4th All Russian Mathematical Olympiad Problems 1964
1. In the triangle ABC, the length of the altitude from A is not less than BC, and the length of the altitude from B is not less than AC. Find the angles.
2. If m, k, n are natural numbers and n>1, prove that we cannot have m(m+1) = kn.
3. Reduce each of the first billion natural numbers (billion = 109) to a single digit by taking its digit sum repeatedly. Do we get more 1s than 2s?
4. Given n odd and a set of integers a1, a2, ... , an, derive a new set (a1 + a2)/2, (a2 + a3)/2, ... , (an-1 + an)/2, (an + a1)/2. However many times we repeat this process for a particular starting set we always get integers. Prove that all the numbers in the starting set are equal. For example, if we started with 5, 9, 1, we would get 7, 5, 3, and then 6, 4, 5, and then 5, 4.5, 5.5. The last set does not consist entirely of integers.
5. (a) The convex hexagon ABCDEF has all angles equal. Prove that AB - DE = EF - BC = CD - FA.
(b) Given six lengths a1, ... , a6 satisfying a1 - a4 = a5 - a2 = a3 - a6, show that you can construct a hexagon with sides a1, ... , a6 and equal angles.
6. Find all possible integer solutions for √(x + √(x ... (x + √(x)) ... )) = y, where there are 1998 square roots.
7. ABCD is a convex quadrilateral. A' is the foot of the perpendicular from A to the diagonal BD, B' is the foot of the perpendicular from B to the diagonal AC, and so on. Prove that A'B'C'D' is similar to ABCD.
8. Find all natural numbers n such that n2 does not divide n!.
9. Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B along the edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least AB/2 in one direction. If it travels exactly AB/2 in one direction, how many edges does it traverse?
10. A circle center O is inscribed in ABCD (touching every side). Prove that ∠AOB + ∠COD = 180o.
11. The natural numbers a, b, n are such that for every natural number k not equal to b, b - k divides a - kn. Prove that a = bn.
12. How many (algebraically) different expressions can we obtain by placing parentheses in a1/a2/ ... /an?
13. What is the smallest number of tetrahedrons into which a cube can be partitioned?
14. (a) Find the smallest square with last digit not 0 which becomes another square by the deletion of its last two digits. (b) Find all squares, not containing the digits 0 or 5, such that if the second digit is deleted the resulting number divides the original one.
15. A circle is inscribed in ABCD. AB is parallel to CD, and BC = AD. The diagonals AC, BD meet at E. The circles inscribed in ABE, BCE, CDE, DAE have radius r1, r2, r3, r4 respectively. Prove that 1/r1 + 1/r3 = 1/r2 + 1/r4.
Problem 1
In the triangle ABC, the length of the altitude from A is not less than BC, and the length of the altitude from B is not less than AC. Find the angles.
Solution
Let k be twice the area of the triangle. Then k≥BC2, k≥AC2 and k≤AC.BC, with equality in the last case only if AC is perpendicular to BC. Hence AC and BC have equal lengths and are perpendicular. So the angles are 90, 45, 45.
Problem 2
If m, k, n are natural numbers and n>1, prove that we cannot have m(m+1) = kn.
Solution
m and m+1 have no common divisors, so each must separately be an nth power. But the difference betwee the two nth powers is greater than 1 (for n>1).
Problem 3
Reduce each of the first billion natural numbers (billion = 109) to a single digit by taking its digit sum repeatedly. Do we get more 1s than 2s?
Solution
Taking digit sums repeatedly gives the remainder after dividing the number by 9, or 9 if the number is exactly divisible by 9. 109 - 1 = 9n, and for any r>=0 the nine consecutive numbers 9r+1, 9r+2, ... , 9r+9 include just one number giving remainder 1 and one number giving remainder 2. Hence the numbers up to 109 - 1 give equal numbers of 1s and 2s. 109 itself gives 1, so there is just one more of the 1s than the 2s.
Problem 4
Given n odd and a set of integers a1, a2, ... , an, derive a new set (a1 + a2)/2, (a2 + a3)/2, ... , (an-1 + an)/2, (an + a1)/2. However many times we repeat this process for a particular starting set we always get integers. Prove that all the numbers in the starting set are equal.
