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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.

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    If you've got a diagram to go with the question, that would be very helpful.2012-01-25
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    Okay, sorry for such a poor presentation of a question, you can leave it till I make a diagram. Thanks for not voting down.2012-01-25
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    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.2012-01-25
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    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 $u$.2012-01-26
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    I added the (analytic-geometry) tag because the accepted answer uses an algebraic technique.2012-02-03

2 Answers 2

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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}.

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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

enter image description here

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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.

enter image description here

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    In your diagram are the scales of the triangle $[DEF]$ and the big circle's equal?2012-01-27
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    No. The scale of the triangle DEF is scaled by a factor u, whereas the radius of the big circle is 14 (absolutely).2012-01-30
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    Thanks. It seems to me that the solution would be very difficult.2012-01-30
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    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.2012-02-02