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.)
Any triangle. Including:
An isosceles right triangle (into smaller right triangles).
Any other right triangle (into smaller right triangles).
An equilateral triangle.
Any other isosceles triangle.
(Are there other special cases?).
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.
- 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).
- Right triangle: 2
BAC = 90o. See the above mentioned solution.
- Isosceles triangle: 5
Let assume A the apex of the triangle with AB=AC.
- BAC is obtuse
Let consider AH perpendicular to BC and HK perpendicular to AC within the
ABH, AHK, CHK are the 3 RRT.
- BAC is acute
Let consider BH the perpendicular height on AC. ABH abd BCH are the 2 RRT.
- Equilateral triangle: 3
Same solution as above (3-1).
- 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
- 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.
- 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
- 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.
- 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.
- Non regular hexagon: 8
Richard Brabant sent the following solution for the square:
| +++ = |
| +++ = | 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.
Mail to Ken