For example, if we started with 5, 9, 1, we would get 7, 5, 3, and then 6, 4, 5, and then 5, 4.5, 5.5. The last set does not consist entirely of integers.
Solution
Let the smallest value be s and suppose it occurs m times (with m<n). Then the values in the next stage are all at least s, and at most m-1 equal s. So after at most m iterations the smallest value is increased.
We can never reach a stage where all the values are equal, because if (a1+a2)/2 = (a2+a3)/2 = ... = (an-1+an)/2 = (an+a1)/2, then a1+a2 = a2+a3 and hence a1 = a3. Similarly, a3 = a5, and so a1 = a3 = a5 = ... = an (n odd). Similarly, a2 = a4 = ... = an-1. But we also have an + a1 = a1 + a2 and so a2 = an, so that all ai are equal. In other words, if all the values are equal at a particular stage, then they must have been equal at the previous stage, and hence at every stage.
Thus if the values do not start out all equal, then the smallest value increases indefinitely. But that is impossible, because the sum of the values is the same at each stage, and hence the smallest value can never exceed (a1 + ... + an)/n.
Note that for n even the argument breaks down because a set of unequal numbers can iterate into a set of equal numbers. For example: 1, 3, 1, 3, ... , 1, 3.
Problem 5
The convex hexagon ABCDEF has all angles equal. Prove that AB - DE = EF - BC = CD - FA.
(b) Given six lengths a1, ... , a6 satisfying a1 - a4 = a5 - a2 = a3 - a6, show that you can construct a hexagon with sides a1, ... , a6 and equal angles.
Solution
(a) Extend AB, CD, EF. We get an equilateral triangle with sides AF + AB + BC, BC + CD + DE, ED + EF + FA. Hence AB - DE = CD - FA = EF - BC, as required.
(b) Take an equilateral triangle with sides s, t, u lengths a2 + a3 + a4, a4 + a5 + a6, and a6 + a1 + a2 respectively. Construct BC length a2 parallel to t with B on u and C on s. Construct DE length a4 parallel to u with D on s and E on t. Construct FA length a6 parallel to s with F on t and A on u. Then ABCDEF is the required hexagon, with AB = a1, BC = a2 etc.
Problem 6
Find all possible integer solutions for sqrt(x + sqrt(x ... (x + sqrt(x)) ... )) = y, where there are 1998 square roots.
Solution
Let s1 = sqrt(x), s2 = sqrt(x + s1), s3 = sqrt(x + s2) and so on. So the equation given is y = s1998. We show first that all sn must be integral for 1 <= n <= 1998. y is integral, so s1998 is integral. Now suppose sn is integral. Then sn-1 = sn2 - x is integral, proving the claim.
So in particular s1 and s2 are integers and s22 = s12 + s1. But if s1 > 0, then s12 < s12 + s1 < (s1 + 1)2, which is impossible. Similarly s1 < 0 is impossible. So the only possible solution is s1 = 0 and hence x = 0 and y = 0.
Problem 7
ABCD is a convex quadrilateral. A' is the foot of the perpendicular from A to the diagonal BD, B' is the foot of the perpendicular from B to the diagonal AC, and so on. Prove that A'B'C'D' is similar to ABCD.
Solution
Let the diagonals meet at O. Then CC'O is similar to AA'O (because CC'O = AA'O = 90, and COC', AOA' are opposite angles), so A'O/C'O = AO/CO. Similarly, B'O/D'O = BO/DO. AA'O is also similar to BB'O, so A'O/B'O = AO/BO. Thus OA':OB':OC':OD' = OA:OB:OC:OD. Hence triangles OA'B' and OAB are similar. Likewise OB'C' and OBC, OC'D' and OCD, and OD'A' and ODA. Hence result.
Problem 8
Find all natural numbers n such that n2 does not divide n!.
Solution
Answer: n = 4 or prime.
