How many obtuse angles can be formed from the 15 rays on a single point on a same plane?
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Consider 15 rays that originate from a point. What is the maximum number of obtuse angles they can form, assuming that the angles between two rays is less than or equal to 180 degrees?
geometry
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
$endgroup$
add a comment |
$begingroup$
Consider 15 rays that originate from a point. What is the maximum number of obtuse angles they can form, assuming that the angles between two rays is less than or equal to 180 degrees?
geometry
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
$endgroup$
add a comment |
$begingroup$
Consider 15 rays that originate from a point. What is the maximum number of obtuse angles they can form, assuming that the angles between two rays is less than or equal to 180 degrees?
geometry
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
$endgroup$
Consider 15 rays that originate from a point. What is the maximum number of obtuse angles they can form, assuming that the angles between two rays is less than or equal to 180 degrees?
geometry
geometry
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
asked Dec 9 '18 at 3:03
Heroic24Heroic24
1577
1577
add a comment |
add a comment |
1 Answer
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Call two rays near if they form a non-obtuse angle. If there are $n$ ordered pairs of near rays, there are exactly ${15choose 2}-frac n2$ obtuse angles among the rays. Hence we want to minimize $n$.
Claim. $nge 60$.
Proof. Suppose one of the rays (wlog the positive $x$ axis) is near $ale 2$ other rays (i.e., we have $a$ other rays within the (closed) first or fourth quadrant). Then there are $b$ rays in the second and $c$ rays in the third quadrant (with the negative $x$ axis being counted as either of these quadrants) where $b+cge12$.
Then we have (e.g., by Jensen's inequality) $$nge b(b-1)+c(c-1)ge 2cdot 6cdot 5=30.$$
Therefore, we need only consider configurations where each ray is near $ge 3$ other rays.
Again, suppose the some ray is near $a=3$ other rays, and define $b$ and $c$ as above, where now $b+c=11$. This time, we have
$ge b(b-1)+c(c-1)=(b-5)^2+(b-6)^2+49ge50$ near pairs within the left half plane. Additionally, each of the $4$ rays in the right half plane is first component of at least $3$ near pairs. Hence,
$$ nge 50+4cdot 3=62.$$
Remains the case that each ray is near at least $4$ other rays. Then clearly,
$$nge 4cdot 15=60.$$
$square$
As we can achieve the lower bound $n=60$ (e.g., with five rays per each of the directions $0^circ$, $120^circ$, $240^circ$), the maximal number of obtuse angles is
$$ 75.$$
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
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1 Answer
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1 Answer
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$begingroup$
Call two rays near if they form a non-obtuse angle. If there are $n$ ordered pairs of near rays, there are exactly ${15choose 2}-frac n2$ obtuse angles among the rays. Hence we want to minimize $n$.
Claim. $nge 60$.
Proof. Suppose one of the rays (wlog the positive $x$ axis) is near $ale 2$ other rays (i.e., we have $a$ other rays within the (closed) first or fourth quadrant). Then there are $b$ rays in the second and $c$ rays in the third quadrant (with the negative $x$ axis being counted as either of these quadrants) where $b+cge12$.
Then we have (e.g., by Jensen's inequality) $$nge b(b-1)+c(c-1)ge 2cdot 6cdot 5=30.$$
Therefore, we need only consider configurations where each ray is near $ge 3$ other rays.
Again, suppose the some ray is near $a=3$ other rays, and define $b$ and $c$ as above, where now $b+c=11$. This time, we have
$ge b(b-1)+c(c-1)=(b-5)^2+(b-6)^2+49ge50$ near pairs within the left half plane. Additionally, each of the $4$ rays in the right half plane is first component of at least $3$ near pairs. Hence,
$$ nge 50+4cdot 3=62.$$
Remains the case that each ray is near at least $4$ other rays. Then clearly,
$$nge 4cdot 15=60.$$
$square$
As we can achieve the lower bound $n=60$ (e.g., with five rays per each of the directions $0^circ$, $120^circ$, $240^circ$), the maximal number of obtuse angles is
$$ 75.$$
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
$endgroup$
add a comment |
$begingroup$
Call two rays near if they form a non-obtuse angle. If there are $n$ ordered pairs of near rays, there are exactly ${15choose 2}-frac n2$ obtuse angles among the rays. Hence we want to minimize $n$.
Claim. $nge 60$.
