Mixing
Understand proofs concerning angles on circles
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I am reading the paper "On zeros of convex combination of polynomials" by Fell. In proving a theorem, the author listed two lemmas (Lemma 2 and Lemma 3) without proof.
How do we prove the two lemmas?
geometry angle
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
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add a comment |
$begingroup$
I am reading the paper "On zeros of convex combination of polynomials" by Fell. In proving a theorem, the author listed two lemmas (Lemma 2 and Lemma 3) without proof.
How do we prove the two lemmas?
geometry angle
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
$endgroup$
add a comment |
$begingroup$
I am reading the paper "On zeros of convex combination of polynomials" by Fell. In proving a theorem, the author listed two lemmas (Lemma 2 and Lemma 3) without proof.
How do we prove the two lemmas?
geometry angle
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
$endgroup$
I am reading the paper "On zeros of convex combination of polynomials" by Fell. In proving a theorem, the author listed two lemmas (Lemma 2 and Lemma 3) without proof.
How do we prove the two lemmas?
geometry angle
geometry angle
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
asked Feb 1 at 23:33
MyCindy2012MyCindy2012
12911
12911
add a comment |
add a comment |
1 Answer
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I'd prove the first one by observing that we can assume (by coordinate change) that $c$ is the origin, and by rotation, that $p$ is the point $(s, 0)$ on the $x$-axis, with $s > 1$. Further, we know that $a = (cos alpha, sin alpha)$ and $b = (cos beta, sin beta)$ for some angles $alpha$ and $beta$. Clearly angle $APB$ will be maximal if $alpha$ and $beta$ have opposite signs and are between $0$ and $pi/2$.
At this point, I'd compute the angle $APB$ --- call it $u$ --- in terms of $alpha$ and $beta$, with a little trig. Then I'd set
$$
frac{partial u(alpha, beta)}{partial alpha} = frac{partial u(alpha, beta)}{partial beta} = 0
$$
and solve them to show that in fact $alpha = beta$.
For the second, I'd again place the circle at the origin, and $a$ and $b$ at opposite angles $pm alpha$, so that the segment $ab$ is vertical. Then $p$ and $p'$ are $(pm 1, 0)$, and the angles to be summed are computable from $alpha$ and a little trig. Differentiate the sum with respect to $alpha$ and set to $0$, and discover that the max occurs when $cos(alpha) = 0$, so that $a$ and $b$ are opposite ends of the vertical diameter.
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
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1 Answer
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1 Answer
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$begingroup$
I'd prove the first one by observing that we can assume (by coordinate change) that $c$ is the origin, and by rotation, that $p$ is the point $(s, 0)$ on the $x$-axis, with $s > 1$. Further, we know that $a = (cos alpha, sin alpha)$ and $b = (cos beta, sin beta)$ for some angles $alpha$ and $beta$. Clearly angle $APB$ will be maximal if $alpha$ and $beta$ have opposite signs and are between $0$ and $pi/2$.
At this point, I'd compute the angle $APB$ --- call it $u$ --- in terms of $alpha$ and $beta$, with a little trig. Then I'd set
$$
frac{partial u(alpha, beta)}{partial alpha} = frac{partial u(alpha, beta)}{partial beta} = 0
$$
and solve them to show that in fact $alpha = beta$.
For the second, I'd again place the circle at the origin, and $a$ and $b$ at opposite angles $pm alpha$, so that the segment $ab$ is vertical. Then $p$ and $p'$ are $(pm 1, 0)$, and the angles to be summed are computable from $alpha$ and a little trig. Differentiate the sum with respect to $alpha$ and set to $0$, and discover that the max occurs when $cos(alpha) = 0$, so that $a$ and $b$ are opposite ends of the vertical diameter.
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
I'd prove the first one by observing that we can assume (by coordinate change) that $c$ is the origin, and by rotation, that $p$ is the point $(s, 0)$ on the $x$-axis, with $s > 1$. Further, we know that $a = (cos alpha, sin alpha)$ and $b = (cos beta, sin beta)$ for some angles $alpha$ and $beta$. Clearly angle $APB$ will be maximal if $alpha$ and $beta$ have opposite signs and are between $0$ and $pi/2$.
At this point, I'd compute the angle $APB$ --- call it $u$ --- in terms of $alpha$ and $beta$, with a little trig. Then I'd set
$$
frac{partial u(alpha, beta)}{partial alpha} = frac{partial u(alpha, beta)}{partial beta} = 0
$$
and solve them to show that in fact $alpha = beta$.
For the second, I'd again place the circle at the origin, and $a$ and $b$ at opposite angles $pm alpha$, so that the segment $ab$ is vertical. Then $p$ and $p'$ are $(pm 1, 0)$, and the angles to be summed are computable from $alpha$ and a little trig. Differentiate the sum with respect to $alpha$ and set to $0$, and discover that the max occurs when $cos(alpha) = 0$, so that $a$ and $b$ are opposite ends of the vertical diameter.
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
I'd prove the first one by observing that we can assume (by coordinate change) that $c$ is the origin, and by rotation, that $p$ is the point $(s, 0)$ on the $x$-axis, with $s > 1$. Further, we know that $a = (cos alpha, sin alpha)$ and $b = (cos beta, sin beta)$ for some angles $alpha$ and $beta$. Clearly angle $APB$ will be maximal if $alpha$ and $beta$ have opposite signs and are between $0$ and $pi/2$.
At this point, I'd compute the angle $APB$ --- call it $u$ --- in terms of $alpha$ and $beta$, with a little trig. Then I'd set
$$
frac{partial u(alpha, beta)}{partial alpha} = frac{partial u(alpha, beta)}{partial beta} = 0
$$
and solve them to show that in fact $alpha = beta$.
For the second, I'd again place the circle at the origin, and $a$ and $b$ at opposite angles $pm alpha$, so that the segment $ab$ is vertical. Then $p$ and $p'$ are $(pm 1, 0)$, and the angles to be summed are computable from $alpha$ and a little trig. Differentiate the sum with respect to $alpha$ and set to $0$, and discover that the max occurs when $cos(alpha) = 0$, so that $a$ and $b$ are opposite ends of the vertical diameter.
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
$endgroup$
I'd prove the first one by observing that we can assume (by coordinate change) that $c$ is the origin, and by rotation, that $p$ is the point $(s, 0)$ on the $x$-axis, with $s > 1$. Further, we know that $a = (cos alpha, sin alpha)$ and $b = (cos beta, sin beta)$ for some angles $alpha$ and $beta$. Clearly angle $APB$ will be maximal if $alpha$ and $beta$ have opposite signs and are between $0$ and $pi/2$.
At this point, I'd compute the angle $APB$ --- call it $u$ --- in terms of $alpha$ and $beta$, with a little trig. Then I'd set
$$
frac{partial u(alpha, beta)}{partial alpha} = frac{partial u(alpha, beta)}{partial beta} = 0
$$
and solve them to show that in fact $alpha = beta$.
For the second, I'd again place the circle at the origin, and $a$ and $b$ at opposite angles $pm alpha$, so that the segment $ab$ is vertical. Then $p$ and $p'$ are $(pm 1, 0)$, and the angles to be summed are computable from $alpha$ and a little trig. Differentiate the sum with respect to $alpha$ and set to $0$, and discover that the max occurs when $cos(alpha) = 0$, so that $a$ and $b$ are opposite ends of the vertical diameter.
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
answered Feb 1 at 23:48
John HughesJohn Hughes
65.4k24293
65.4k24293
add a comment |
add a comment |
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