this picture:
shows a way to construct the inverse of a number $a\ge1$. but how can we construct for a number that is less than 1?
My try::
- Q1: is my try correct?
- Q2: how to prove them both?
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2$\begingroup$ The original diagram actually shows the process for $a \leq 1$ as well; simply relabel the points $a \leftrightarrow 1/a$. (Your version effectively duplicates this logic.) For proof, merely consider "long leg over short leg" proportion for the red right triangle in its two extreme positions: the big right triangle has "long-over-short" = $\frac{a}{1}$; the small right triangle has "long-over-short" = $\frac{1}{1/a}$. $\endgroup$– BlueCommented Jan 4, 2015 at 13:13
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$\begingroup$ By the way: Where did you find the original diagram? It's pretty neat, so the creator deserves some credit. $\endgroup$– BlueCommented Jan 4, 2015 at 13:17
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$\begingroup$ @Blue okay write an answer then so that i acccept it (also why are those two triangles similar), i found it in tumblr $\endgroup$– user153330Commented Jan 4, 2015 at 14:26
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$\begingroup$ I believe the image's creator may be Tumblr user curiosamathematica, aka Jens Bossaert. $\endgroup$– BlueCommented Jan 4, 2015 at 15:21
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1 Answer
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Given the diagram as labeled ...
... we consider that construction occurred as follows:
Starting with point $P$ (or point $Q$) on $\overrightarrow{OR}$, construct $\overleftrightarrow{AP}$ (or $\overleftrightarrow{BQ}$) and let $C$ be the point where this line meets the unit circle. Then $\overleftrightarrow{BC}$ (or $\overleftrightarrow{AC}$) determines the point $Q$ (or $P$) on $\overrightarrow{OR}$.
Now, because $\angle ACB$ is inscribed in a semi-circle, it is a right angle by Thales' Theorem. Consequently, $\angle P \cong \angle B$ (as each is the complement of $\angle A$), so that $\triangle POA \sim \triangle BOQ$ and we can write $$\frac{|\overline{OP}|}{|\overline{OA}|} = \frac{|\overline{OB}|}{|\overline{OQ}|} \qquad\to\qquad \frac{|\overline{OP}|}{1} = \frac{1}{|\overline{OQ}|}$$
This proves the reciprocal relation. $\square$
Note: Even when the circle doesn't have unit radius, the relationship involves the geometric mean $$|\overline{OR}|^2 \;=\; |\overline{OP}|\;|\overline{OQ}|$$ which is important for the study of inversive geometry and such. The construction given is a nice companion to the more-common (to me) one involving the chord between points of tangency.
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2$\begingroup$ your answer is truly a gem, as always : ) !! (btw thanks for your trig addition formulas images, they helped me a lot) $\endgroup$ Commented Jan 4, 2015 at 15:26
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2$\begingroup$ Glad to be of service. "Trigonography" (my name for diagrams that make trig relations clear) is a bit of a passion of mine. This reciprocal construction will make a nice addition to my collection. :) $\endgroup$– BlueCommented Jan 4, 2015 at 15:34
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2$\begingroup$ you should most probably write a book where you collection all of your gems it will be awesome !! ; ) $\endgroup$ Commented Jan 4, 2015 at 15:36
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1$\begingroup$ @BillyRubina: The relation follows from the similarity of $\triangle POA$ and $\triangle BOQ$. Because the triangles happen to be right triangles, the relation matches an equality of $\tan\angle P$ and $\tan\angle Q$, so it's fine to approach the relation that way. (Note: If you use cotangent instead of tangent, then you get the relation without an extra multiplication step. But that amounts to a trade-off between a very-slightly-less-familiar trig function vs very-slightly-more algebra, so neither is really better or worse.) $\endgroup$– BlueCommented May 2, 2021 at 14:48
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1$\begingroup$ @BillyRubina: Similarity is about having the same "shape" and not (necessarily) the same "size". Same-shape-ness amounts to (1) corresponding angles being congruent, and (2) corresponding sides being proportional. Conveniently for triangles, if you know (1), then you get (2) for free, and vice-versa. (Other polygons don't do that. For instance, a square and a (non-square) rectangle have (1), but not (2); a square and a (non-square) rhombus have (2), but not (1).) (continued) $\endgroup$– BlueCommented May 2, 2021 at 15:46