MHB Solve the BEASTly Trapezoids Puzzle: Find the 4 Heights

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The discussion revolves around solving a geometric puzzle involving four isosceles trapezoids inscribed in a circle with a radius of 725 and shorter parallel sides measuring 666. Participants explore the relationship between the trapezoids' heights, other parallel sides, and equal sides, all of which must be integers. Using the Pythagorean theorem and identities for expressing sums of squares, they derive potential coordinates for the trapezoids. After some calculations and corrections, the final heights determined are 144, 280, 1008, and 1144. The conversation emphasizes the importance of accurate equations in deriving these values.
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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?
 
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Wilmer said:
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.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.
 

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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?
 
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Wilmer said:
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.
 
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