If n = rs, with 1 < r < s, then r < s < n, and hence rsn = n2 divides n!. Similarly, if n = r2 with r > 2, then r < 2r < n, and hence n2 divides n!. This covers all possibilities except n = 4 or n = prime, and it is easy to see that in these cases n2 does not divide n!.
Problem 9
Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B along edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least AB/2 in one direction. If it travels exactly AB/2 in one direction, how many edges does it traverse?
Solution
If B is on the same lattice as A, then we can give B coordinates (m,n) relative to A and the shortest path from A to B must move m units east and n units east of north. The shortest path between a lattice point and the next lattice point east is evidently one edge in direction 3 followed by one edge in direction 2. Similarly, the shortest path between a lattice point and the next lattice point east of north is one edge in direction 1, followed by one edge in direction 2. So a shortest path from A to B must have m+n edges in direction 2.
B is a distance √3(m+n/2) east of A and a distance 3n/2 north of A, so AB2 = (3m2+3mn+3n2) < (4m2+8mn+4n2) = 4 (m+n)2. So in this case the bug must travel more than AB/2 in direction 2.
Now suppose B is on the other lattice. Let C be the lattice point immediately north of A and D the lattice point in direction 3 from A. Then a shortest path from A to B must either be A to C and then a shortest path from C to B, or A to D and then a shortest path from D to B. Take B to have coordinates (m, n) relative to C or D.
In the first case, AB2 = (√3(m+n/2))2 + (3n/2 + 1)2 = (3m2 + 3mn + 3n2) + 3n + 1 and a shortest path has m+n units in direction 2. But 4(m + n)2 > (3m2 + 3mn + 3n2) + 3n + 1, if m2 + n2 + 5mn > 3n + 1, which is true for m, n at least 1. If m=0 and n=1, then a shortest path has 2 units in direction 1 and AB = √7 < 4. If m=1 and n=0, then AB=2 and a shortest path has 1 unit in each direction. So in this case (the only one so far) we have equality.
It remains to consider the case where the path starts out towards D. In this case AB2 = (√3(m+n/2) + √3/2)2 + (3n/2 -1/2)2 = (3m2+3mn+3n2) + 3m + 1 and a path has m + n units in direction 2. But 4(m + n)2 > (3m2 + 3mn + 3n2) + 3m + 1 for m2+n2 + 5mn > 3m + 1, which is true for m, n at least 1. If m=1, n=0, then a shortest path has 2 units in direction 3 and AB = √7 < 4. Finally, if m=0 and n=1, then a shortest path has 1 unit in each direction and AB = 2.
Thus the answer to the final question is 3, because the only cases where the bug travels exactly AB/2 in one direction are where it goes to the opposite vertex of a hexagon it is on.
Problem 10
A circle center O is inscribed in ABCD (touching every side). Prove that angle AOB + angle COD equals 180 degrees.
Solution
Let AB touch the circle at W, BC at X, CD at Y, and DA at Z. Then AO bisects angle ZOW and BO bisects angle XOW. So angle AOB is half angle ZOX. Similarly angle COD is half angle XOZ and hence angle AOB + angle COD equals 180.
Problem 11
The natural numbers a, b, n are such that for every natural number k not equal to b, b - k divides a - kn. Prove that a = bn.
Solution
We have kn - a = bn - a (mod b - k). Hence bn - a = 0 (mod b- k) for every k not equal to b. But if bn does not equal a, then by taking k - b > bn - a we could render the equation false.
Problem 12
How many (algebraically) different expressions can we obtain by placing parentheses in a1/a2/ ... /an?
Solution
Answer 2n-2. a1 must be in the numerator, and a2 must be in the denominator, but the other symbols can be in either. This is easily proved by induction.
Problem 13
What is the smallest number of tetrahedrons into which a cube can be partitioned?
Solution
Answer: 5.
Tetrahedral faces are triangular, so each cube face requires at least two tetrahedral faces. So at least 12 tetrahedral faces are needed in all. At most three faces of a tetrahedron can be mutually orthogonal (and no two faces can be parallel), so at most 3 faces from each tetrahedron can contribute towards these 12. So we require at least 4 tetrahedra to provide the cube faces. But these tetrahedra each have volume at most 1/6 (1/3 x face area x 1, and face area is at most 1/2). So if we have only 4 tetrahedra in total then their total volume is less than the cube's volume. Contradiction. Hence we need at least 5 tetrahedra.
