3rd Canadian Mathematical Olympiad Problems 1971
1. A diameter and a chord of a circle intersect at a point inside the circle. The two parts of the chord are length 3 and 5 and one part of the diameter is length 1. What is the radius of the circle?
2. If two positive real numbers x and y have sum 1, show that (1 + 1/x)(1 + 1/y) ≥ 9.
3. ABCD is a quadrilateral with AB = CD and ∠ABC > ∠BCD. Show that AC > BD.
4. Find all real a such that x2 + ax + 1 = x2 + x + a = 0 for some real x.
5. A polynomial with integral coefficients has odd integer values at 0 and 1. Show that it has no integral roots.
6. Show that n2 + 2n + 12 is not a multiple of 121 for any integer n.
7. Find all five digit numbers such that the number formed by deleting the middle digit divides the original number.
8. Show that the sum of the lengths of the perpendiculars from a point inside a regular pentagon to the sides (or their extensions) is constant. Find an expression for it in terms of the circumradius.
9. Find the locus of all points in the plane from which two flagpoles appear equally tall. The poles are heights h and k and are a distance 2a apart.
10. n people each have exactly one unique secret. How many phone calls are needed so that each person knows all n secrets? You should assume that in each phone call the caller tells the other person every secret he knows, but learns nothing from the person he calls.
Solutions
Problem 1
A diameter and a chord of a circle intersect at a point inside the circle. The two parts of the chord are length 3 and 5 and one part of the diameter is length 1. What is the radius the circle?
Solution
Let the chord be AB, the diameter be CD and the point of intersection X. Then XA.XB = XC.XD, so the other part of the diameter has length 15. So the diameter has length 16 and the radius is 8.
Problem 2
If two positive real numbers x and y have sum 1, show that (1 + 1/x)(1 + 1/y) ≥ 9.
Solution
8(x - 1/2)2 ≥ 0, so 8x2 - 8x + 2 ≥ 0, so -x2 + x + 2 ≥ 9x - 9x2, or (x+1)(2-x) ≥ 9x(1-x). Putting y = 1-x, this becomes (x+1)(1+y) ≥ 9xy. Since x and y are positive, we may divide by xy to get the result.
Problem 3
ABCD is a quadrilateral with AB = CD and angle ABC > angle BCD. Show that AC > BD.
Solution
We have AC2 = AB2 + BC2 - 2AB·BC cos ABC, BD2 = CD2 + BC2 - 2CD·BC cos BCD. But AB = CD and cos ABC < cos BCD, so the result follows.
Problem 4
Find all real a such that x2 + ax + 1 = x2 + x + a = 0 for some real x.
Solution
Subtracting, (a - 1)(x - 1) = 0, so a = 1 or x = 1. If a = 1, then the two quadratics are the same and have two complex roots. So we cannot have a = 1. If a = -2, the first quadratic has roots 1, 1 and the second has roots 1, -2. So a = -2 is the unique solution.
Problem 5
A polynomial with integral coefficients has odd integer values at 0 and 1. Show that it has no integral roots.
Solution
Let the polynomial be p(x). If x is an even integer, then every term of p(x) involving x is an even integer, so p(x) must have the same parity as p(0), which is odd, so it cannot be zero. If x is an odd integer, then every term of p(x) has the same parity as the corresponding term of p(1). Hence p(x) has the same parity as p(1), so it must be odd and hence cannot be zero.
Problem 6
Show that n2 + 2n + 12 is not a multiple of 121 for any integer n.
Solution
It is easy to check that n(n+2) + 1 = 1, 3, 9, 5, 3, 3, 5, 9, 4, 1, 0 mod 11 for n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 mod 11. So if n2 + 2n + 12 is divisible by 11, then we must have n = -1 mod 11. But in that case n = 11k - 1 for some k, and so n2 + 2n + 12 = 121k2 - 1 + 12 = 11 mod 121. So we can never have n2 + 2n + 12 a multiple of 121.
