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### Reflection Points on a Circle-Shaped Mirror

```Date: 09/30/2003 at 06:46:16
From: Kees
Subject: reflection points on a circle-shaped mirror

Dear Dr. Math,

I cannot construct the location of the reflection points P on a
circle-shaped mirror for two points A and B.  A and B are located
within the circle, the reflection points are located on the circle.
If A were a light emitting point and B a light receiving point, then
B would receive light from points on the circle where the angle of
the tangent of the circle in P with the incoming light ray would be
equal to the reflected light beam with the tangent in P.

Is it possible to construct these points?

Is it possible to derive an equation that describes the location of
these points?

By computer-aided trying I found that there are 2, 3, or 4 points P
for a given set of A and B.

Constructing reflection points for a straight line mirror is so
simple, yet this is seems to be so complex!

I found out that the bisection of the angle APB intersects with the
centre of the circle.  I tried to solve it by using a cosine formula
for the sharp angles between APM and BPM (M:circle centre) and
formulating px of P(px,py) as sqrt(radius^2-py^2) but then it becomes
a very complex equation that I cannot solve.
```

```
Date: 10/01/2003 at 07:36:11
From: Doctor Floor
Subject: Re: reflection points on a circle-shaped mirror

Hi, Kees,

I assume that the ray of light is only reflected once.

I am quite unsure whether there is a construction (by ruler and

In computational sense: If you take the origin at the circumcenter M,
then the lines MP are given by their slopes. The reflection of A
through MP has to be on BP. It shouldn't be too difficult to compute
this. This condition should give you a key to find a general
description of points P, but it may be tedious. Good luck!

If you have more questions, just write back.

Best regards,

- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/01/2003 at 12:03:31
From: Kees
Subject: Thank you (reflection points on a circle-shaped mirror)

It is indeed only single reflections that are taken in account.  I
wrote a little program in Delphi that finds the points P by
"traveling" along the circle and calculating the angles to A, B and P.
It works but it is slow and not very elegant; if you think a
construction is not likely I am still hoping for an algebraic solution...
```

```
Date: 10/02/2003 at 09:03:52
From: Doctor Floor
Subject: Re: reflection points on a circle-shaped mirror

Hi, again, Kees,

Let's say that the circle is the unit circle and M is located at the
origin (0,0). The point P can be given by (t,u) with t^2 + u^2 = 1. Of
course, for some angle alpha we have t = cos(alpha) and u = sin(alpha).

Let B be on the x-axis, say (x_b,0). Let the coordinates of A be
(x_a,y_a).

Then the reflection of B through the line MP is

B'( (t^2 - u^2) * x_b, 2 * t * u * x_b ).

The points P, A and B' are collinear if the following determinant is
equal to zero:

|      t                          u        1 |
|     x_a                        y_a       1 | = 0
| (t^2 - u^2) * x_b     2 * t * u * x_b    1 |

We can expand this to

0 =   t * y_a
- u * t^2 * x_b
+ 2 * x_a * u * t * x_b
- x_a * u
- x_b * t^2 * y_a
- x_b * u^3
+ x_b * u^2 * y_a

Since we also have t^2 + u^2 = 1 we can substitute in the above
t^2 = 1 - u^2. Having done so we have an equation that is linear in t,
giving

t = -(-y_a * x_b - u * x_b - u * x_a + 2 * u^2 * y_a * x_b)/(y_a + 2
* u * x_a * x_b)

Substituting t back into

0 =   t * y_a
- u * t^2 * x_b
+ 2 * x_a * u * t * x_b
- x_a * u
- x_b * t^2 * y_a
- x_b * u^3
+ x_b * u^2 * y_a

and doing some rewriting (getting rid of denominators for instance)
we get the quartic equation:

0 =   4 * (x_b)^2 * ((y_a)^2 + (x_a)^2) * u^4
- 4(x_b)^2 * y_a * u^3
+ (2 * x_a * x_b - 4 * (x_b)^2 * (y_a)^2 + (x_b)^2 + (y_a)^2 +
(x_a)^2 - 4 * (x_a)^2 * (x_b)^2) * u^2
- 2 * x_b * y_a * (x_a - x_b) * u
+ (y_a)^2 * ((x_b)^2 - 1)

This shows that there must be four solutions for u, possibly not real,
and thus for sin(alpha). The accompanying t can be found by substition
into the above expression for t.

For u = 0, the value of the left hand side equals (y_a)^2((x_b)^2 -
1), which is negative because x_b < 1 (B lies in the interior of the
circle).

For u = 1, the value of the left hand side can be written as (x_b +
x_a - x_b * y_a)^2, which is nonnegative.

We find the same value for u = -1.

This shows there must be at least two real solutions.

This is as much as I can do for you now analytically.

If you have more questions, just write back.

Best regards,

- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Conic Sections/Circles
High School Euclidean/Plane Geometry

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