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I have the following exercise to solve:

A (golf) player hits the ball at B, the ball touches line a exactly once and goes into the whole L. How does the ball's way look like?

My idea has been the following (and I have forgotten most of my basic school geometry): The incoming angle of the ball must be equal to the outgoing one, hence I was just dividing the distance from B to L to determine where the ball must hit a. But somehow it feels that I am missing something, but I don't know what...any help appreciated.

enter image description here

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HINT:

draw $L'$, reflection of $L$ across $a$, and join $B$ with $L'$.

To see why that works, see the diagram below: right triangles $ALH$ and $AL'H$ are congruent (SAS), whence $\angle LAH \cong \angle L'AH$. On the other hand we have $\angle L'AH \cong \angle BAa$ for they are vertical angles. It follows that $\angle LAH \cong \angle BAa$, as required.

enter image description here

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  • $\begingroup$ Many thanks. I have seen this solution, but why is my approach wrong, and also I would like to understand why your solution is the suggested one - what are the underlying rules and laws applied? I mean I assume that connecting B with L' will ensure that the incoming and outgoing angle of the ball hiting a are equal, but from what does that follow? Many thanks $\endgroup$ – Pegah Nov 20 '16 at 9:17
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    $\begingroup$ I'm only suggesting a practical method to construct the bouncing point. To understand why it works only a little geometry is needed: mainly vertical angles and congruence criteria. I can provide a full explanation if needed. $\endgroup$ – Aretino Nov 20 '16 at 11:06
  • $\begingroup$ I would really appreciate that! $\endgroup$ – Pegah Nov 20 '16 at 14:50
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    $\begingroup$ I updated my answer. $\endgroup$ – Aretino Nov 20 '16 at 16:39
  • $\begingroup$ ..now I don't even know what I did not get:) Maaany thanks! $\endgroup$ – Pegah Nov 20 '16 at 17:21

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