7th USA Mathematical Olympiad 1978 Problems
1. The sum of 5 real numbers is 8 and the sum of their squares is 16. What is the largest possible value for one of the numbers?
2. Two square maps cover exactly the same area of terrain on different scales. The smaller map is placed on top of the larger map and inside its borders. Show that there is a unique point on the top map which lies exactly above the corresponding point on the lower map. How can this point be constructed?
3. You are told that all integers from 33 to 73 inclusive can be expressed as a sum of positive integers whose reciprocals sum to 1. Show that the same is true for all integers greater than 73.
4. Show that if the angle between each pair of faces of a tetrahedron is equal, then the tetrahedron is regular. Does a tetrahedron have to be regular if five of the angles are equal?
5. There are 9 delegates at a conference, each speaking at most three languages. Given any three delegates, at least 2 speak a common language. Show that there are three delegates with a common language.
Solution
7th USA Mathematical Olympiad 1978
Problem 1
The sum of 5 real numbers is 8 and the sum of their squares is 16. What is the largest possible value for one of the numbers?
Solution
Answer: 16/5.
Let the numbers be v, w, x, y, z. We have (v -6/5)2 + (w - 6/5)2 + ... + (z - 6/5)2 = (v2 + ... + z2) - 12/5 (v + ... + z) + 36/5 = 16 - 96/5 + 36/5 = 4. Hence |v - 6/5| ≤ 2, so v ≤ 16/5. This value can be realized by putting v = 16/5 and setting the other numbers to 6/5. Problem 2
Two square maps cover exactly the same area of terrain on different scales. The smaller map is placed on top of the larger map and inside its borders. Show that there is a unique point on the top map which lies exactly above the corresponding point on the lower map. How can this point be constructed?
Solution
The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps).
Let the small map square be A'B'C'D' and the large be ABCD, where X and X' are corresponding points. We deal first with the special case where A'B' is parallel to AB. In this case let AA' and BB' meet at O. Then triangles OAB and OA'B' are similar, so O must represent the same point. So assume A'B' is not parallel to AB.
Let the lines A'B' and AB meet at W, the lines B'C' and BC meet at X, the lines C'D' and CD meet at Y, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between A' and B' (or one of the vertices A', B' of the smaller square would lie outside the larger square). If it lies on the opposite side of A' to B', then Y must lie on the opposite side of C' to D'. Thus the segment WY must cut the side A'D' at some point Z' and the side B'C' at some point X'. The same conclusion holds if W lies on the opposite side of B' to A', because then Y must lie on the opposite side of D' to C'. Similarly, the segment XZ must cut the side A'B' at some point W' and the side C'D' at some point Y'. But now the segments X'Z' and W'Y' join pairs of points on opposite sides of the small square and so they must meet at some point O inside the small square.
Now the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY = OW'/OY'. So if we set up coordinate systems with AB as the x-axis and AD as the y-axis (for the large square) and A'B' as the x'-axis and A'D' as the y'-axis (for the small square) so that corresponding points have the same coordinates, then the y coordinate of O equals the y' coordinate of O. Similarly, XOX' and ZOZ' are similar, so OX/OZ = OX'/OZ', so the x-coordinate of O equals its x'-coordinate. In other words, O represents the same point on both maps.
You are told that all integers from 33 to 73 inclusive can be expressed as a sum of positive integers whose reciprocals sum to 1. Show that the same is true for all integers greater than 73.
Solution
The trick is consider the integers 2a1, 2a2, ... , 2am given that a1, a2, ... , am is a solution for n. The sum of their reciprocals is 1/2. So if we throw in two 4s, we get a solution for 2n + 8. Similarly, adjoining 3 and 6 gives a solution for 2n + 9. It is now a simple induction. For the starter set gives 74 thru 155, then those give 156 thru 319, and so on. In general, n thru 2n+7 gives 2n+8 thru 4n+23 = 2(2n+8) + 7.
Show that if the angle between each pair of faces of a tetrahedron is equal, then the tetrahedron is regular. Does a tetrahedron have to be regular if five of the angles are equal?
Solution
Answer: no.
Let the tetrahedron be ABCD. Let the insphere have center O and touch the sides at W, X, Y, Z. The OW, OX, OY, OZ are the normals to the faces. But the angle between each pair of normals is equal. So OW, OX, OY and OZ are vectors of equal length at equal angles. Hence each side of WXYZ is equal (eg WX = 2 OW sin(WOX/2) ). So WXYZ is a regular tetrahedron. But the faces of ABCD are just the tangent planes at W, X, Y, Z. So if we rotate through an angle 120o about the line OW, then X goes to Y, Y to Z and Z to X. Hence AB = AC, AC = AD, AD = AB and BC = CD = DB (assuming appropriate labeling). Similarly, for rotation about the other axes. So ABCD has equal edges and hence is regular.
Consider the four normals OW, OX, OY, OZ. We can move X, Y, Z slightly closer together so that XYZ remains an equilateral triangle. Then move W so that WX = WY = XY. So five of the distances are equal, but the sixth is unequal. The reason for slightly is that all the angles between pairs of normals must remain less than 180o. The corresponding tetrahedron will now have the angles between only five pairs of faces equal.
There are 9 delegates at a conference, each speaking at most three languages. Given any three delegates, at least 2 speak a common language. Show that there are three delegates with a common language.
Solution
Suppose not. Then A can shares a language with at most 3 delegates, because if he shared a language with 4, he would have to share the same language with 2 of them (since he can only speak 3 languages). So there are 5 delegates who do not share a language with A. Let one of them be B. By the same argument, there must be at least one of the other 4 (call her C) who does not share a language with B. But now no two of A, B, C share a language. Contradiction.
Comment. The result is best possible. Suppose each pair of A, B, C, D share a different language and each pair of A', B', C', D' share a different language (12 languages in total). No other pairs share a language. Then given any three delegates two are primed and share a language or two are unprimed and share a language. But no three delegates share a language.
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