Two congruent squares inside a circle. The extended diagonals meet at the circle circumference. What’s their angle?
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The angle is 30°.
This proof was kindly provided by Ben Akkus. It folows these steps:
- Construct the circle centre O and draw the radius to point I. From reflection symmetry it bisects the requested angle in two angles α.
- Draw another radius to vertex A and from the isosceles triangle AOI deduce that angle OAI is α as well.
- Constructing the right triangle ABK and applying the converse of Thales, O must be the midpoint of AK. Angle AKB is found from the interior angle sum as 45-α.
- Then consider right triangle OMK. From the inscribed angle theorem, angle MOK is found to be 4α.
- Finally, again applying the interior angle sum, one gets 4α+90+45-α=180. This gives α=15°.
A solution that makes extensive use of the inscribed angle theorem is the one provided by Galip. It goes as follows.
- The diameter through I is drawn as well, again from reflection symmetry bisecting the requested angle in two angles α.
- Now the circular arc angles are evaluated of the semicircle from K to I. The 2α follows from the inscribed angle theorem. Equally, the 90 comes from looking at angle BAI. Therefore AB must be 90-2α.
- The triangle HIB is then considered. Since M is midpoint of base BH and MI is perpendicular to it, MI bisects angle HIB.
- Again applying the inscribed angle theorem, angle MIB must be 45-α and it is equal to MIH, which is 2α. This gives α=15°.
We were honoured to receive a unique and visual solution for The complicated jet from David Andriana:
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