Find $DF$ in a triangle $DEF$

Consider we have a triangle $ABC$ where there are three points $D$, $E$ & $F$ such as point $D$ lies on the segment $AE$, point $E$ lies on $BF$, point $F$ lies on $CD$. We also know that center of a circle over ABC is also a center of a circle inside $DEF$. $DFE$ angle is $90^\circ$, $DE/EF = 5/3$, radius of circle around $ABC$ is $14$ and $S$ (area of $ABC$), K (area of DEF), $S/K=9.8$. I need to find $DF$. Help me please, I'd be very grateful if you could do it as fast as you can. Sorry for inconvenience.

• If you've got a diagram to go with the question, that would be very helpful. – Isaac Jan 25 '12 at 20:00
• Okay, sorry for such a poor presentation of a question, you can leave it till I make a diagram. Thanks for not voting down. – user1131662 Jan 25 '12 at 20:23
• In many cases, the text alone would have been fine; I was just having enough trouble visualizing how the things fit together that a diagram seemed like a good idea—not a big deal and certainly nothing I'd down-vote over. – Isaac Jan 25 '12 at 20:26
• Given the picture from Victor Liu's answer, letting the center of the circles be $O$, I'm convinced that $m\angle AOB=m\angle AEB$, $m\angle AOC=m\angle ADC$, and $m\angle BOC=m\angle BFC$, but I don't have a proof of it yet. If that's provably true, then we could use the central angles and the radii to compute the area of $\triangle ABC$ in terms of $u$, then use the given ratio of areas to solve for $u$. – Isaac Jan 26 '12 at 6:10
• I added the (analytic-geometry) tag because the accepted answer uses an algebraic technique. – Américo Tavares Feb 3 '12 at 18:33

I refer to the diagram of Victor Liu's answer. This is an analytical verification that $DF=8$ (which means that $u=2$) but omits some details. Using the equation of the circle centered at $(0,0)$ with radius $14$, the equation of $AE$ (tangent to the small circle at the point $(x,y)=(-8/5,6/5)$) $$y=\frac{4}{3}\left( x+\frac{8}{5}\right) +\frac{6}{5}$$ and the equations of $BF$ ($y=-2)$ and $CD$ ($x=2$), we get the coordinates of the vertices of triangle $ABC$: $$A\left( -\frac{8}{5}+\frac{24}{5}\sqrt{3},\frac{32}{5}\sqrt{3}+\frac{6}{5} \right) ,\qquad B(-8\sqrt{3},-2),\qquad C(2,-8\sqrt{3}).$$ The coordinates of the vertices of the right triangle's $DEF$ are $$D(2,8),\qquad E(-4,-2),\qquad F(2,-2).$$

The lengths of the sides of $ABC$ computed by the distance formula are $$a =BC=14\sqrt{2}, \qquad b =AC=\frac{42}{5}\sqrt{10}, \qquad c =AB=\frac{56}{5} \sqrt{5}.$$

The semi-perimeter $p$ of $ABC$ is thus $$p=\frac{a+b+c}{2}=7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}.$$

By Heron's formula the area of $ABC$ is $$S=S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)}.$$

Since $$\begin{eqnarray*} &&p(p-a)(p-b)(p-c) \\ &=&\left( 7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right) \left( -7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right) \\ &&\times \left( 7\sqrt{2}-\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right) \left( 7\sqrt{2}+\frac{21}{5}\sqrt{10}-\frac{28}{5}\sqrt{5}\right) \\ &=&\frac{1382976}{25}, \end{eqnarray*}$$

we get $$S=\sqrt{\frac{1382976}{25}}=\frac{1176}{5}.$$

The area of $DEF$ is $$K=S_{DEF}=\frac{EF\times DF}{2}=\frac{6\times 8}{2}=24$$ and the ratio $$\frac{S}{K}=\frac{1176/5}{24}=\frac{49}{5}=9.8,$$ as given.

Added: The scale is uniform throughout the following diagram drawn with the calculated equations Here is a diagram. I may or may not post a solution later.

Edit I will not post a solution since it appears to be quite messy. Please direct votes towards an actual solution. • In your diagram are the scales of the triangle $[DEF]$ and the big circle's equal? – Américo Tavares Jan 27 '12 at 19:47
• No. The scale of the triangle DEF is scaled by a factor u, whereas the radius of the big circle is 14 (absolutely). – Victor Liu Jan 30 '12 at 3:11
• Thanks. It seems to me that the solution would be very difficult. – Américo Tavares Jan 30 '12 at 11:10
• The scales of your diagram are uniform for both triangles and the big circle. At least the factor $u=2$ is compatible with my verification. – Américo Tavares Feb 2 '12 at 23:03