Proof. Suppose one of the rays (wlog the positive $x$ axis) is near $ale 2$ other rays (i.e., we have $a$ other rays within the (closed) first or fourth quadrant). Then there are $b$ rays in the second and $c$ rays in the third quadrant (with the negative $x$ axis being counted as either of these quadrants) where $b+cge12$.
Then we have (e.g., by Jensen's inequality) $$nge b(b-1)+c(c-1)ge 2cdot 6cdot 5=30.$$
Therefore, we need only consider configurations where each ray is near $ge 3$ other rays.
Again, suppose the some ray is near $a=3$ other rays, and define $b$ and $c$ as above, where now $b+c=11$. This time, we have
$ge b(b-1)+c(c-1)=(b-5)^2+(b-6)^2+49ge50$ near pairs within the left half plane. Additionally, each of the $4$ rays in the right half plane is first component of at least $3$ near pairs. Hence,
$$ nge 50+4cdot 3=62.$$
Remains the case that each ray is near at least $4$ other rays. Then clearly,
$$nge 4cdot 15=60.$$
$square$
As we can achieve the lower bound $n=60$ (e.g., with five rays per each of the directions $0^circ$, $120^circ$, $240^circ$), the maximal number of obtuse angles is
$$ 75.$$
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
$endgroup$
add a comment |
$begingroup$
Call two rays near if they form a non-obtuse angle. If there are $n$ ordered pairs of near rays, there are exactly ${15choose 2}-frac n2$ obtuse angles among the rays. Hence we want to minimize $n$.
Claim. $nge 60$.
Proof. Suppose one of the rays (wlog the positive $x$ axis) is near $ale 2$ other rays (i.e., we have $a$ other rays within the (closed) first or fourth quadrant). Then there are $b$ rays in the second and $c$ rays in the third quadrant (with the negative $x$ axis being counted as either of these quadrants) where $b+cge12$.
Then we have (e.g., by Jensen's inequality) $$nge b(b-1)+c(c-1)ge 2cdot 6cdot 5=30.$$
Therefore, we need only consider configurations where each ray is near $ge 3$ other rays.
Again, suppose the some ray is near $a=3$ other rays, and define $b$ and $c$ as above, where now $b+c=11$. This time, we have
$ge b(b-1)+c(c-1)=(b-5)^2+(b-6)^2+49ge50$ near pairs within the left half plane. Additionally, each of the $4$ rays in the right half plane is first component of at least $3$ near pairs. Hence,
$$ nge 50+4cdot 3=62.$$
Remains the case that each ray is near at least $4$ other rays. Then clearly,
$$nge 4cdot 15=60.$$
$square$
As we can achieve the lower bound $n=60$ (e.g., with five rays per each of the directions $0^circ$, $120^circ$, $240^circ$), the maximal number of obtuse angles is
$$ 75.$$
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
$endgroup$
Call two rays near if they form a non-obtuse angle. If there are $n$ ordered pairs of near rays, there are exactly ${15choose 2}-frac n2$ obtuse angles among the rays. Hence we want to minimize $n$.
Claim. $nge 60$.
Proof. Suppose one of the rays (wlog the positive $x$ axis) is near $ale 2$ other rays (i.e., we have $a$ other rays within the (closed) first or fourth quadrant). Then there are $b$ rays in the second and $c$ rays in the third quadrant (with the negative $x$ axis being counted as either of these quadrants) where $b+cge12$.
Then we have (e.g., by Jensen's inequality) $$nge b(b-1)+c(c-1)ge 2cdot 6cdot 5=30.$$
Therefore, we need only consider configurations where each ray is near $ge 3$ other rays.
Again, suppose the some ray is near $a=3$ other rays, and define $b$ and $c$ as above, where now $b+c=11$. This time, we have
$ge b(b-1)+c(c-1)=(b-5)^2+(b-6)^2+49ge50$ near pairs within the left half plane. Additionally, each of the $4$ rays in the right half plane is first component of at least $3$ near pairs. Hence,
$$ nge 50+4cdot 3=62.$$
Remains the case that each ray is near at least $4$ other rays. Then clearly,
$$nge 4cdot 15=60.$$
$square$
As we can achieve the lower bound $n=60$ (e.g., with five rays per each of the directions $0^circ$, $120^circ$, $240^circ$), the maximal number of obtuse angles is
$$ 75.$$
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
answered Dec 9 '18 at 5:10
Hagen von EitzenHagen von Eitzen
278k23269501
278k23269501
add a comment |
add a comment |
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