JB
Lv 7
JB asked in Science & MathematicsMathematics · 1 decade ago

# Can a square be dissected into a finite number of obtuse, isosceles triangles?

(1) Show that a square can be dissected into a finite number of OBTUSE, ISOSCELES triangles.

(2) What is the minimum number of OBTUSE, ISOSCELES triangles needed to accomplish this?

[based loosely on a recent question of Scythian]

Relevance
• 1 decade ago

I have 10 triangles.

Let's ABP is triangle with angles (15, 15, 150) respectively.

BPQ is triangle with angles (30, 30, 120).

BQR has angles (30, 120, 30).

BRC is (15, 150, 15).

Then AQC is also (30, 120, 30) and ABC is a half of the square ABCD. Another half is on the same way, and it's 10 triangles.

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http://s524.photobucket.com/albums/cc321/Krejakovi...

• Duke
Lv 7

(1) Your question reminded me a famous problem (Coxeter, Geometry Revisited, Chapter 1): given a square ABCD and internal point P, such that triangle ABP has angels 15°, 15° and 150° - prove that triangle CDP is equilateral. The most artistic solution uses a second triangle ADQ with angels 15°, 15° and 150° - connecting then P with Q we get an equilateral triangle APQ and the rest is easy /follow the link below to see a picture - #1/:

http://farm4.static.flickr.com/3337/3308125501_883...

That leads to a partition of 14 triangles: 5 of them congruent with 15°, 15° and 150° (yellow) and 9 with 30°, 30° and 120° (green).

I also thought of another partition into 11 triangles (#2 on the picture). In an acute triangle with angles α, β, γ the center of the circumscribed circle dissects it into 3 isosceles with angles 2α, 2β and 2γ around the circumcenter. Take the midpoint of AB and connect with C and D. The angle at the midpoint is 2arctan(1/2) > 45°, so the circumcenter dissects it into 3 obtuse. The 2 remaining right triangles are dissected by medians to the hypotenuses as shown on the picture, and the 2 acute triangles are further dissected into 3 obtuse (2 different kinds - same color similar) each.

(2) This may turn out difficult, I'll give it another thought.

P.S. Excellent solution by Dragan K! A very strong candidate for optimal solution, congratulations!

• 1 decade ago

Well, we know that dissecting a square by its diagonals will result in 4 "obtuse" isosceles triangles, but that depends if a right isosceles triangle can be called "obtuse". I'll assume that it can't be.

Er, make that 2 right isosceles triangles.

Edit: I have a preliminary figure of 12 obtuse isosceles that make up a square. First, 2 parallel lines are cut from opposite corners such it forms a parallogram of sides 2 and 1, but is done inside a square of sides of about 1.9. This parallogram is first cut in half into rhombuses of sides 1, which are then further divided into obtuse triangles. The remaining triangles are dissected into obtuse triangles of sides 1-1.9-1, and acute triangles of sides 1-0.9-1. These acute triangles can turther be dissected into 3 obtuse triangles each. It is for this reason that 0.9 is chosen, so that all three isosceles triangles from this last dissection are obtuse.

Is this the minimum? I don't know yet.

Edit 2 (Early Morning Edition): It's time for my morning coffee, and already other answerers here have beat me to it. Figuring out Dragan K's "answer" was a problem enough in just deciphering it, but he is absolutely right. It can be done with 10 obtuse isosceles triangles.