# Show that if the centres of the circumscribed circles of the triangles $DEF$ and $ABC$ coincide, then $ABC$ is an equilateral triangle.

Let $$ABC$$ be a triangle, and $$D,E,F$$ points on the segments $$BC, CA, AB$$ respectively such that $$\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}$$

Show that if the centres of the circumscribed circles of the triangles $$DEF$$ and $$ABC$$ coincide, then $$ABC$$ is an equilateral triangle.

I can’t seem to find anything useful given the known information I just know that if $$O$$ is the circumcentre for the two triangles then $$OC=OB=OA$$ and $$OF=OD=OE$$

Hints ,suggestions and solutions would be appreciated.

Taken from the 2019 Pan African Maths Olympiad

http://pamo-official.org/problemes/PAMO_2019_Problems_En.pdf

• Hint: Stewart's theorem seems useful. Nov 10, 2019 at 21:47

Let $$\lambda\in(0,1)$$ be the equal value for the proportions $$BD:BC$$, $$CE:CA$$, and $$AF:AB$$. (So the "denominators" correspond to the sides.)

(We implicitly assume that $$D,E,F$$ do not coincide with $$A,B,C$$, which is not explicitly claimed in the OP, but must be claimed.)

Assume that the centers of the two circles $$(ABC)$$ and $$(DEF)$$ coincide, and let $$R$$ and $$d$$ be their radii. We start now the...

Proof: The power of the point $$D$$ with respect to the circle $$(ABC)$$ is $$R^2-d^2=BD\cdot DC=\lambda(1-\lambda)\; BC^2\ .$$ The same applies also for the other points, so $$BC=CA=AB$$.

$$\square$$

• [+1] You should just say that $d=r$. Nov 11, 2019 at 8:13
• @JeanMarie Merci beaucoup, yes, i've changed the notation in the last second, since $r$ my be confused with the radius of the incircle, but i have to do it at all places. Nov 11, 2019 at 21:06

Here's another proof. Denote $$\lambda = B D / D C = D E / E A = A F / F B, \\ R = O A = O B = O C, \\ r = O D = O E = O F.$$ Applying Stewart's theorem to the triangle $$\triangle A O B$$ and its cevian $$O F$$, we obtain $$R^2 \cdot A F + R^2 \cdot F B = A B (r^2 + A F \cdot F B)$$. This simplifies to $$A F = \sqrt{\frac{R^2 - r^2}{\lambda}}$$, and then $$A B = (1 + \lambda) \sqrt{\frac{R^2 - r^2}{\lambda}}$$. The analogous argument shows that $$B C = C A = (1 + \lambda) \sqrt{\frac{R^2 - r^2}{\lambda}}$$ and it follows that $$\triangle A B C$$ is equilateral.

$$\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}$$

This relation results in: DE||AB, EF||BC and FD||AC, so in parallelogram FECD,DC= FE also in parallelogram AFED we have FE=BD that is DC=BD.

Similarly CE=EA and AF=FB. So we have:

$$\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=1$$

That is $$BD=DC$$, $$CE=EA$$ and $$AF=FB$$, which means AD, BE and CF are medians of ABC . Since MA=MB=MC,also MF=MD=ME, then AD=BE=CF. That is triangle ABC has equal medians so it is equilateral.

• I don't think it follows from the assumptions that $A D$, $C F$ and $B E$ are concurrent. Consider the situation where $\triangle A B C$ is equilateral and $B D / D C = D E / E A = A F / F B = 2$. Nov 11, 2019 at 9:47