Problem 7
Find all five digit numbers such that the number formed by deleting the middle digit divides the original number.
Solution
Let the number be n = AhB, in other words 1000A + 100h + B, with 10 ≤ A < 100, 0 ≤ h ≤ 10, 0 ≤ B < 100. Then we have 1000A + 100h + B = k(100A + B) for some integer k. If k > 10, then k(100A + B) - 1000A ≥ 100A + B > 100h + B, which is not possible. If k < 10, then 1000A + 100h + B - k.100A ≥ 100A + B ≥ 1000 > kB, which is not possible. So we must have k = 10. Hence 100h = 9B. But the that implies that 9 divides h, so h = 0 or 9. It cannot be 9, because 9B < 900, so the only solution is h = B = 0. Hence n = 10000, 11000, 12000, ... , or 99000. [It is obvious that they are solutions.]
Problem 8
Show that the sum of the lengths of the perpendiculars from a point inside a regular pentagon to the sides (or their extensions) is constant. Find an expression for it in terms of the circumradius.
Solution
Use vectors. Let the pentagon be ABCDE. Let the point be K. Take an origin O and let the vectors OA, OB, OC, OD, OE, OK be a, b, c, d, e, k respectively. Then (a - k) x (a - b) = is a vector normal to the plane of the pentagon and with length AB times the length of the perpendicular from K to AB. Hence (a - k) x (a - b) + (b - k) x (b - c) + (c - k) x (c - d) + (d - k) x (d - e) + (e - k) x (e - a) (*) is a vector normal to the plane of the pentagon with length AB times the sum of the lengths of the perpendiculars from K to the sides. But after expansion the terms involving k cancel in pairs and hence sum to zero. So the sum of the perpendicular lengths is independent of K. [Note that for K inside the pentagon all five cross products vectors in (*) point in the same direction, but for K outside the pentagon, some of them point in opposite direction, so the argument no longer holds.]
Problem 9
Find the locus of all points in the plane from which two flagpoles appear equally tall. The poles are heights h and k and are a distance 2a apart.
Solution
Let the flagpole height h be at H, and the flagpole height k be at K. Let A point on the segment HK such that AH/AK = h/k and let B be the other point on the line HK such that BH/BK = h/k. Then the locus is the circle with diameter AB. If h = k, then there is no point B and the locus is the perpendicular bisector of AB.
This is a well-known result (the circle of Apollonius) which scarcely needs proving. The case h = k is obvious. So assume h > k. Suppose P is such that HP/PK = h/k. Let the internal and external bisectors of ∠HPK meet the line HK at A' and B' respectively. Let the line through K parallel to PB' meet the line HP at X. Then HA'/A'K = HP/PK = h/k and HB'/KB' = HP/XP (HPB' and HXK similar) = HP/PK = h/k. So A' = A and B' = B. But ∠A'PB' = 90o, so P lies on the circle diameter AB. Conversely, if P lies on that circle, then let the line through K parallel to PB meet PH at X, and let the line through K parallel to PA meet the line PH at Y. Then HP/PY = HA/AK = h/k and HP/PX = HB/KB = h/k. So PX = PY. But ∠XKY = 90o, so P is the center of the circle XKY, so PY = PK. Hence HP/KP = HP/PY = h/k, so P is part of the locus.
Problem 10
n people each have exactly one unique secret. How many phone calls are needed so that each person knows all n secrets? You should assume that in each phone call the caller tells the other person every secret he knows, but learns nothing from the person he calls.
Solution
Everyone must receive a call after n-2 calls have been made. For calls up to and including the (n-2)th can communicate in total at most n-2 secrets, which is less than the n-1 which each person needs to receive. Thus the minimum number of calls is 2n-2. This can be achieved as follows. Let the people be A1, A2, ... , An. A1 calls A2, then A2 calls A3, and so on up to An-1 calls An. After that call An knows all the secrets. He then calls everyone else.
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