## Ken's POTW

Right Triangles
Show how to divide each of the following figures into a minimum number of unique right triangles (such that each triangle is different from the others in both rotation and reflection.)
1. Any triangle. Including:
1. An isosceles right triangle (into smaller right triangles).
2. Any other right triangle (into smaller right triangles).
3. An equilateral triangle.
4. Any other isosceles triangle.
5. (Are there other special cases?).
2. A square.
3. A pentagon.
4. A hexagon.

Source: Original.

Solution received from Philippe Fondanaiche, Paris, France:

Geometrical figures would be more appropriate than a long chatter but the universal and easy software to produce them is not yet available.

1. Any triangle ABC: 2
Let assume BAC the greatest angle: acute or obtuse. With AH the perpendicular height on BC, ABH and ACH are the 2 required right triangles (RRT).
2. Right triangle: 2
BAC = 90o. See the above mentioned solution.
3. Isosceles triangle: 5
Let assume A the apex of the triangle with AB=AC. Two cases:
1. BAC is obtuse Let consider AH perpendicular to BC and HK perpendicular to AC within the triangle ACH. ABH, AHK, CHK are the 3 RRT.
2. BAC is acute Let consider BH the perpendicular height on AC. ABH abd BCH are the 2 RRT.
4. Equilateral triangle: 3
Same solution as above (3-1).
5. Right isosceles triangle: 5
Let assume that the right angle is BAC with AB=AC.Let consider any point on BC distinct from the middle of the side BC. In ABD, DH is the perpendicular height drawn from D, DK is the perpendicular height drawn from D in ACD and in AHD, HP is the perpendicular from H on AD. BHD, AHP, DHP, ADK, CDK are the 5 RRT.
6. Square: 6
On the side AB of the square ABCD, let consider E distinct from the middle of AB. The parallel drawn from E to AD and BC cuts CD on F. In the triangle ADE, let consider the perpendicular height AH and in the triangle BCE BI is the perpendicular height on CE. The 6 RRT are AEH,ADH,DEF, CEF, BCI and BEI.
7. Regular pentagon: 7
Let consider the regular pentagon ABCDE. In the triangle ABE, let draw the perpendicular AH on BE and within the triangle ABH, HI is perpendicular to AB. Let consider DJ perpendicular to BE and within BCJ, BK is the perpendicular height on CJ. AHI, BHI, AEH, BCK, BJK, CDJ and DEJ are the 7 RRT.
8. Non regular pentagon: 6
In the triangle ADE, let consider EH perpendicular to AD. Let draw BI perpendicular on AD and CJ perpendicular on BD. AEH, DEH, ABI, BDI, BCJ and CDJ are the 6 RRT.
9. Regular hexagon: 8
Let consider the regular hexagon ABCDEF. Let draw BE and BF. In the triangle ABF, AH is the perpendicular height on BF whereas HI is the perpendicular height from H on AB within the triangle ABH. Let draw DJ perpendicular to BE and BK perpendicular to the line CJ. AHI, BHI, AFH, BCK, BJK, CDJ and DEJ are the 8 RRT.
10. Non regular hexagon: 8
As above.

Richard Brabant sent the following solution for the square:
```--------------------
|R              ++=|
|            +++ = |
|         +++   =  |  The square can be broken into four unique
|      +++     =   |  right triangles.
|   +++       =    |
|+++         =     |
|xxxR       =      |
|  xxx     =       |
|    xxx  =        |
|R     xxx        R|
--------------------
```

Some of these are obviously minimum, but can any be improved upon? Please feel free to also send a diagram or description.
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