### 12th APMO 2000
**Problem 1**
Find a_{1}^{3}/(1 - 3a_{1} + 3a_{1}^{2}) + a_{2}^{3}/(1 - 3a_{2} + 3a_{2}^{2}) + ... + a_{101}^{3}/(1 - 3a_{101} + 3a_{101}^{2}), where a_{n} = n/101.
**Solution**
Answer: 51.
The nth term is a_{n}^{3}/(1 - 3a_{n} + 3a_{n}^{2}) = a_{n}^{3}/( (1 - a_{n})^{3} + a_{n}^{3}) = n^{3}/( (101 - n)^{3} + n^{3}). Hence the sum of the nth and (101-n)th terms is 1. Thus the sum from n = 1 to 100 is 50. The last term is 1, so the total sum is 51.
**Problem 2**
Find all permutations a_{1}, a_{2}, ... , a_{9} of 1, 2, ... , 9 such that a_{1} + a_{2} + a_{3} + a_{4} = a_{4} + a_{5} + a_{6} + a_{7} = a_{7} + a_{8} + a_{9} + a_{1} and a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} = a_{4}^{2} + a_{5}^{2} + a_{6}^{2} + a_{7}^{2} = a_{7}^{2} + a_{8}^{2} + a_{9}^{2} + a_{1}^{2}.
**Solution**
We may start by assuming that a_{1} < a_{4} < a_{7} and that a_{2} < a_{3}, a_{5} < a_{6}, a_{8} < a_{9}.
Note that 1 + ... + 9 = 45 and 1^{2} + ... + 9^{2} = 285. Adding the three square equations together we get (a_{1}^{2} + ... + a_{9}^{2}) + a_{1}^{2} + a_{4}^{2} + a_{7}^{2} = 285 + a_{1}^{2} + a_{4}^{2} + a_{7}^{2}. The total must be a multiple of 3. But 285 is a multiple of 3, so a_{1}^{2} + a_{4}^{2} + a_{7}^{2} must be a multiple of 3. Now 3^{2}, 6^{2} and 9^{2} are all congruent to 0 mod 3 and the other squares are all congruent to 1 mod 3. Hence either a_{1}, a_{4} and a_{7} are all multiples of 3, or none of them are. Since 45 is also a multiple of three a similar argument with the three linear equations shows that a_{1} + a_{4} + a_{7} is a multiple of 3. So if none of a_{1}, a_{4}, a_{7} are multiples of 3, then they are all congruent to 1 mod 3 or all congruent to 2 mod 3. Thus we have three cases: (1) a_{1} = 3, a_{4} = 6, a_{7} = 9, (2) a_{1} = 1, a_{4} = 4, a_{7} = 7, and (3) a_{1} = 2, a_{4} = 5, a_{7} = 8.
In case (1), we have that each sum of squares equals 137. Hence a_{8}^{2} + a_{9}^{2} = 47. But 47 is not a sum of two squares, so this case gives no solutions.
In case (2), we have that each sum of squares is 117. Hence a_{5}^{2} + a_{6}^{2} = 52. But the only way of writing 52 as a sum of two squares is 4^{2} + 6^{2} and 4 is already taken by a_{4}, so this case gives no solutions.
In case (3), we have that each sum of squares is 126 and each linear sum 20. We quickly find that the only solution is 2, 4, 9, 5, 1, 6, 8, 3, 7.
Obviously, this generates a large number of equivalent solutions. We can interchange a_{2} and a_{3}, or a_{5} and a_{6}, or a_{8} and a_{9}. We can also permute a_{1}, a_{4} and a_{7}. So we get a total of 2 x 2 x 2 x 6 =48 solutions.
**Problem 3**
ABC is a triangle. The angle bisector at A meets the side BC at X. The perpendicular to AX at X meets AB at Y. The perpendicular to AB at Y meets the ray AX at R. XY meets the median from A at S. Prove that RS is perpendicular to BC.
**Solution**
Let the line through C parallel to AX meet the ray BA at C'. Let the perpendicular from B meet the ray C'C at T and the ray AX at U. Let the line from C parallel to BT meet BA at V and let the perpendicular from V meet BT at W. So CVWT is a rectangle.
AU bisects ∠CAV and CV is perpendicular to AU, so U is the midpoint of WT. Hence the intersection N of AU and CW is the center of the rectangle and, in particular, the midpoint of CW. Let M be the midpoint of BC. Then since M, N are the midpoints of the sides CB and CW of the triangle CBW, MN = BW/2.
Since CC' is parallel to AX, ∠CC'A = ∠BAX = ∠CAX = ∠C'CA, so AC' = AC. Let A' be the midpoint of CC'. Then AU = C'T - C'A'. But N is the center of the rectangle CTWV, so NU = CT/2 and AN = AU - NU = C'T - C'A' - CT/2 = C'T/2. Hence MN/AN = BW/C'T. But MN is parallel to BW and XY, so SX/AX = MN/AN = BW/C'T.
