I just made this question up...

Let ABC be a triangle with circumcentre O such that the ratio of the areas of the triangles $ABO: ACO: BCO$ is $2:3:4$. Find the ratios of the sides of the triangle $AB: BC: CA$.

Of course the numbers $2:3:4$ are arbitrarily chosen. I'm just interested in how to solve this problem.


What I tried: I let the radius of the circumcentre be $OA = OB = OC = r$, and let angle $AOB = \alpha_1$, angle $AOC = \alpha_2$, angle $BOC = \alpha_3$

I tried using the sine area rule for each triangle ABO, ACO, BCO, and got:

$$r^2 \sin\alpha_1 = 4 \qquad (1)$$ $$ r^2 \sin\alpha_2 = 6 \qquad (2)$$ $$r^2 \sin\alpha_3 = 8 \qquad(3)$$

and of course

$$ \alpha_1 + \alpha_2 + \alpha_3 = 360 ^\circ \qquad (4)$$

I tried a few things here but was not successful in making much progress. We can of course let r=1 WLOG.

The "R addition formulas" didn't seem to help.

Also, there are the "factor formulas" for $\sin:$ $ \quad sin x + \sin y \equiv 2 \sin(\frac{x+y}{2}) \cos(\frac{x-y}{2})$ ,

but this is for two terms only, and we have three terms in the above equations.

Edit: I also discovered this about arcsin: https://www.math-only-math.com/arcsin-x-plus-arcsin-y.html

We could apply it to $(1), (2) $ and $(3)$ (i.e. $\ \alpha_1 = \arcsin \frac {4} {r^2}, \ \alpha_2 = \arcsin \frac {6} {r^2}, and \ \alpha_3 = \arcsin \frac {8} {r^2}$ and then add these) and I tried this, but the expression we get at the end doesn't seem particularly helpful. It seems like a dead end to me.

So can we make progress using my current equations, or are there easier ways to do it? (Of course there are: I'm just interested in seeing the different methods).

  • $\begingroup$ Recall $\sin\frac{\alpha_1}{2}=\frac{\frac12 AB}{r}$ and the sums of the areas should give the area of the triangle. However, the answer is not that simple. Maybe I have mistaken somewhere) $\endgroup$ – Alexey Burdin Jun 19 '20 at 18:02

enter image description here

The area ratios are $\frac12r^2\sin\alpha_1 : \frac12r^2\sin\alpha_2 : \frac12r^2\sin\alpha_3=2:3:4$. Then, let

$$\sin\alpha_1 = 2x,\>\>\>\>\>\sin\alpha_2 = 3x,\>\>\>\>\>\sin\alpha_3 = 4x $$

Note that $\sin(\alpha_1+\alpha_2) = \sin(2\pi- \alpha_3)$, which is

$$2\sqrt{1-9x^2}+3\sqrt{1-4x^2}=4 $$

where it is assumed that $\alpha_1, \alpha_2 > \frac\pi2$. Solve to get $x=\frac{\sqrt{15}}{16}$. Then,

$$\sin\alpha_1 = \frac{\sqrt{15}}{8},\>\>\>\>\>\sin\alpha_2 = \frac{3\sqrt{15}}{16},\>\>\>\>\>\sin\alpha_3 = \frac{\sqrt{15}}{4} $$

and the side ratios are

\begin{align} AB:BC:CA & = 2r\sin\frac{\alpha_1}2:2r\sin\frac{\alpha_3}2:2r\sin\frac{\alpha_2}2 \\ & = \sqrt{\frac{1-\cos\alpha_1}2}:\sqrt{\frac{1-\cos\alpha_3}2}:\sqrt{\frac{1-\cos\alpha_2}2} \\ & = \sqrt{\frac{1+{\frac78}}2}:\sqrt{\frac{1-\frac{1}{4}}2}:\sqrt{\frac{1+\frac{11}{16}}2} \\ & = \sqrt{\frac{15}{16}}:\sqrt{\frac38}:\sqrt{\frac{27}{32}} \end{align}

In the case of the circumcenter O outside the triangle ABC, we have $\alpha_1 = \alpha_2+\alpha_3 $, which, assuming $\alpha_3>\frac\pi2$, leads to

$$4\sqrt{1-9x^2}-3\sqrt{1-16x^2}=2\implies x=\frac{\sqrt{15}}{16} $$ and the corresponding side ratios are

\begin{align} AB:BC:CA & = \sqrt{\frac{15}{16}}:\sqrt{\frac 58}:\sqrt{\frac{5}{32}} \end{align}

enter image description here

  • $\begingroup$ Great methods and tricks used here. I'm wondering if there's other solutions, i.e. for when $\alpha_1 \leq \frac{\pi}{2} $ or $ \alpha_2 \leq \frac{\pi}{2} $. I guess you can use similar methods. $\endgroup$ – Adam Rubinson Jun 20 '20 at 12:15
  • $\begingroup$ I think you have to split it up into different cases, and it gets ugly. $\endgroup$ – Adam Rubinson Jun 20 '20 at 12:37
  • $\begingroup$ @AdamRubinson - Another possibility is that O is outside ABC, which is added in the answer $\endgroup$ – Quanto Jun 20 '20 at 14:33
  • $\begingroup$ Why does $\alpha_1 = \alpha_2 + \alpha_3$ ? Wouldn't this imply $\alpha_1 = \pi $? $\endgroup$ – Adam Rubinson Jun 20 '20 at 15:54
  • $\begingroup$ @AdamRubinson - See the added graph in the answer $\endgroup$ – Quanto Jun 20 '20 at 16:11

I would use Heron's formula of the area of a triangle: $$A=\sqrt{s(s-a)(s-b)(s-c)}$$where$$s=\frac{a+b+c}2$$ Then for $\triangle ABO$, $s=\frac{2r+c}2=r+\frac c2$ and the area is $$A_{ABO}=\sqrt{(r+\frac c2)\cdot \frac c2\cdot \frac c2\cdot (r-\frac c2)}$$ You can write similar expressions for the other triangles. In case you need $r$, the formula for that is $$r=\frac{abc}{4A_{ABC}}$$

  • $\begingroup$ Did you mean the area of $A_{ABO}=\sqrt{(r+\frac c2)\cdot \frac c2\cdot \frac c2\cdot (r-\frac c2)}$, or am I mis-applying the formula? $\endgroup$ – Adam Rubinson Jun 19 '20 at 21:39
  • $\begingroup$ @AdamRubinson Thanks. My mistake $\endgroup$ – Andrei Jun 20 '20 at 3:06
  • $\begingroup$ I've tried using this method (I've got 3 equations - one for each triangle using Heron's formula), but I still don't see how to get to an answer... The formula for r doesn't help either $\endgroup$ – Adam Rubinson Jun 21 '20 at 14:49

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