2nd All Russian Mathematical Olympiad Problems 1962
1. ABCD is any convex quadrilateral. Construct a new quadrilateral as follows. Take A' so that A is the midpoint of DA'; similarly, B' so that B is the midpoint of AB'; C' so that C is the midpoint of BC'; and D' so that D is the midpoint of CD'. Show that the area of A'B'C'D' is five times the area of ABCD.
2. Given a fixed circle C and a line L throught the center O of C. Take a variable point P on L and let K be the circle center P through O. Let T be the point where a common tangent to C and K meets K. What is the locus of T?
3. Given integers a0, a1, ... , a100, satisfying a1>a0, a1>0, and ar+2=3 ar+1 - 2 ar for r=0, 1, ... , 98. Prove a100 > 299
4. Prove that there are no integers a, b, c, d such that the polynomial ax3+bx2+cx+d equals 1 at x=19 and 2 at x=62.
5. Given an n x n array of numbers. n is odd and each number in the array is 1 or -1. Prove that the number of rows and columns containing an odd number of -1s cannot total n.
6. Given the lengths AB and BC and the fact that the medians to those two sides are perpendicular, construct the triangle ABC.
7. Given four positive real numbers a, b, c, d such that abcd=1, prove that a2 + b2 + c2 + d2 + ab + ac + ad + bc + bd + cd ≥ 10.
8. Given a fixed regular pentagon ABCDE with side 1. Let M be an arbitary point inside or on it. Let the distance from M to the closest vertex be r1, to the next closest be r2 and so on, so that the distances from M to the five vertices satisfy r1 ≤ r2 ≤ r3 ≤ r4 ≤ r5. Find (a) the locus of M which gives r3 the minimum possible value, and (b) the locus of M which gives r3 the maximum possible value.
9. Given a number with 1998 digits which is divisible by 9. Let x be the sum of its digits, let y be the sum of the digits of x, and z the sum of the digits of y. Find z.
10. AB=BC and M is the midpoint of AC. H is chosen on BC so that MH is perpendicular to BC. P is the midpoint of MH. Prove that AH is perpendicular to BP.
11. The triangle ABC satisfies 0 ≤ AB ≤ 1 ≤ BC ≤ 2 ≤ CA ≤ 3. What is the maximum area it can have?
12. Given unequal integers x, y, z prove that (x-y)5 + (y-z)5 + (z-x)5 is divisible by 5(x-y)(y-z)(z-x).
13. Given a0, a1, ... , an, satisfying a0 = an = 0, and and ak-1 - 2ak + ak+1 ≥ 0 for k=0, 1, ... , n-1. Prove that all the numbers are negative or zero.
14. Given two sets of positive numbers with the same sum. The first set has m numbers and the second n. Prove that you can find a set of less than m+n positive numbers which can be arranged to part fill an m x n array, so that the row and column sums are the two given sets.
Solutions
Problem 1
ABCD is any convex quadrilateral. Construct a new quadrilateral as follows. Take A' so that A is the midpoint of DA'; similarly, B' so that B is the midpoint of AB'; C' so that C is the midpoint of BC'; and D' so that D is the midpoint of CD'. Show that the area of A'B'C'D' is five times the area of ABCD.
Solution
Compare the triangles A'B'A and ADB. The base of A'B'A can be taken as A'A, which is the same length as AD. The height of A'B'A is AB' times sin B'AA', which is twice AB times sin BAD. So area A'B'A = 2 area ADB. Similarly, area B'C'B = 2 area BAC, area C'D'C = 2 area CBD, and area D'A'D = 2 area DCA. So adding, the area A'B'A + area C'D'C = 2 area ABCD, and area B'C'B + area D'A'D = 2 area ABCD. But ABCD = A'B'A + B'C'B + C'D'C + D'A'D + ABCD. Hence result.
ABCD is any convex quadrilateral. Construct a new quadrilateral as follows. Take A' so that A is the midpoint of DA'; similarly, B' so that B is the midpoint of AB'; C' so that C is the midpoint of BC'; and D' so that D is the midpoint of CD'. Show that the area of A'B'C'D' is five times the area of ABCD.
Solution
Compare the triangles A'B'A and ADB. The base of A'B'A can be taken as A'A, which is the same length as AD. The height of A'B'A is AB' times sin B'AA', which is twice AB times sin BAD. So area A'B'A = 2 area ADB. Similarly, area B'C'B = 2 area BAC, area C'D'C = 2 area CBD, and area D'A'D = 2 area DCA. So adding, the area A'B'A + area C'D'C = 2 area ABCD, and area B'C'B + area D'A'D = 2 area ABCD. But ABCD = A'B'A + B'C'B + C'D'C + D'A'D + ABCD. Hence result.
