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The four shown circles have the same radius and each one is tangent to one side or two sides of the triangle. Each circle is tangent to the segment which is inside the triangle ABC. Besides, the central lower circle is tangent to its neighbor circles. If AC = 12 cm, what is the value of the area of the triangle ABC?

I tried to assign angle variables to the triangle to compute the sides of the triangle, and so to use Heron's formula to calculate the area, but after all, it appears that some data is missing. I haven't found any book or article which treats these kind of problems.

Many thanks in advance enter image description here

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  • $\begingroup$ Is the smaller line parallel to the base line? $\endgroup$ – Aditya Jul 16 '17 at 1:10
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    $\begingroup$ @ADITYA: if the depicted circles have the same radius it has to be. $\endgroup$ – Jack D'Aurizio Jul 16 '17 at 1:25
  • $\begingroup$ @JackD'Aurizio I just drew this figure in a CAD software (solidworks) and I found that, although the radius is variable, the altitude to AC is always 6. Hence the answer is 36. However, I don't know how to solve it mathematically. $\endgroup$ – Huang Jul 16 '17 at 1:37
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    $\begingroup$ @LuoKaisa: I just used Geogebra (www.geogebra.org) and the export-to-png feature. $\endgroup$ – Jack D'Aurizio Jul 16 '17 at 2:20
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    $\begingroup$ @LuoKaisa, I hope my solution will help too. $\endgroup$ – Seyed Jul 16 '17 at 23:23
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Let us construct a similar configuration by choosing the radius $r$ of the involved circles and the "critical distance" $d$:

enter image description here

The area of the innermost triangle is $4r^2$ and the innermost triangle is similar to $ABC$. The angle bisectors of $ABC$ are also the angle bisectors of the innermost triangle, so the ratio between a side length of $ABC$ and the length of the parallel side in the innermost triangle is $\frac{i+r}{i}$, with $i$ being the inradius of the innermost triangle. It follows that the area of $ABC$ is $$ 4r^2\left(\frac{i+r}{i}\right)^2 $$ and $\frac{i+r}{i}=\frac{AC}{4r}$. But... wait! This gives that the area of $ABC$ is just $\color{red}{\frac{1}{4}AC^2}$, we do not need to know neither $r$, or $i$, or $d$!!!


In particular, the distance of $B$ from $AC$ is exactly $\frac{AC}{2}$.

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    $\begingroup$ +1 for your nice answer. You always shows great talents in mathematics. $\endgroup$ – Huang Jul 16 '17 at 1:47
  • $\begingroup$ I was a little bit late with my solution :) $\endgroup$ – Seyed Jul 16 '17 at 23:24
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By a little modification we can change the figure to a more easier form. We can move the upper circle to the right and exactly above the middle circle without changing the area of the triangle. enter image description here

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  • $\begingroup$ If you move the upper circle you also change the length of $AC$, so you should prove it is enough to consider this symmetric configuration. $\endgroup$ – Jack D'Aurizio Jul 16 '17 at 23:31
  • $\begingroup$ @JackD'Aurizio, No we don't change the length of $AC$. Imagine the $AC$ is fixed and we have only those 3 circles tangent to $AC$, and now if we move all those 3 circles to the right or left, $B$ will stay at the same height and the area remains unchanged. By this we only change the length of $AB$ and $BC$ which they don't have any effect to the area. $\endgroup$ – Seyed Jul 16 '17 at 23:51
  • $\begingroup$ You have to convince me better. Assume that $AC$ is fixed and $B$ is given. In order to bring $B$ on the perpendicular bisector of $AC$ and preserving the area of $ABC$, you have to change the radii of the involved circles, not only their positions. How do you do that? $\endgroup$ – Jack D'Aurizio Jul 17 '17 at 0:03
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    $\begingroup$ @JackD'Aurizio, Of course the radii of the circles wouldn't be the same, something which is not $given$ in the problem. But no matter where the $B$ is located at that level above $AC$, we can still have four equal circles tangents to each other, sides and that line segment with no effect to the area on the triangle. $\endgroup$ – Seyed Jul 17 '17 at 0:18

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