This is a question for olmpiad

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Yashwanth

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C and D are two points on the same side of straight line AB. Find a point X on AB such that the angles CXA and DXB are equal.
 
Draw a picture of what a solution would look like, so we can have something to talk about.

Then you might think about whatever you might know about optics. Does the picture remind you of something about light rays?
 
Yashwanth said:
Please explain the solution
Click to expand...
That would not be what we consider to be helpful to you. We prefer that you explain the solution to us with our guidance. So please post a drawing and state what you think you need to do. Thanks.
 
Why not draw the circle so it includes point x as well? See where that gets you.
 
The circle is just to indicate where the reflection of [MATH]D[/MATH] is.
 
I don't know what you mean, please explain it in this new diagram. How can we include point [MATH]x[/MATH] in that circle
1584635373638.png
 
I think the key is just the word "reflection", which I was hoping to elicit. The only problem I see with the drawing is that it suggests that the circle plays some specific role in solving the problem, which I don't think it does. Of course, we don't want to give an actual solution yet, so showing a complete construction wouldn't have been appropriate. I would just have drawn D' and omitted the circle, if I wanted to go that far.

Unfortunately, it isn't clear from the question how you are required to "find" X (e.g. with a compass and straightedge construction, or something else).

@Yashwanth, do you have any ideas yet for solving the problem?
 
The circle is just to help us to find the point [MATH]D'[/MATH] and nothing else. To find the reflection of [MATH]D[/MATH] we need to drop a perpendicular from [MATH]D[/MATH] and continue it with the same distance. (two radii). The rest is trivial.
 
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