1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Hint: Stewart's theorem seems useful. $\endgroup$ Nov 10, 2019 at 21:47

3 Answers 3

3
$\begingroup$

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 $BC=CA=AB$.

$\square$

$\endgroup$
2
  • $\begingroup$ [+1] You should just say that $d=r$. $\endgroup$
    – Jean Marie
    Nov 11, 2019 at 8:13
  • $\begingroup$ @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. $\endgroup$
    – dan_fulea
    Nov 11, 2019 at 21:06
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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 $B D / D C = D E / E A = A F / F B = 2$. $\endgroup$ Nov 11, 2019 at 9:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .