IMO 2016 Problem 3
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Let $P = A_1 A_2 cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, ldots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.
I tried using the formula for area of a polygon in cartesian plane, after assuming coordinates, but to no avail.
geometry elementary-number-theory contest-math analytic-geometry polygons
edited Jul 2 '18 at 7:45
rabota
14.3k32782
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add a comment |
$begingroup$
Let $P = A_1 A_2 cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, ldots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.
I tried using the formula for area of a polygon in cartesian plane, after assuming coordinates, but to no avail.
geometry elementary-number-theory contest-math analytic-geometry polygons
edited Jul 2 '18 at 7:45
rabota
14.3k32782
$endgroup$
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the solutions I have seen do it for $n=p^a$ with $p$ an odd prime and induction over $k$.
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– Jorge Fernández Hidalgo
Jul 11 '16 at 20:15
3
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Duplicate of math.stackexchange.com/questions/1855759/…
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– Arthur
Jul 11 '16 at 20:18
$begingroup$
link
$endgroup$
– Ricardo Largaespada
Nov 8 '16 at 5:02
add a comment |
$begingroup$
Let $P = A_1 A_2 cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, ldots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.
I tried using the formula for area of a polygon in cartesian plane, after assuming coordinates, but to no avail.
geometry elementary-number-theory contest-math analytic-geometry polygons
edited Jul 2 '18 at 7:45
rabota
14.3k32782
$endgroup$
Let $P = A_1 A_2 cdots A_k$ be a convex polygon in the plane. The vertices $A_1, A_2, ldots, A_k$ have integral coordinates and lie on a circle. Let $S$ be the area of $P$. An odd positive integer $n$ is given such that the squares of the side lengths of $P$ are integers divisible by $n$. Prove that $2S$ is an integer divisible by $n$.
I tried using the formula for area of a polygon in cartesian plane, after assuming coordinates, but to no avail.
geometry elementary-number-theory contest-math analytic-geometry polygons
geometry elementary-number-theory contest-math analytic-geometry polygons
edited Jul 2 '18 at 7:45
rabota
14.3k32782
edited Jul 2 '18 at 7:45
rabota
14.3k32782
edited Jul 2 '18 at 7:45
rabota
14.3k32782
edited Jul 2 '18 at 7:45
rabota
14.3k32782
edited Jul 2 '18 at 7:45
rabota
14.3k32782
14.3k32782
asked Jul 11 '16 at 19:59
user313479
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the solutions I have seen do it for $n=p^a$ with $p$ an odd prime and induction over $k$.
$endgroup$
– Jorge Fernández Hidalgo
Jul 11 '16 at 20:15
3
$begingroup$
Duplicate of math.stackexchange.com/questions/1855759/…
$endgroup$
– Arthur
Jul 11 '16 at 20:18
$begingroup$
link
$endgroup$
– Ricardo Largaespada
Nov 8 '16 at 5:02
add a comment |
$begingroup$
the solutions I have seen do it for $n=p^a$ with $p$ an odd prime and induction over $k$.
$endgroup$
– Jorge Fernández Hidalgo
Jul 11 '16 at 20:15
3
$begingroup$
Duplicate of math.stackexchange.com/questions/1855759/…
$endgroup$
– Arthur
Jul 11 '16 at 20:18
$begingroup$
link
$endgroup$
– Ricardo Largaespada
Nov 8 '16 at 5:02
$begingroup$
the solutions I have seen do it for $n=p^a$ with $p$ an odd prime and induction over $k$.
$endgroup$
– Jorge Fernández Hidalgo
Jul 11 '16 at 20:15
$begingroup$
the solutions I have seen do it for $n=p^a$ with $p$ an odd prime and induction over $k$.
$endgroup$
– Jorge Fernández Hidalgo
Jul 11 '16 at 20:15
3
3
$begingroup$
Duplicate of math.stackexchange.com/questions/1855759/…
$endgroup$
– Arthur
Jul 11 '16 at 20:18
$begingroup$
Duplicate of math.stackexchange.com/questions/1855759/…
$endgroup$
– Arthur
Jul 11 '16 at 20:18
$begingroup$
link
$endgroup$
– Ricardo Largaespada
Nov 8 '16 at 5:02
$begingroup$
link
$endgroup$
– Ricardo Largaespada
Nov 8 '16 at 5:02
add a comment |
1 Answer
1
active
oldest
votes
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Incomplete proof
Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.
Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.
If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.
Please figure out the case with $P$ truncated by some non-reducible triangles.
edited Jan 25 '17 at 18:55
nosyarg
352215
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Incomplete proof
Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.
Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.
If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.
Please figure out the case with $P$ truncated by some non-reducible triangles.
edited Jan 25 '17 at 18:55
nosyarg
352215
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
$endgroup$
add a comment |
$begingroup$
Incomplete proof
Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.
Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.
If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.
Please figure out the case with $P$ truncated by some non-reducible triangles.
edited Jan 25 '17 at 18:55
nosyarg
352215
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
$endgroup$
add a comment |
$begingroup$
Incomplete proof
Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.
Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.
If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.
Please figure out the case with $P$ truncated by some non-reducible triangles.
edited Jan 25 '17 at 18:55
nosyarg
352215
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
$endgroup$
Incomplete proof
Since $P$ is convex, we can uniquely truncate $P$ with right-angled triangles with hypotenuses equal to the side lengths of $P$ and the remaining rectilinears inside.
Let reducible triangle be a right-angled triangle with Pythagorean triple $(a,b,c)$ where both $a$,$b$ and $c$ are divisible by $n$.
If all triangles above truncating $P$ is reducible triangles, area of each triangle and rectilinear is divisible by $n^2$. Thus, $2S$ must be divisible by $n$.
Please figure out the case with $P$ truncated by some non-reducible triangles.
edited Jan 25 '17 at 18:55
nosyarg
352215
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
edited Jan 25 '17 at 18:55
nosyarg
352215
edited Jan 25 '17 at 18:55
nosyarg
352215
edited Jan 25 '17 at 18:55
nosyarg
352215
352215
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
answered Jul 19 '16 at 2:33
Jacky NgaiJacky Ngai
1
1
add a comment |
add a comment |
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$begingroup$
the solutions I have seen do it for $n=p^a$ with $p$ an odd prime and induction over $k$.
$endgroup$
– Jorge Fernández Hidalgo
Jul 11 '16 at 20:15
3
$begingroup$
Duplicate of math.stackexchange.com/questions/1855759/…
$endgroup$
– Arthur
Jul 11 '16 at 20:18
$begingroup$
link
$endgroup$
– Ricardo Largaespada
Nov 8 '16 at 5:02