, so that


2 ×10n+1 + 125 = 125(16 ×10n-2 + 1) = 53 (16 ×10n-2 + 1) .
If such a number is a kth power, it must be divisible by 5k. Since 16 ×10n-2 + 1 is not a multiple of 5, no number in the sequence can be a kth power for k 4.

Let un = 16 ×10n-2 + 1, for n 5. Suppose that (10xn + 1)3 = un. Then


1000xn3 + 300xn2 + 30xn = 16 ×10n-2
so that xn = 10n-3yn for some yn and we find that


16 = 103n-6yn3 + 3×102n-4yn2 + 3yn yn2 + 3yn
so that yn = 1 or 2. It is straightfoward to check that neither works, so that un can never be a cube. Hence no number in the given sequence can be a cube.



93.
For any natural number n, prove the following inequalities:


2(n-1)/(2n-2) 2 [4 ]4 [8 ]8[2n ]2n < 4 .
Solution. The middle member of the inequality is 2 raised to the power s 1/2 + 2/4 ++ [n/(2n)]. Note that


s - 1
2
s = 1
2
+ 1
4
+ + 1
2n
- n
2n+1
= 1 - 1
2n
- n
2n+1
= 1 - n+2
2n+1
whence s = 2 - (n+2)2-n. Clearly, s < 2.

On the other hand,


s - n-1
2n-2
= 2n+1 - (n+2) - 4(n-1)
2n
= 2n+1 - (5n - 2)
2n
 .
When n = 1, 2n+1 - (5n-2) = 1 > 0; when n = 2, 2n+1 - (5n-2) = 0, and when n = 3, 2n+1 -(5n - 2) = 3 > 0. Suppose, as an induction hypothesis, that 2k+1 > 5k - 2 for some k 3. Then


2k+2 > 10k - 4 = 5(k+1) - 2 + (5k - 7) > 5(k+1) - 2 ,
so that, for each positive integer n, 2n+1 5n - 2, with equality if and only if n = 2. The desired result follows.



94.
ABC is a right triangle with arms a and b and hypotenuse c = |AB |; the area of the triangle is s square units and its perimeter is 2p units. The numbers a, b and c are positive integers. Prove that s and p are also positive integers and that s is a multiple of p.
Solution. Since a2 + b2 = c2 and squares are congruent to 0 or 1, modulo 4, it is not possible for a and b to be both odd. Since, at least one of them is even, s = 1/2ab is an integer. Also, since either two or none of a, b, c are odd, p = 1/2(a + b + c) is an integer. Let r be the inradius of the triangle, so that s = rp. It remains to find r and show that it is an integer. In any triangle with sides a, b, c, the length of the tangents from vector to incircle are 1/2(a+b-c), 1/2(b+c-a) and 1/2(b+c-a). For a right triangle with hypotenuse c, one of the lengths is r, namely r = 1/2(a+b-c). Since one of the legs of the triangle is even, the other leg and the hypotenuse must have the same parity, so that a + b - c is even and r is an integer.



95.
The triangle ABC is isosceles is isosceles with equal sides AC and BC. Two of its angles measure 40. The interior point M is such that MAB = 10 and MBA = 20. Determine the measure of CMB.
Solution. Let BM be produced to meet AC at N. Since AMB = 150, we have that NMA = 30 = NAM so that NA = NM. Let a = |BC | = |AC |, c = |AB |, u = |CN | and v = |NA | = |NM |. Since BN bisects angle CBA, we have that v/u = c/a.

By the Law of Sines,


c
a
= sin100
sin40
= sin80
sin40
= 2cos40
and


v
u
= sin(a- 60)
sina
where a = CMB. Hence


sin(a- 60) = 2sinacos40 = sin(a+ 40) + sin(a- 40)
whence


2cos(a- 50)sin10 = sin(a- 60) - sin(a- 40) = sin(a+ 40circ) .
Since a+ 40 = (a- 50) + 90, sin(a+ 40) = cos(a- 50). Therefore, we find that 2cos(a- 50)sin10 = cos(a- 50). Since sin10 1/2, we have that cos(a- 50) = 0 and a = 140.



96.
Find all prime numbers p for which all three of the numbers p2 - 2, 2p2 - 1 and 3p2 + 4 are also prime.
Solution. Modulo 7, we find that p2 - 2 0 when p 3, 4, 2p2 - 1 0 when p 2, 5 and 3p2 + 4 0 when p 1, 6. Thus, when p > 7, at least one of the four numbers is a proper multiple of 7. The only primes p for which all numbers are prime at 3 and 7, and we get the quadruples (3, 7, 17, 31) and (7, 47, 97, 151).


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