Problem 2
Given a fixed circle C and a line L throught the center O of C. Take a variable point P on L and let K be the circle center P through O. Let T be the point where a common tangent to C and K meets K. What is the locus of T?
Given a fixed circle C and a line L throught the center O of C. Take a variable point P on L and let K be the circle center P through O. Let T be the point where a common tangent to C and K meets K. What is the locus of T?
Solution
Let the common tangent meet C at S. Let X be the intersection of C and OP lying between O and P. PT = PO, hence ∠POT = ∠PTO, so ∠OPT = 180o - 2 ∠POT. But PT and OS are parallel, because both are perpendicular to the common tangent. Hence ∠POS = 2 ∠POT, so ∠SOT = ∠XOT. Hence TX is tangent to C, in other words T lies on the (fixed) tangent to C at X. Conversely, it is easy to see that any such point can be obtained (just take P such that PO = PT). Thus the required locus is the pair of tangents to C which are perpendicular to L.
Let the common tangent meet C at S. Let X be the intersection of C and OP lying between O and P. PT = PO, hence ∠POT = ∠PTO, so ∠OPT = 180o - 2 ∠POT. But PT and OS are parallel, because both are perpendicular to the common tangent. Hence ∠POS = 2 ∠POT, so ∠SOT = ∠XOT. Hence TX is tangent to C, in other words T lies on the (fixed) tangent to C at X. Conversely, it is easy to see that any such point can be obtained (just take P such that PO = PT). Thus the required locus is the pair of tangents to C which are perpendicular to L.
Problem 3
Given integers a0, a1, ... , a100, satisfying a1>a0, a1>0, and ar+2=3 ar+1 - 2 ar for r=0, 1, ... , 98. Prove a100 > 299.
Given integers a0, a1, ... , a100, satisfying a1>a0, a1>0, and ar+2=3 ar+1 - 2 ar for r=0, 1, ... , 98. Prove a100 > 299.
Solution
An easy induction gives ar = (2r - 1)a1 - (2r - 2)a0 for r = 2, 3, ... , 100. Hence, in particular, a100 = (2100 - 2)(a1 - a0) + a1. But a1 and (a1 - a0) are both at least 1. Hence result.
An easy induction gives ar = (2r - 1)a1 - (2r - 2)a0 for r = 2, 3, ... , 100. Hence, in particular, a100 = (2100 - 2)(a1 - a0) + a1. But a1 and (a1 - a0) are both at least 1. Hence result.
Problem 4
Prove that there are no integers a, b, c, d such that the polynomial ax3+bx2+cx+d equals 1 at x=19 and 2 at x=62.
Prove that there are no integers a, b, c, d such that the polynomial ax3+bx2+cx+d equals 1 at x=19 and 2 at x=62.
Solution
If there were such values, then subtract the equation with x = 19 from the equation with x = 62 to get: a(623 - 193) + b(622 - 192) + c(62 - 19) = 1. But the left hand side is divisible by 62 -19 = 43, contradiction.
If there were such values, then subtract the equation with x = 19 from the equation with x = 62 to get: a(623 - 193) + b(622 - 192) + c(62 - 19) = 1. But the left hand side is divisible by 62 -19 = 43, contradiction.
Problem 5
Given an n x n array of numbers. n is odd and each number in the array is 1 or -1. Prove that the number of rows and columns containing an odd number of -1s cannot total n.
Given an n x n array of numbers. n is odd and each number in the array is 1 or -1. Prove that the number of rows and columns containing an odd number of -1s cannot total n.
Solution
If we change a -1 to 1, we affect the total number of rows and columns (containing an odd number of -1s) by 0, 2 or -2. After changing all the -1s we have total of 0. Hence the starting total must be even. So it cannot be n.
If we change a -1 to 1, we affect the total number of rows and columns (containing an odd number of -1s) by 0, 2 or -2. After changing all the -1s we have total of 0. Hence the starting total must be even. So it cannot be n.
Problem 6
Given the lengths AB and BC and the fact that the medians to those two sides are perpendicular, construct the triangle ABC.
Given the lengths AB and BC and the fact that the medians to those two sides are perpendicular, construct the triangle ABC.