It can be done with 5: lop off 4 non-adjacent corners to leave a tetrahedron. More precisely, take the cube as ABCDA'B'C'D' with ABCD horizontal, A' directly under A, B' directly under B and so on. Then the five tetrahedra are AA'BD, CC'BC, DD'A'C', BB'A'C', BDA'C'.
Problem 14
a) Find the smallest square with last digit not 0 which becomes another square (not zero) by the deletion of its last two digits. #(b) Find all squares, not containing the digits 0 or 5, such that if the second digit is deleted the resulting number divides the original one.
Solution
(a) This one must have slipped through: 121!
(b) Answer: 16,36,121,484. Suppose the number has more than 2 digits. Write it as (10m + n)10r + s, where 1 <= m <= 9, 0 <= n <= 9, 0 <= s < 10r. Then we have k(m.10r + s) = (10m +n)10r + s, for some k > 1.
s does not contain the digits 0 or 5, so 5 does not divide s. Hence 5 divides k-1, and so k must be 6, 11, or 16 (if k was 21 or more, then the rhs would be negative). Since 25 does not divide k-1, we must have r=1 and s is a single digit.
We look at each possibility for k in turn. k = 6 gives no solutions. k = 11 gives about two dozen multiples of 11 from 121 to 891. By inspection the only squares are 121 and 484. k = 16 gives 192, which is not a square.
In addition, there is the possibility of 2 digit solutions, which I had overlooked. It is easiest to check each of the 2 digit squares, thus finding the additional solutions 16, 36.
Problem 15
A circle is inscribed in ABCD. AB is parallel to CD, and BC = AD. The diagonals AC, BD meet at E. The circles inscribed in ABE, BCE, CDE, DAE have radius r1, r2, r3, r4 respectively. Prove that 1/r1 + 1/r3 = 1/r2 + 1/r4.
Solution
A necessary and sufficient condition for ABCD to have an inscribed circle is AB + CD = BC + AD. So we have AB + CD = 2AD, which we use repeatedly. Extend DC to X so that BX is parallel to EC. Then DX = AB + CD = 2AD and the triangles DEC, AEB, DBX are similar. Let h be the perpendicular distance from AB to CD. The similar triangles give us the heights of DEC and AEB in terms of h.
1/r1 = perimeter ABE/(2 area ABE) = (AB + 2EB)/(AB.height) = (AB + 2.BD.AB/(AB+CD))/(AB.h.AB/(AB+CD)) = 2(AD + BD)/(AB.h). Similarly, 1/r3 = 2(AD + BD)/(CD.h).
The area of AED = area ABD - area ABE = 1/2 AB.h.CD/(2AD), so 1/r2 = 1/r4 = perimeter ADE/(2 area ADE) = (AD + BD)/(h.AB.CD/2AD), and 1/r2 + 1/r4 = 2(AD + BD)/h 2AD/(AB.CD) = 2(AB + BD)/h (AB + CD)/(AB.CD) = 1/r1 + 1/r3.
Solutions
Problem 1
In the triangle ABC, the length of the altitude from A is not less than BC, and the length of the altitude from B is not less than AC. Find the angles.
Solution
Let k be twice the area of the triangle. Then k≥BC2, k≥AC2 and k≤AC.BC, with equality in the last case only if AC is perpendicular to BC. Hence AC and BC have equal lengths and are perpendicular. So the angles are 90, 45, 45.
Problem 2
If m, k, n are natural numbers and n>1, prove that we cannot have m(m+1) = kn.
Solution
m and m+1 have no common divisors, so each must separately be an nth power. But the difference betwee the two nth powers is greater than 1 (for n>1).
Problem 3
Reduce each of the first billion natural numbers (billion = 109) to a single digit by taking its digit sum repeatedly. Do we get more 1s than 2s?
