Problem 1. We are given \[2\le a\le b\le c\le d\le16,\]\[(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2,\text{ and}\]\[(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2\]and we want to check if there is one unique integer solution. Except there is not, as \((a,b,c,d)=(3,6,15,16),(3,7,10,12)\) are both solutions.

Problem 3. \(c=6\). Clearly \(y=\pm1\) cannot have a finite solution. For \(y=0\), \(x=-6\) is a solution. For \(x\ge5\), \(1<|y|<2\), so it suffices just to check \(1\le x\le4\). This gives only the solutions \((1,7)\) and \((4,-2)\), for a total of 3 solutions as desired.

Multiple solutions exist. It just asked you to find a solution. Also, 10 is incorrect, as it passes through (-10,0). Also, I'm posting solutions at my leisure.

@Calvin Lin
–
Yes!! This is exactly what I was thinking. Thanks
I found out 18 to be okay. Is there any sequence of numbers which satisfy the given conditions?

I think you should buy "mathematical olympiad primer" if you want to prepare for BMO1. I have it and this BMO1 paper is included there with complete solutions.

Problem 3 (Yes I know Cody already did a solution to this problem. Bad Cody.)

We claim that \(c = 10\) works.

Manipulating the given equation, we get \[\begin{aligned} xy^2-y^2-x+y &= 10 \\ (xy^2-x) - (y^2-y) &= 10 \\ x(y-1)(y+1) - y(y-1) &= 10 \\ (y-1)(xy+x-y) &= 10 \end{aligned}\] Since \(y \ge 1,\) we must have \(y - 1 \in \{1, 2, 5, 10\}.\)

Case 1: \(y - 1 = 1\) and \(xy+x-y = 10\)

From the first EQ, we get \(y = 2.\) substituting into the second EQ and solving gives \(x = 4,\) giving one solution in this case.

Case 2: \(y - 1 = 2\) and \(xy+x-y = 5\)

From the first EQ, we get \(y = 3.\) substituting into the second EQ and solving gives \(x = 2,\) giving one solution in this case.

Case 3: \(y - 1 = 5\) and \(xy+x-y = 2\)

From the first EQ, we get \(y = 6.\) substituting into the second EQ and solving gives \(x = \dfrac87,\) which is not an integer; thus there are no solutions in this case.

Case 4: \(y - 1 = 10\) and \(xy+x-y = 1\)

From the first EQ, we get \(y = 11.\) substituting into the second EQ and solving gives \(x = 1,\) giving one solution in this case.

In conclusion, setting \(c = 10\) gives three solutions, as requested. \(\square\)

Can anybody give me the hint how I should proceed to solve this question?
Q. In a very hotly fought battle, at least 70% of the soldiers lost an eye, at least 75% lost an ear ,at least 80% lost an arm ,at least 85% lost a leg.How many lost all four limbs?

Hints:
1) Bounding.
2) Similar triangles. In fact, first find an expression for tan(<APX)/tan(<PAX) to see what you need to prove is constant.
3) Experiment: First see if $c=p$ for a prime $p$ works. Then try $c=pq$ (p,q primes), etc. You will find $(p,q)=(2,5)$ works.
4) Bound! It boils down to showing that such $r$ exists for all $n\le 27$ (if i remember correctly)
5) What can $f(1)$ be? Once you've got that sorted, you'll need a clever insight, but once you find it the problem is just computation.

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## Comments

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TopNewestProblem 1. We are given \[2\le a\le b\le c\le d\le16,\]\[(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2,\text{ and}\]\[(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2\]and we want to check if there is one unique integer solution. Except there is not, as \((a,b,c,d)=(3,6,15,16),(3,7,10,12)\) are both solutions.

Problem 3. \(c=6\). Clearly \(y=\pm1\) cannot have a finite solution. For \(y=0\), \(x=-6\) is a solution. For \(x\ge5\), \(1<|y|<2\), so it suffices just to check \(1\le x\le4\). This gives only the solutions \((1,7)\) and \((4,-2)\), for a total of 3 solutions as desired.

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I have found it to be 10 @ (4,2) (2,3) and (1,11), is it correct?? And what have you found in Questions 1, 2, 4 and 5?

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Multiple solutions exist. It just asked you to find

asolution. Also, 10 is incorrect, as it passes through (-10,0). Also, I'm posting solutions at my leisure.Log in to reply

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nupur, you are 21 years old and you are allowed to sit for olympiads?? which olympiad by the way??

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I am training my Brother, 15, so I am posting on behalf of him. Good observation.....

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wew.... I think i cant reach this level.. I wish I can...

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Never mind...

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I think you should buy "mathematical olympiad primer" if you want to prepare for BMO1. I have it and this BMO1 paper is included there with complete solutions.

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What is BMO1?

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British Mathematical Olympiad Round 1

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No, I am Indian and I am going to give RMO(Indian exam similar to BMO) but just for practice I solved a BMO paper.

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Problem 3 (Yes I know Cody already did a solution to this problem. Bad Cody.)

We claim that \(c = 10\) works.

Manipulating the given equation, we get \[\begin{aligned} xy^2-y^2-x+y &= 10 \\ (xy^2-x) - (y^2-y) &= 10 \\ x(y-1)(y+1) - y(y-1) &= 10 \\ (y-1)(xy+x-y) &= 10 \end{aligned}\] Since \(y \ge 1,\) we must have \(y - 1 \in \{1, 2, 5, 10\}.\)

Case 1:\(y - 1 = 1\) and \(xy+x-y = 10\)From the first EQ, we get \(y = 2.\) substituting into the second EQ and solving gives \(x = 4,\) giving one solution in this case.

Case 2:\(y - 1 = 2\) and \(xy+x-y = 5\)From the first EQ, we get \(y = 3.\) substituting into the second EQ and solving gives \(x = 2,\) giving one solution in this case.

Case 3:\(y - 1 = 5\) and \(xy+x-y = 2\)From the first EQ, we get \(y = 6.\) substituting into the second EQ and solving gives \(x = \dfrac87,\) which is not an integer; thus there are no solutions in this case.

Case 4:\(y - 1 = 10\) and \(xy+x-y = 1\)From the first EQ, we get \(y = 11.\) substituting into the second EQ and solving gives \(x = 1,\) giving one solution in this case.

In conclusion, setting \(c = 10\) gives three solutions, as requested. \(\square\)

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This is exactly how I solved.

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Can anybody give me the hint how I should proceed to solve this question? Q. In a very hotly fought battle, at least 70% of the soldiers lost an eye, at least 75% lost an ear ,at least 80% lost an arm ,at least 85% lost a leg.How many lost all four limbs?

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It can't be determined exactly...

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Hints: 1) Bounding. 2) Similar triangles. In fact, first find an expression for tan(<APX)/tan(<PAX) to see what you need to prove is constant. 3) Experiment: First see if $c=p$ for a prime $p$ works. Then try $c=pq$ (p,q primes), etc. You will find $(p,q)=(2,5)$ works. 4) Bound! It boils down to showing that such $r$ exists for all $n\le 27$ (if i remember correctly) 5) What can $f(1)$ be? Once you've got that sorted, you'll need a clever insight, but once you find it the problem is just computation.

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Which book are you referring for RMO? I need help for my olympiad.

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