Solution
Let M be the midpoint of AB and X the midpoint of MB. Construct the circle center B, radius BC/2 and the circle diameter AX. If they do not intersect (so BC<AB/2 or BC>AB) then the construction is not possible. If they intersect at N, then take C so that N is the midpoint of BC. Let CM meet AN at O. Then AO/AN = AM/AX =2/3, so the triangles AOM and ANX are similar. Hence ∠AOM = ∠ANX = 90o.
Let M be the midpoint of AB and X the midpoint of MB. Construct the circle center B, radius BC/2 and the circle diameter AX. If they do not intersect (so BC<AB/2 or BC>AB) then the construction is not possible. If they intersect at N, then take C so that N is the midpoint of BC. Let CM meet AN at O. Then AO/AN = AM/AX =2/3, so the triangles AOM and ANX are similar. Hence ∠AOM = ∠ANX = 90o.
Problem 7
Given four positive real numbers a, b, c, d such that abcd=1, prove that a2 + b2 + c2 + d2 + ab + ac + ad + bc + bd + cd >= 10.
Given four positive real numbers a, b, c, d such that abcd=1, prove that a2 + b2 + c2 + d2 + ab + ac + ad + bc + bd + cd >= 10.
Solution
Applying the arithmetic/geometric mean result to the 10 numbers gives the result immediately.
Applying the arithmetic/geometric mean result to the 10 numbers gives the result immediately.
Problem 8
Given a fixed regular pentagon ABCDE with side 1. Let M be an arbitary point inside or on it. Let the distance from M to the closest vertex be r1, to the next closest be r2 and so on, so that the distances from M to the five vertices satisfy r1 ≤ r2 ≤ r3 ≤ r4 ≤ r5. Find (a) the locus of M which gives r3 the minimum possible value, and (b) the locus of M which gives r3 the maximum possible value.
Given a fixed regular pentagon ABCDE with side 1. Let M be an arbitary point inside or on it. Let the distance from M to the closest vertex be r1, to the next closest be r2 and so on, so that the distances from M to the five vertices satisfy r1 ≤ r2 ≤ r3 ≤ r4 ≤ r5. Find (a) the locus of M which gives r3 the minimum possible value, and (b) the locus of M which gives r3 the maximum possible value.
Solution
Let X be the midpoint of AB and O the center of ABCDE. Suppose M lies inside AXO. Then ME=r3. So we maximise r3 by taking M at X, with distance 1.5590, and we minimise r3 by taking M as the intersection of AO and EB with distance 0.8090. AXO is one of 10 congruent areas, so the required loci are (a) the 5 midpoints of the diagonals, and (b) the 5 midpoints of the sides.
Let X be the midpoint of AB and O the center of ABCDE. Suppose M lies inside AXO. Then ME=r3. So we maximise r3 by taking M at X, with distance 1.5590, and we minimise r3 by taking M as the intersection of AO and EB with distance 0.8090. AXO is one of 10 congruent areas, so the required loci are (a) the 5 midpoints of the diagonals, and (b) the 5 midpoints of the sides.
Problem 9
Given a number with 1998 digits which is divisible by 9. Let x be the sum of its digits, let y be the sum of the digits of x, and z the sum of the digits of y. Find z.
Given a number with 1998 digits which is divisible by 9. Let x be the sum of its digits, let y be the sum of the digits of x, and z the sum of the digits of y. Find z.
Solution
x ≤ 9·1998 = 17982. Hence y ≤ the greater of 1+7+9+9+9=35 and 9+9+9+9=36. But 9 divides the original number and hence also x, y and z. Hence z=9.
x ≤ 9·1998 = 17982. Hence y ≤ the greater of 1+7+9+9+9=35 and 9+9+9+9=36. But 9 divides the original number and hence also x, y and z. Hence z=9.
Problem 10
AB=BC and M is the midpoint of AC. H is chosen on BC so that MH is perpendicular to BC. P is the midpoint of MH. Prove that AH is perpendicular to BP.
AB=BC and M is the midpoint of AC. H is chosen on BC so that MH is perpendicular to BC. P is the midpoint of MH. Prove that AH is perpendicular to BP.
Solution
Take X on AH so that BX is perpendicular to AH. Extend to meet HM at P'. Let N be the midpoint of AB. A, B, M and X are on the circle center N radius NA (because angles AMB and AXB are 90). Also MN is parallel to BC (because AMN, ACB are similar), so NM is perpendicular to MH, in other words HM is a tangent to the circle. hence P'M·P'M = P'X·P'B. Triangles P'XH and P'HB are similar (angles at P' same and both have a right angle), so P'H/P'X = P'B/P'H, so P'H·P'H = P'X·P'B. Hence P'H = P'M and P' coincides with P.