Solution
Taking digit sums repeatedly gives the remainder after dividing the number by 9, or 9 if the number is exactly divisible by 9. 109 - 1 = 9n, and for any r>=0 the nine consecutive numbers 9r+1, 9r+2, ... , 9r+9 include just one number giving remainder 1 and one number giving remainder 2. Hence the numbers up to 109 - 1 give equal numbers of 1s and 2s. 109 itself gives 1, so there is just one more of the 1s than the 2s.
Problem 4
Given n odd and a set of integers a1, a2, ... , an, derive a new set (a1 + a2)/2, (a2 + a3)/2, ... , (an-1 + an)/2, (an + a1)/2. However many times we repeat this process for a particular starting set we always get integers. Prove that all the numbers in the starting set are equal.
For example, if we started with 5, 9, 1, we would get 7, 5, 3, and then 6, 4, 5, and then 5, 4.5, 5.5. The last set does not consist entirely of integers.
Solution
Let the smallest value be s and suppose it occurs m times (with m<n). Then the values in the next stage are all at least s, and at most m-1 equal s. So after at most m iterations the smallest value is increased.
We can never reach a stage where all the values are equal, because if (a1+a2)/2 = (a2+a3)/2 = ... = (an-1+an)/2 = (an+a1)/2, then a1+a2 = a2+a3 and hence a1 = a3. Similarly, a3 = a5, and so a1 = a3 = a5 = ... = an (n odd). Similarly, a2 = a4 = ... = an-1. But we also have an + a1 = a1 + a2 and so a2 = an, so that all ai are equal. In other words, if all the values are equal at a particular stage, then they must have been equal at the previous stage, and hence at every stage.
Thus if the values do not start out all equal, then the smallest value increases indefinitely. But that is impossible, because the sum of the values is the same at each stage, and hence the smallest value can never exceed (a1 + ... + an)/n.
Note that for n even the argument breaks down because a set of unequal numbers can iterate into a set of equal numbers. For example: 1, 3, 1, 3, ... , 1, 3.
Problem 5
The convex hexagon ABCDEF has all angles equal. Prove that AB - DE = EF - BC = CD - FA.
(b) Given six lengths a1, ... , a6 satisfying a1 - a4 = a5 - a2 = a3 - a6, show that you can construct a hexagon with sides a1, ... , a6 and equal angles.
Solution
(a) Extend AB, CD, EF. We get an equilateral triangle with sides AF + AB + BC, BC + CD + DE, ED + EF + FA. Hence AB - DE = CD - FA = EF - BC, as required.
(b) Take an equilateral triangle with sides s, t, u lengths a2 + a3 + a4, a4 + a5 + a6, and a6 + a1 + a2 respectively. Construct BC length a2 parallel to t with B on u and C on s. Construct DE length a4 parallel to u with D on s and E on t. Construct FA length a6 parallel to s with F on t and A on u. Then ABCDEF is the required hexagon, with AB = a1, BC = a2 etc.
Problem 6
Find all possible integer solutions for sqrt(x + sqrt(x ... (x + sqrt(x)) ... )) = y, where there are 1998 square roots.
Solution
Let s1 = sqrt(x), s2 = sqrt(x + s1), s3 = sqrt(x + s2) and so on. So the equation given is y = s1998. We show first that all sn must be integral for 1 <= n <= 1998. y is integral, so s1998 is integral. Now suppose sn is integral. Then sn-1 = sn2 - x is integral, proving the claim.
So in particular s1 and s2 are integers and s22 = s12 + s1. But if s1 > 0, then s12 < s12 + s1 < (s1 + 1)2, which is impossible. Similarly s1 < 0 is impossible. So the only possible solution is s1 = 0 and hence x = 0 and y = 0.
Problem 7
ABCD is a convex quadrilateral. A' is the foot of the perpendicular from A to the diagonal BD, B' is the foot of the perpendicular from B to the diagonal AC, and so on. Prove that A'B'C'D' is similar to ABCD.
Solution
Let the diagonals meet at O. Then CC'O is similar to AA'O (because CC'O = AA'O = 90, and COC', AOA' are opposite angles), so A'O/C'O = AO/CO. Similarly, B'O/D'O = BO/DO. AA'O is also similar to BB'O, so A'O/B'O = AO/BO. Thus OA':OB':OC':OD' = OA:OB:OC:OD. Hence triangles OA'B' and OAB are similar. Likewise OB'C' and OBC, OC'D' and OCD, and OD'A' and ODA. Hence result.
