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Mathematical proof regarding angle mirroring


Triangle and parallel lines, find angle xWould $angle{BAC} $ become $180^circ $ if we go on increasing the length of $AB$?Help Drawing Figure and Understanding ProofHelp Finishing Proof Involving Collinearity and Cyclic QuadsHelp Finishing Proof ?!In the following diagram of a triangle, AB = BC = CD and AD = BD. Find the measure of angle D.Find the area of the parallelogram given perimeter, one angle and the proportion of heights$P$ is a point on the angular bisector of $angle A$. Show that $frac{1}{AB}+frac{1}{AC}$ doesn't depend on the line through $P$Solving for the value of $ angle CEB$ - $frac{1}{4}$ $angle CBA$ where $E$ is an exterior point of $triangle ABC$Trisecting an angle $theta$ equally via applying trigonometry













0












$begingroup$


I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):




Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.




Let me show some examples to make this clearer.



Example 1



$beta = 45^circ$, $theta = 30^circ$



enter image description here



The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.



Example 2



$beta = 50^circ$, $theta = 15^circ$



enter image description here



In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.



I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
    $endgroup$
    – dmtri
    Mar 18 at 14:08












  • $begingroup$
    @dmtri Yes, $theta$ must lie between $0$ and $beta$.
    $endgroup$
    – Tendero
    Mar 18 at 14:10
















0












$begingroup$


I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):




Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.




Let me show some examples to make this clearer.



Example 1



$beta = 45^circ$, $theta = 30^circ$



enter image description here



The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.



Example 2



$beta = 50^circ$, $theta = 15^circ$



enter image description here



In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.



I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
    $endgroup$
    – dmtri
    Mar 18 at 14:08












  • $begingroup$
    @dmtri Yes, $theta$ must lie between $0$ and $beta$.
    $endgroup$
    – Tendero
    Mar 18 at 14:10














0












0








0





$begingroup$


I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):




Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.




Let me show some examples to make this clearer.



Example 1



$beta = 45^circ$, $theta = 30^circ$



enter image description here



The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.



Example 2



$beta = 50^circ$, $theta = 15^circ$



enter image description here



In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.



I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?










share|cite|improve this question











$endgroup$




I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):




Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.




Let me show some examples to make this clearer.



Example 1



$beta = 45^circ$, $theta = 30^circ$



enter image description here



The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.



Example 2



$beta = 50^circ$, $theta = 15^circ$



enter image description here



In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.



I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?







geometry proof-writing polar-coordinates angle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 18 at 15:38







Tendero

















asked Mar 18 at 13:58









TenderoTendero

378214




378214












  • $begingroup$
    Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
    $endgroup$
    – dmtri
    Mar 18 at 14:08












  • $begingroup$
    @dmtri Yes, $theta$ must lie between $0$ and $beta$.
    $endgroup$
    – Tendero
    Mar 18 at 14:10


















  • $begingroup$
    Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
    $endgroup$
    – dmtri
    Mar 18 at 14:08












  • $begingroup$
    @dmtri Yes, $theta$ must lie between $0$ and $beta$.
    $endgroup$
    – Tendero
    Mar 18 at 14:10
















$begingroup$
Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
$endgroup$
– dmtri
Mar 18 at 14:08






$begingroup$
Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
$endgroup$
– dmtri
Mar 18 at 14:08














$begingroup$
@dmtri Yes, $theta$ must lie between $0$ and $beta$.
$endgroup$
– Tendero
Mar 18 at 14:10




$begingroup$
@dmtri Yes, $theta$ must lie between $0$ and $beta$.
$endgroup$
– Tendero
Mar 18 at 14:10










1 Answer
1






active

oldest

votes


















2












$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
    $endgroup$
    – Tendero
    Mar 18 at 16:13














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1 Answer
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1 Answer
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$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
    $endgroup$
    – Tendero
    Mar 18 at 16:13


















2












$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
    $endgroup$
    – Tendero
    Mar 18 at 16:13
















2












2








2





$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.






share|cite|improve this answer









$endgroup$



Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 18 at 15:54









Christian BlatterChristian Blatter

176k8115327




176k8115327












  • $begingroup$
    Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
    $endgroup$
    – Tendero
    Mar 18 at 16:13




















  • $begingroup$
    Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
    $endgroup$
    – Tendero
    Mar 18 at 16:13


















$begingroup$
Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
$endgroup$
– Tendero
Mar 18 at 16:13






$begingroup$
Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
$endgroup$
– Tendero
Mar 18 at 16:13




















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