Take X on AH so that BX is perpendicular to AH. Extend to meet HM at P'. Let N be the midpoint of AB. A, B, M and X are on the circle center N radius NA (because angles AMB and AXB are 90). Also MN is parallel to BC (because AMN, ACB are similar), so NM is perpendicular to MH, in other words HM is a tangent to the circle. hence P'M·P'M = P'X·P'B. Triangles P'XH and P'HB are similar (angles at P' same and both have a right angle), so P'H/P'X = P'B/P'H, so P'H·P'H = P'X·P'B. Hence P'H = P'M and P' coincides with P.
Problem 11
The triangle ABC satisfies 0 ≤ AB ≤ 1 ≤ BC ≤ 2 ≤ CA ≤ 3. What is the maximum area it can have?
The triangle ABC satisfies 0 ≤ AB ≤ 1 ≤ BC ≤ 2 ≤ CA ≤ 3. What is the maximum area it can have?
Solution
If we ignore the restrictions of CA, then the maximum area is 1, achieved when AB is perpendicular to BC. But in this case CA satisfies the restrictions.
If we ignore the restrictions of CA, then the maximum area is 1, achieved when AB is perpendicular to BC. But in this case CA satisfies the restrictions.
Problem 12
Given unequal integers x, y, z prove that (x-y)5 + (y-z)5 + (z-x)5 is divisible by 5(x-y)(y-z)(z-x).
Given unequal integers x, y, z prove that (x-y)5 + (y-z)5 + (z-x)5 is divisible by 5(x-y)(y-z)(z-x).
Solution
Put x-y=r, y-z=s. Then z-x = -(r+s), and (x-y)5 + (y-z)5 + (z-x)5 = r5 + s5 - (r+s)5 = -5r4s - 10r3s2 - 10r2s3 - 5rs4 = -5rs(r+s)(r2 + rs + s2).
Put x-y=r, y-z=s. Then z-x = -(r+s), and (x-y)5 + (y-z)5 + (z-x)5 = r5 + s5 - (r+s)5 = -5r4s - 10r3s2 - 10r2s3 - 5rs4 = -5rs(r+s)(r2 + rs + s2).
Problem 13
Given a0, a1, ... , an, satisfying a0 = an = 0, and and ak-1 - 2ak + ak+1 ≥ 0 for k=0, 1, ... , n-1. Prove that all the numbers are negative or zero.
Given a0, a1, ... , an, satisfying a0 = an = 0, and and ak-1 - 2ak + ak+1 ≥ 0 for k=0, 1, ... , n-1. Prove that all the numbers are negative or zero.
Solution
The essential point is that if we plot the values ar against r, then the curve formed by joining the points is cup shaped. Its two endpoints are on the axis, so the other points cannot be above it.There are many ways of turning this insight into a formal proof. Barry Paul's was neater than mine: ar+1-ar ≥ ar-ar-1. Hence (easy induction) if as - as-1> 0, then an > as. Take as to be the first positive, then certainly as > as-1, so an > 0. Contradiction.
The essential point is that if we plot the values ar against r, then the curve formed by joining the points is cup shaped. Its two endpoints are on the axis, so the other points cannot be above it.There are many ways of turning this insight into a formal proof. Barry Paul's was neater than mine: ar+1-ar ≥ ar-ar-1. Hence (easy induction) if as - as-1> 0, then an > as. Take as to be the first positive, then certainly as > as-1, so an > 0. Contradiction.
Problem 14
Given two sets of positive numbers with the same sum. The first set has m numbers and the second n. Prove that you can find a set of less than m+n positive numbers which can be arranged to part fill an m x n array, so that the row and column sums are the two given sets.
Example: row sums 1, 5, 3; column sums 2, 7. Array is:
Given two sets of positive numbers with the same sum. The first set has m numbers and the second n. Prove that you can find a set of less than m+n positive numbers which can be arranged to part fill an m x n array, so that the row and column sums are the two given sets.
Example: row sums 1, 5, 3; column sums 2, 7. Array is:
x5
x1
21
Solution
Induction on m+n. Trivial for m+n=2.
Let x be the largest number in the two given sets. Suppose it is a row total; let y be the largest column total. If y<x, then replace x by x-y in the set of row totals and remove y from the col totals. By induction find <=m+n-2 positive numbers in an m x (n-1) array with the new totals. Adding a col empty except for y in the row totalling x-y gives the required original set.
If y=x, then drop x from the row totals and y from the col totals and argue as before.
If x was a col total we interchange rows and cols in the argument.
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