Isosceles Triangles in a Circle

If we draw the first isosceles triangle, then use both lateral sides as the bases for isosceles triangles with lateral sides length $a$ , (with sides external to the first triangle), then the convex hull we have formed is a regular pentagon with sides length $a$ .

The radius of the circle in which this regular pentagon is inscribed is

$R = \dfrac{a}{2\sin(36^{\circ})}$ ,

and the chord that corresponds to side/base $b$ has length

$b = \dfrac{a}{2\cos(72^{\circ})}$ .

Thus $\dfrac{ab}{R^{2}} = \dfrac{\frac{a^{2}}{2\cos(72^{\circ})}}{\frac{a^{2}}{4\sin^{2}(36^{\circ})}} = \dfrac{2\sin^{2}(36^{\circ})}{\cos(72^{\circ})} = \sec(72^{\circ}) - 1 = \sqrt{5} = \boxed{2.236}$

to $3$ decimal places.

The formula of a triangle's circumradius given its sides $x, y, z$ can be expressed as: $\frac{xyz}{\sqrt{2*(x^{2}y^{2} + x^{2}z^{2} + y^{2}z^{2}) - x^{4} - y^{4} - z^{4}}}$

Let $R_{1}$ be the radius of the circumcircle of the triangle with sides $(a, b, b)$ , and define $R_{2}$ analogously for the triangle of sides $(a, a, b)$ . Assume, WLOG, that $a > b$ ; the result that will be derived from here doesn't depend on whether $a > b$ because it will be, in a sense, symmetrical.

Using the formula I've shown above, let us calculate $R_{1}$ and $R_{2}$ .

$R_{1} = \frac {b^{2}}{\sqrt{4b^{2} - a^{2}}}$

$R_{2} = \frac {a^{2}}{\sqrt{4a^{2} - b^{2}}}$

Since $R_{1} = R_{2}$ , we can write:

$\frac {b^{2}}{\sqrt{4b^{2} - a^{2}}} = \frac {a^{2}}{\sqrt{4a^{2} - b^{2}}}$

Squaring both sides and multiplying the factors yields:

$4a^{2}b^{4} - b^{6} = 4a^{4}b^{2} - a^{6}$

Rearranging the terms and factoring conveniently:

$a^{6} - b^{6} = 4a^{2}b^{2}*(a^{2} - b^{2})$

$(a^{2} - b^{2})*(a^{4} + a^{2}b^{2} + b^{4}) = 4a^{2}b^{2}*(a^{2} - b^{2})$

Since we know that $a \neq b$ , we can divide both terms by $(a^{2} - b^{2})$ to obtain the following result:

$a^{4} + a^{2}b^{2} + b^{4} = 4a^{2}b^{2}$

$a^{4} - 3a^{2}b^{2} + b^{4} = 0$

Divide both sides by $b^{4}$ and let us define $t = \frac {a}{b}$ .

Then, we will obtain the following equation:

$t^{2} - 3t + 1 = 0$

Solving for $t$ yields the solutions $\frac{3 \pm \sqrt{5}}{2}$ , which happen to be the square of the golden ratio and its inverse (namely $\phi^{2}$ and $\phi^{-2}$ ). Since we've assumed that $a > b$ , we can conclude that $a = \phi b$ . Had we assumed the opposite, we would have used the other root we've found.

Substituting $a = \phi b$ back in the formula for $R$ :

$R = \frac {b^{2}}{\sqrt{4b^{2} - \phi^{2} b^{2}}}$

Thus:

$R^{2} = \frac {b^{4}}{4b^{2} - \phi^{2} b^{2}} = \frac {b^{2}}{4 - \phi^{2}}$ .

Now we have all of the information we need in order to calculate $\frac {ab} {R^{2}}$ :

$\frac {ab} {R^{2}} = \frac{b*(\phi b)}{\frac {b^{2}}{4 - \phi^{2}}} = \phi *(4 - \phi^{2}) = \sqrt(5) \approx 2.236$

You probably can figure this out from looking at this graphic

Define b>a, when we rotate so that base b coincides with side b, recognize this as a regular pentagon in a circle radius 1, with side length a. The product ab is (2cos54)(2/sec18) or 2.2359

a b/ R^ 2 = Sqrt (5) = 2.2360679774997896964091736687313

Sketch {a, a, b} of side a longer than side b on a circle. Then, sketch {b, b, a} on one side of side a. Label angle B and then angle A according to most usual convention of opposite to b and opposite to a respectively.

