For $A$, $B$, $C$, $X$ on a circle of radius $r$, is it possible to find the area of $\triangle ABC$ given distances $AX$, $BX$, $CX$?

Question

Construct a circle with some radius, $r$, and centre, $O$, and pick any four points on its circumference: $A$, $B$, $C$, and $X$. Is it possible to find the area of $\triangle ABC$, given $AX$, $BX$, $CX$, and $r$? If so, what is the formula to do so? If not, why not?

Answer 1

No. The reason is that there are several triangles, all with the same $AX$, $BX$, $CX$, with different areas. Draw a circle of radius $r$. For simplicity, choose $X$ at the bottom. Where are all the points at distance $AX$ from $X$?

Answer 2

Let XA = a, XB = b, XC = c.

$F(b, r) \rightarrow \angle yellow$ ;when F is the cosine function.

$F(a, r) \rightarrow \angle (red + yellow)$. Then, $\angle red$ is known.

$F(c, r) \rightarrow \angle green$.

$G(a, b, \angle red) \rightarrow [\triangle XAB]$; where G is the area function.

Similarly, $G(b, c, \angle (green + yellow)) \rightarrow [\triangle XBC]$.

Therefore, [quad ABCD] is known.

$G(a, c, \angle (red + yellow + green)) \rightarrow [\triangle XAC]$.

$[\triangle ABC]$ is found by subtraction.

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In general, If we let $\angle red = \angle 1$, $\angle yellow = \angle 2$, $\angle green = \angle 3$,

then $[\triangle ABC] = \dfrac {ab \sin \angle 1}{2} + \dfrac {bc \sin \angle (2 + 3)}{2} - \dfrac {ca \sin \angle (1 + 2 + 3)}{2} $

A slight variation will occur depending how XB is inclined.

Answer 3

Given the radius $r$ and the lengths XA, XB and XC, the triangle ABC can have four configurations, as shown in the diagram. (XA $\le$ XB $\le$ XC without loss of generality.)

The area of the triangle is given by

$$K = 2r^2\sin\alpha\sin\beta\sin\gamma$$

where $\alpha$, $\beta$ and $\gamma$ are the three angles of the triangle, to be expressed in terms of the given lengths. Let $\theta_a$, $\theta_b$ and $\theta_c$ be the angles formed between the diameter XY and XA, XB and XC, respectively,

$$\theta_a=\cos^{-1}\frac {XA}{2r},\>\>\>\>\> \theta_b=\cos^{-1}\frac {XB}{2r},\>\>\>\>\> \theta_c=\cos^{-1}\frac {XC}{2r}$$

Then, for case 1, we have $\alpha = \theta_b-\theta_c$, $\beta= \pi - (\theta_a-\theta_c)$ and $\gamma= \theta_a-\theta_b$. Thus, its area is,

$$K_1 = 2r^2\sin(\theta_b-\theta_c)\sin(\theta_a-\theta_c)\sin( \theta_a-\theta_b)$$

Similarly,

$$K_2 = 2r^2\sin(\theta_b+\theta_c)\sin(\theta_a+\theta_c)\sin( \theta_a-\theta_b)$$ $$K_3 = 2r^2\sin(\theta_b+\theta_c)\sin(\theta_a-\theta_c)\sin( \theta_a+\theta_b)$$ $$K_4 = 2r^2\sin(\theta_b-\theta_c)\sin(\theta_a+\theta_c)\sin( \theta_a+\theta_b)$$