Now AX is parallel to VW and XY is parallel to BW, so AXY and VWB are similar and AX/XY = VW/BW = CT/BW. Hence SX/XY = (SX/AX) (AX/XY) = CT/C'T.
YX is an altitude of the right-angled triangle AXR, so AXY and YXR are similar. Hence XY/XR = XA/XY. But AXY and C'TB are similar, so XA/XY = C'T/BT. Hence SX/XR = (SX/XY) (XY/XR) = (CT/C'T) (C'T/BT) = CT/BT. But angles CTB and SXR are both right angles, so SXR and CTB are similar. But XR is perpendicular to BT, so SR is perpendicular to BC.
**Problem 4**
If m < n are positive integers prove that n^{n}/(m^{m} (n-m)^{n-m}) > n!/( m! (n-m)! ) > n^{n}/( m^{m}(n+1) (n-m)^{n-m}).
**Solution**
The key is to consider the binomial expansion (m + n-m)^{n}. This is a sum of positive terms, one of which is nCm m^{m}(n-m)^{n-m}, where nCm is the binomial coefficient n!/( m! (n-m)! ). Hence nCm m^{m}(n-m)^{n-m} < n^{n}, which is one of the required inequalities.
We will show that nCm m^{m}(n-m)^{n-m} is the largest term in the binomial expansion. It then follows that (n+1) nCm m^{m}(n-m)^{n-m} > n^{n}, which is the other required inequality.
Comparing the rth term nCr m^{r}(n-m)^{n-r} with the r+1th term nCr+1 m^{r+1}(n-m)^{n-r-1} we see that the rth term is strictly larger for r ≥ m and smaller for r < m. Hence the mth term is larger than the succeeding terms and also larger than the preceding terms.
**Problem 5**
Given a permutation s_{0}, s_{2}, ... , s_{n} of 0, 1, 2, .... , n, we may *transform* it if we can find i, j such that s_{i} = 0 and s_{j} = s_{i-1} + 1. The new permutation is obtained by transposing s_{i} and s_{j}. For which n can we obtain (1, 2, ... , n, 0) by repeated transformations starting with (1, n, n-1, .. , 3, 2, 0)?
**Solution**
Experimentation shows that we can do it for n=1 (already there), n = 2 (already there), 3, 7, 15, but not for n = 4, 5, 6, 8, 9, 10, 11, 12, 13, 14. So we conjecture that it is possible just for n = 2^{m} - 1 and for n = 2.
Notice that there is at most one transformation possible. If n = 2m, then we find that after m-1 transformations we reach
1 n 0 n-2 n-1 n-4 n-3 ... 4 5 2 3
and we can go no further. So n even fails for n > 2.
If n = 15 we get successively:
1 15 14 13 12 11 10 9 8 7 6 5 4 3 2 0 start
1 0 14 15 12 13 10 11 8 9 6 7 4 5 2 3 after 7 moves
1 2 3 0 12 13 14 15 8 9 10 11 4 5 6 7 after 8 more moves
1 2 3 4 5 6 7 0 8 9 10 11 12 13 14 15 after 8 more moves
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 after 8 more moves
This pattern is general. Suppose n = 2^{m} - 1. Let P_{0} be the starting position and P_{r} be the position:
1 2 3 ... R-1 0, n-R+1 n-R+2 n-R+3 ... n, n-2R+1 n-2R+2 ... n-R, ... , R R+1 ... 2R-1
Here R denotes 2^{r} and the commas highlight that, after the initial 1 2 ... R-1 0, we have increasing runs of R terms. If we start from P_{r}, then the 0 is transposed successively with R, 3R, 5R, ... , n-R+1, then with R+1, 3R+1, ... , n-R+2, and so on up to 2R-1, 4R-1, ... , n. But that gives P_{r+1}. It is also easy to check that P_{0} leads to P_{1} and that P_{m} is the required finishing position. Thus the case n = 2^{m} - 1 works.
Now suppose n is odd but not of the form 2^{m} - 1. Then we can write n = (2a + 1)2^{b} - 1 (just take 2^{b} as the highest power of 2 dividing n + 1). We can now define P_{0}, P_{1}, ... , P_{b} as before. As before we will reach P_{b}:
1 2 ¼ B-1 0, 2aB 2aB+1¼ (2a+1)B-1, (2a-1)B ¼ 2aB-1, ¼ , 3B, 3B+1, ¼ 4B-1, 2B, 2B+1, ¼ , 3B-1, B, B+1, ¼ , 2B-1
where B = 2^{b} - 1. But then the 0 is transposed successively with B, 3B, 5B, ... , (2a-1)B, which puts it immediately to the right of (2a+1)B-1 = n, so no further transformations are possible and n = (2a+1)2^{b} - 1 fails.
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