Problem about 5 integer sided right triangles whose larger acute angles add exactly to 360 degrees?

I confirm Uncle68's answer analytically:

Let tan α = 4/3, tan β = 12/5, tan γ = 24/7, tan δ = 40/9 and tan ε be the tangent of the unknown right-angled triangle, which is the ratio of its legs.

It must hold that

tan(α+β+γ+δ) = -tan ε.

Using tan(x+y) = (tan x + tan y)/(1 - tan x tan y) repeatedly, we find that

tan(α+β) = -56/33

tan(α+β+γ) = 16/63

tan(α+β+γ+δ) = -2664/73

Note that the numerator and denominator always form, up to a sign, legs of some Pythagorean triangles, which can be easily proven.

From this, we can directly see that (73, 2664, √(73^2+2664^2) = 2665) is the missing triangle.

Edit: The proof:

Let a, b, c, d, e, f be integers such that

a^2 + b^2 = c^2 and d^2 + e^2 = f^2,

let tan x = b/a and tan y = e/d. Then

tan(x±y) = (b/a ± e/d) / (1 ∓ be/ad) = (bd ± ae) / (ad ∓ be).

We want to show that there is an integer g such that (bd ± ae)^2 + (ad ∓ be)^2 = g^2. By straightforward expansion and factorisation,

(bd ± ae)^2 + (ad ∓ be)^2 = b^2d^2 + a^2e^2 ± 2abde + a^2d^2 + b^2e^2 ∓ 2abde = b^2(d^2+e^2) + a^2(d^2+e^2) = b^2f^2 + a^2f^2 = (a^2+b^2)f^2 = c^2f^2.

Obviously, g = cf is the product of the original hypotenuses.

How about the right-angled triangle with sides (73, 2664 and 2665)?

I haven't been able to show that the solution is exact, but it fits within the 9-digit resolution of my calculator. Perhaps I can find something a little more precise ... hang on!

Added: Ok, I persuaded a friend with Mathematica on his computer to evaluate the expression for the difference between 2.pi and the sum of the 5 ArcSin terms using exact calculation. He claimed that the result came back as zero, which would imply that the relationship is correct and exact, as claimed.

While waiting for him to reply, I did a quick brute force calculation based on trig identities. If a, b, c, d and e are the larger acute angles, with e the unknown, then

a + b + c + d = 2.π - e, and sin(a + b + c + d) = sin(2.π - e)

With the angle sum and periodicity identities, the LHS can be expanded into products of the sines and cosines of the known larger acute angles, and the RHS gives the sine of the unknown angle. To cut a rather messy story short, the final result gave sin(e) = 66600/66625 = 2664/2665. Pythagoras then confirmed the third side (not personally!).

Added : In fact, the simplest and quickest way of showing that my answer was exact would have been to type the sum of the ArcTans for the 5 angles into WolframAlpha, and the result would have been 2.π. Unfortunately, I used the ArcSins, which didn't even give the right answer, let alone recognise it as 2.π.

I came up with Vašek's approach, except I got there via complex numbers.

e^iα * e^iβ * e^iγ * e^iδ * e^iε = 1

...where α,β,γ,δ,ε are as in Vašek's version. Since cos and sin values are integer ratios in the four given triangles, then the cos and sin of ε are rational too.

1/(cos ε + i sin ε) = (cos α + i sin α)(cos β + i sin β)(cos γ + i sin γ)(cos δ + i sin δ)

cos ε - i sin ε = (cos α + i sin α)(cos β + i sin β)(cos γ + i sin γ)(cos δ + i sin δ)

= (3 + 4i)/5 * (5 + 12i)/13 * (7 + 24i)/25 * (9 + 40i)/41

= (-33 + 56i)/65 * (-897 + 496i)/1025

cos ε - i sin ε = (1825 - 66600i)/66625

Obviously 25 is a common factor, remove that and conjugate both sides:

cos ε + i sin ε = (73 + 2664i)/2665

That's in the 1st quadrant, so e is indeed acute. sin ε = 2664/2665 ~~ 1 so ε is indeed nearly right.

That corresponds to the (73, 2664, 2665) right triangle, just as most everyone else found.

Edit: Hmm. Skipped a step, reasoning from the drawing. Because each is the larger acute angle of a right triangle, each is in (π/4,π/2):

π/4 < α,β,γ,δ < π/8

π < < 2π

That means that there's an ε in (0,π) satisfying the original equation...the one that I left out:

α + β + γ + δ + ε = 2π

Anything else I missed still escapes me.

The large acute angle of the fifth triangle is

2π - [arctan(4/3) + arctan(12/5) + arctan(24/7) + arctan(40/9)]

= arctan (2664/73) using the following Wolfram Alpha link:

http://www.wolframalpha.com/input/?i=arctan%284%2F...

=> the fifth triangle has sides, 2664, 73 and 2665.

Edit:

The rigorous calculations posted by me earlier has been deleted and is replaced by the following simpler calculations.

Sum of the large acute angles of the first and the third right triangles

= arctan(4/3) + arctan(24/7)

= π + arctan[(4/3 + 24/7) / (1 - 96/21)

= π - arctan(4/3)

Sum of the large acute angles of the second and the fourth right traingles

= arctan(12/5) + arctan(4/9)

= π + arctan[(12/5 + 40/9)/(1 - 480/45)]

= π - arctan(308/435)

=> sum of the four large acute angles

= 2π - [tan^-1 (4/3) + tan^-1(308/435)]

= 2π - tan^-1 [(4/3+ 308/435) / (1 - 1232/1305)]

= 2π - tan^-1 [(1740 + 924) / (1305 - 1232)]

= 2π - tan^-1 (2664/73)

=> The fifth right triangle has sides 2664, 73 and 2665.

That's the same answer I got within the resolution of Excel -- 73 - 2664 - 2665

(and it's where the pattern of the "difference" function I was using changed)

I'll go with 73-2664-2665 (to give an arcsin of 88.430350794016243 ...... for the larger acute angle)

Hey,

I easily got for free Geogebra here http://j.mp/Ufouq4

It works very well.

Cheers ;)

To solve, we can simply sum the current angles and subtract them from 360°

Using the trig function tan(θ) = opposite/adjacent

360° - arctan(4/3) - arctan(12/5) - arctan(24/7) - arctan(40/9) ≈ 88.43°

We therefore know that tan(88.43) = opposite/adjacent of the final triangle.

tan(88.43) = 2664/73

Therefore 2664 is the length of one side and 73 is the length of the other.

By Pythagoras

41^2 = 3^2 + b^2

Where 'b' is the side that is not drawn in. The angle between 'b' aand '3' would be 90 degrees.

b^2 = 41^2 - 3^2

b^2 = (41 - 3)(41 + 3)

b^2 = 38* 44

b^2 = 1672

b = sqrt(1672)

b = 40.890....