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BEASTly Trapezoids!

Wilmer

In Memoriam
Mar 19, 2012
376
A circle radius = 725 contains 4 isosceles
trapezoids, length of shorter parallel sides = 666.

Heights, other parallel sides, and equal sides are all integers.

What are the 4 heights?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
A circle radius = 725 contains 4 isosceles
trapezoids, length of shorter parallel sides = 666.

Heights, other parallel sides, and equal sides are all integers.

What are the 4 heights?
Best to start with a picture.
trap.png

The radius is 725, and AB = 666, so HB = 333 and by Pythagoras OH = 644. Let OK = $x$ and KC = $y$. Then $x^2+y^2=725^2$. So we need to find several ways to express $725^2$ as the sum of two squares. The way to do that is to use the identity $$(a^2+b^2)(c^2+d^2) = (ac+bd)(ad-bc) = (ac-bd)(ad+bc).$$ Applying that identity repeatedly to the fact that $$725^2 = 5^4\times 29^2 = (2^2+1^2)(2^2+1^2)(2^2+1^2)(2^2+1^2)(5^2+2^2)(5^2+2^2),$$ you find that $$ 725^2 = 725^2+0^2 = 720^2+85^2 = 715^2+120^2 = 696^2+203^2 = 644^2 + 333^2 = 627^2+364^2 = 580^2+435^2 = 525^2+500^2. $$

To find possible coordinates for the point C = $(x,y)$, we must have $y>333$ and $|x|<644$ (to ensure that AB is the shorter of the parallel sides). We also need BC to be an integer, which (from the triangle BLC) means that $(y-333)^2 + (644-x)^2 = \Box$, where $\Box$ means a square. Since $x^2+y^2 = 725^2$, that relation simplifies to $725^2-644x-333y = \Box.$

Checking through the possible values of $(x,y)$, namely $$ (\pm85,720),\ (\pm120,715),\ (\pm203,696),\ (\pm364,627),\ (\pm435,580),\ (\pm500,525),\ (\pm525,500),\ (\pm580,435),\ (\pm627,364), $$ you can verify that there are exactly four values of $x$ that satisfy $725^2-644x-333y = \Box.$ The corresponding heights are the numbers $644-x.$

The values of $x$ are $\pm525,\ \pm627$, and the heights are 17, 119, 1169, 1271.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Nope...with your 4, the 2 equal sides AD and BC are not integers.


These are my 4 (height, AD/BC, CD, x):
144, 240, 1050, -500
1144, 1160, 1050, 500
280, 406, 1254, -364
1008, 1050, 1254, 364

Using your diagram:
AB = a, CD = b, height HK = h, OK = x, radius = r
I came up with this formula to derive:
r = SQRT(4x^2 + b^2) / 2 where x = (a^2 - b^2 + 4h^2) / (8h)

I don't know why you listed all the triangles you did;
we need triangle 333-644-725 for all cases;
then we need the triangles with a>333:
364-627-725, 435-580-725 and 500-525-725.
That's it, that's all: right?
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Nope...with your 4, the 2 equal sides AD and BC are not integers.


These are my 4 (height, AD/BC, CD, x):
144, 240, 1050, -500
1144, 1160, 1050, 500
280, 406, 1254, -364
1008, 1050, 1254, 364

Using your diagram:
AB = a, CD = b, height HK = h, OK = x, radius = r
I came up with this formula to derive:
r = SQRT(4x^2 + b^2) / 2 where x = (a^2 - b^2 + 4h^2) / (8h)

I don't know why you listed all the triangles you did;
we need triangle 333-644-725 for all cases;
then we need the triangles with a>333:
364-627-725, 435-580-725 and 500-525-725.
That's it, that's all: right?
Yes, you are right. My mistake was that I dropped a factor of 2 in the equation $725^2-644x-333y = \Box.$ It should have read $2(725^2-644x-333y) = \Box.$ I then get the values of $x$ to be $\pm364$ and $\pm500$, giving the heights as 144, 280, 1008, 1144.