Problem 8
Find all natural numbers n such that n2 does not divide n!.
Solution
Answer: n = 4 or prime.
If n = rs, with 1 < r < s, then r < s < n, and hence rsn = n2 divides n!. Similarly, if n = r2 with r > 2, then r < 2r < n, and hence n2 divides n!. This covers all possibilities except n = 4 or n = prime, and it is easy to see that in these cases n2 does not divide n!.
Problem 9
Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B along edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least AB/2 in one direction. If it travels exactly AB/2 in one direction, how many edges does it traverse?
Solution
1 2Suppose vertex A is that marked * at the bottom left. Without loss of generality, B is in a 60 degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked ° and . in the diagram). Each set forms a skewed lattice with axes at 60 degrees. Any path must alternate between the two lattices.
./ \./ \. directions | /
| | | * * *
./°\./°\./°\. \
| | | | 3
./°\./°\./°\./
| | | |
*\./°\./°\./°
If B is on the same lattice as A, then we can give B coordinates (m,n) relative to A and the shortest path from A to B must move m units east and n units east of north. The shortest path between a lattice point and the next lattice point east is evidently one edge in direction 3 followed by one edge in direction 2. Similarly, the shortest path between a lattice point and the next lattice point east of north is one edge in direction 1, followed by one edge in direction 2. So a shortest path from A to B must have m+n edges in direction 2.
B is a distance √3(m+n/2) east of A and a distance 3n/2 north of A, so AB2 = (3m2+3mn+3n2) < (4m2+8mn+4n2) = 4 (m+n)2. So in this case the bug must travel more than AB/2 in direction 2.
Now suppose B is on the other lattice. Let C be the lattice point immediately north of A and D the lattice point in direction 3 from A. Then a shortest path from A to B must either be A to C and then a shortest path from C to B, or A to D and then a shortest path from D to B. Take B to have coordinates (m, n) relative to C or D.
In the first case, AB2 = (√3(m+n/2))2 + (3n/2 + 1)2 = (3m2 + 3mn + 3n2) + 3n + 1 and a shortest path has m+n units in direction 2. But 4(m + n)2 > (3m2 + 3mn + 3n2) + 3n + 1, if m2 + n2 + 5mn > 3n + 1, which is true for m, n at least 1. If m=0 and n=1, then a shortest path has 2 units in direction 1 and AB = √7 < 4. If m=1 and n=0, then AB=2 and a shortest path has 1 unit in each direction. So in this case (the only one so far) we have equality.
It remains to consider the case where the path starts out towards D. In this case AB2 = (√3(m+n/2) + √3/2)2 + (3n/2 -1/2)2 = (3m2+3mn+3n2) + 3m + 1 and a path has m + n units in direction 2. But 4(m + n)2 > (3m2 + 3mn + 3n2) + 3m + 1 for m2+n2 + 5mn > 3m + 1, which is true for m, n at least 1. If m=1, n=0, then a shortest path has 2 units in direction 3 and AB = √7 < 4. Finally, if m=0 and n=1, then a shortest path has 1 unit in each direction and AB = 2.
Thus the answer to the final question is 3, because the only cases where the bug travels exactly AB/2 in one direction are where it goes to the opposite vertex of a hexagon it is on.
Problem 10
A circle center O is inscribed in ABCD (touching every side). Prove that angle AOB + angle COD equals 180 degrees.
Solution
Let AB touch the circle at W, BC at X, CD at Y, and DA at Z. Then AO bisects angle ZOW and BO bisects angle XOW. So angle AOB is half angle ZOX. Similarly angle COD is half angle XOZ and hence angle AOB + angle COD equals 180.
Problem 11
The natural numbers a, b, n are such that for every natural number k not equal to b, b - k divides a - kn. Prove that a = bn.
