This seems to defy the logic. One may ask why not. Looks perfectly possible.

So the claim may very well be wrong.

So the claim may very well be wrong.

If you believe so, you may try to find a counter-example.

This seems to defy the logic. One may ask why not. Looks perfectly possible.

So the claim may very well be wrong.

So the claim may very well be wrong.

If you believe so, you may try to find a counter-example.

Labels:
Aint gonna happen,
Bisector,
chords,
circle,
similarity of triangles

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Labels:
proof,
Rectangle,
semi-circle,
Solutions,
Tangent,
Trigonometry

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According to the order of submission...

by**Bob Ryden**: (**Ajit** commented a very similar one)

by**Bleaug**:

Two tangents to a given circle define a symmetric 'kite' figure. By construction, the two kites defined by OUV and OTU are isometric because they have identical short legs. Since angles in T, U, V are right angles, blue and green angles are supplemental. Hence, angle a is angle(OUV). From Thales theorem we get cos(a) = 1/3

by**Yu**:

Several steps remain implicit in the figure. More words may be needed for a formal proof. But the solution holds.

by**Binary Descartes**:

by

by

Two tangents to a given circle define a symmetric 'kite' figure. By construction, the two kites defined by OUV and OTU are isometric because they have identical short legs. Since angles in T, U, V are right angles, blue and green angles are supplemental. Hence, angle a is angle(OUV). From Thales theorem we get cos(a) = 1/3

by

Several steps remain implicit in the figure. More words may be needed for a formal proof. But the solution holds.

by

Common tangents from the same point have equal lengths.

See the two identical deltoids (kites) sharing the same side.

Then cos(a) = e/3e = 1/3.

http://www.8foxes.com/

Labels:
Deltoid,
Rectangle,
semi-circle,
Solutions,
Tangent,
Thales,
Trigonometry

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Let's continue to mess around with this configuration...

Labels:
Rectangle,
semi-circle,
Tangent,
Trigonometry

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Bob Ryden provided this solution to Fox 124.

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The options are corrected below.

There are about 5 different solutions, that we can post in coming days.

There are about 5 different solutions, that we can post in coming days.

Labels:
Rectangle,
semi-circle,
Tangent,
Trigonometry

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Labels:
Pythagoras,
semi-circle,
Square

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Not much time to update the blog. But, let's throw this to the public knowledge.

Note: when 3 concentric arcs are given, their center can be identified easily.

(which is a nice exercise and easy by itself).

Let us know if you have a solution.

Note: when 3 concentric arcs are given, their center can be identified easily.

(which is a nice exercise and easy by itself).

Let us know if you have a solution.

Aaaah, one more detail. More than one equilaterals should be drawn on 3 concentric circles. Take the figure as it looks and do not go for the "other" equilateral. Let the simplicity ring over the land, in the morning... and during the night...

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