Quadrilateral with angle A implies for two angles of (180 d - A) on the first isosceles triangle sketched. Therefore, 2 (180 d - A) + B = 180 d. Rearrange equation:

180 d - A = A - B

Taking Cosine onto both sides,

Cos (180 d - A) = Cos (A - B)

-Cos A = Cos A Cos B + Sin A Sin B

Cos A (1 + Cos B) + Sin A Sin B = 0

Cos B = (a^2 + a^2 - b^2)/ (2 a^2) = (2 a^2 - b^2)/ (2 a^2) while

Cos A = (b^2 + b^2 - a^2)/ (2 b^2) = (2 b^2 - a^2)/ (2 b^2).

Forming right angle triangles for finding values of Sin B and Sin A,

[2 a^2 - b^2, b Sqrt (4 a^2 - b^2), 2 a^2] for B and

[2 b^2 - a^2, a Sqrt (4 b^2 - a^2), 2 b^2] for A all by Pythagoras' theorem.

Sin A = a Sqrt (4 b^2 - a^2)/ (2 b^2) and Sin B = b Sqrt (4 a^2 - b^2)/ (2 a^2)

Substitutes Cos A, Cos B, Sin A and Sin B into Cos A (1 + Cos B) + Sin A Sin B = 0:

(1 - 0.5 a^2/ b^2)(2 - 0.5 b^2/ a^2) + [a Sqrt (4 b^2 - a^2)/ (2 b^2)][b Sqrt (4 a^2 - b^2)/ (2 a^2)]

Denominators (2 a^2)(2 b^2) can be eliminated:

(a^2 - 2 b^2)(4 a^2 - b^2) = a b Sqrt [(4 b^2 - a^2)(4 a^2 - a^2)]

Since (4 a^2 - b^2) = Sqrt (4 a^2 - b^2) Sqrt (4 a^2 - b^2),

(a^2 - 2 b^2) Sqrt (4 a^2 - b^2) = a b Sqrt (4 b^2 - a^2)

(a^4 - 4 a^2 b^2 + 4 b^2)(4 a^2 - b^2) = (a b)^2 (4 b^2 - a^2)

4 a^6 - 16 a^4 b^2 + 16 a^2 b^4 - 4 b^6 = 0

a^6 - 4 a^4 b^2 + 4 a^2 b^4 - b^6 = 0 {The most critical equation of this question.}

(a^2 - b^2)(a^4 - 3 a^2 b^2 + b^4) = 0

(a^2 - b^2)[a^2 - (3 + Sqrt (5))/ 2 b^2] [a^2 - (3 - Sqrt (5))/ 2 b^2]= 0

For a to b of positive ratios only,

a = b OR

a = Sqrt [(3 + Sqrt (5))/2] b = [(Sqrt (5) + 1)/ 2] b = 1.6180339887498948482045868343656 b OR

a = Sqrt [(3 - Sqrt (5))/2] b = [(Sqrt (5) - 1)/ 2] b = 0.6180339887498948482045868343656 b

Since a <>b and a > b, while only two biggest isosceles triangles are concerned,

a = [(Sqrt (5) + 1)/ 2] b = 1.6180339887498948482045868343656 b is the only one.

Cos B= 1 - 0.5 (b/ a)^2

B = 36 d

Triangle with vertex center O and same side angles as {a, a, b} is 72 d. Half of it forming two right angle triangles with angle 36 d at center O.

Let R = 1 unit for convenience of finding ratios,

R Sin 36 d = 0. 5 b

b = 2 R Sin 36 d = 2 Sin 36 = 1.1755705045849462583374119092781

a = (1.6180339887498948482045868343656) (2 Sin 36)

a b / R^2 = (1.6180339887498948482045868343656) (2 Sin 36)^2

a b / R^2 = 2.2360679774997896964091736687313 = Sqrt (5)