Solution
We have kn - a = bn - a (mod b - k). Hence bn - a = 0 (mod b- k) for every k not equal to b. But if bn does not equal a, then by taking k - b > bn - a we could render the equation false.
Problem 12
How many (algebraically) different expressions can we obtain by placing parentheses in a1/a2/ ... /an?
Solution
Answer 2n-2. a1 must be in the numerator, and a2 must be in the denominator, but the other symbols can be in either. This is easily proved by induction.
Problem 13
What is the smallest number of tetrahedrons into which a cube can be partitioned?
Solution
Answer: 5.
Tetrahedral faces are triangular, so each cube face requires at least two tetrahedral faces. So at least 12 tetrahedral faces are needed in all. At most three faces of a tetrahedron can be mutually orthogonal (and no two faces can be parallel), so at most 3 faces from each tetrahedron can contribute towards these 12. So we require at least 4 tetrahedra to provide the cube faces. But these tetrahedra each have volume at most 1/6 (1/3 x face area x 1, and face area is at most 1/2). So if we have only 4 tetrahedra in total then their total volume is less than the cube's volume. Contradiction. Hence we need at least 5 tetrahedra.
It can be done with 5: lop off 4 non-adjacent corners to leave a tetrahedron. More precisely, take the cube as ABCDA'B'C'D' with ABCD horizontal, A' directly under A, B' directly under B and so on. Then the five tetrahedra are AA'BD, CC'BC, DD'A'C', BB'A'C', BDA'C'.
Problem 14
a) Find the smallest square with last digit not 0 which becomes another square (not zero) by the deletion of its last two digits. #(b) Find all squares, not containing the digits 0 or 5, such that if the second digit is deleted the resulting number divides the original one.
Solution
(a) This one must have slipped through: 121!
(b) Answer: 16,36,121,484. Suppose the number has more than 2 digits. Write it as (10m + n)10r + s, where 1 <= m <= 9, 0 <= n <= 9, 0 <= s < 10r. Then we have k(m.10r + s) = (10m +n)10r + s, for some k > 1.
s does not contain the digits 0 or 5, so 5 does not divide s. Hence 5 divides k-1, and so k must be 6, 11, or 16 (if k was 21 or more, then the rhs would be negative). Since 25 does not divide k-1, we must have r=1 and s is a single digit.
We look at each possibility for k in turn. k = 6 gives no solutions. k = 11 gives about two dozen multiples of 11 from 121 to 891. By inspection the only squares are 121 and 484. k = 16 gives 192, which is not a square.
In addition, there is the possibility of 2 digit solutions, which I had overlooked. It is easiest to check each of the 2 digit squares, thus finding the additional solutions 16, 36.
Problem 15
A circle is inscribed in ABCD. AB is parallel to CD, and BC = AD. The diagonals AC, BD meet at E. The circles inscribed in ABE, BCE, CDE, DAE have radius r1, r2, r3, r4 respectively. Prove that 1/r1 + 1/r3 = 1/r2 + 1/r4.
Solution
A necessary and sufficient condition for ABCD to have an inscribed circle is AB + CD = BC + AD. So we have AB + CD = 2AD, which we use repeatedly. Extend DC to X so that BX is parallel to EC. Then DX = AB + CD = 2AD and the triangles DEC, AEB, DBX are similar. Let h be the perpendicular distance from AB to CD. The similar triangles give us the heights of DEC and AEB in terms of h.
1/r1 = perimeter ABE/(2 area ABE) = (AB + 2EB)/(AB.height) = (AB + 2.BD.AB/(AB+CD))/(AB.h.AB/(AB+CD)) = 2(AD + BD)/(AB.h). Similarly, 1/r3 = 2(AD + BD)/(CD.h).
The area of AED = area ABD - area ABE = 1/2 AB.h.CD/(2AD), so 1/r2 = 1/r4 = perimeter ADE/(2 area ADE) = (AD + BD)/(h.AB.CD/2AD), and 1/r2 + 1/r4 = 2(AD + BD)/h 2AD/(AB.CD) = 2(AB + BD)/h (AB + CD)/(AB.CD) = 1/r1 + 1/r3.
Labels:
ARMO,
Russian Mathematical Olympiad
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