Another mathematical puzzle:

Find the area of $\triangle FGH$, given that $FG=FH$ and the radii of the circles shown are $2$ and $1$

Made with GeoGebra

  • $\begingroup$ please provide the source if this is not your own work. else it will likely be closed. thanks $\endgroup$ – Omega Krypton Aug 12 '20 at 14:17
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    $\begingroup$ The question says FG=GH but the diagram is drawn as if FG=FH and GH is very different. Is there a typo in the question, or is the diagram deliberately very misleading? $\endgroup$ – Gareth McCaughan Aug 12 '20 at 14:38
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    $\begingroup$ It somewhat looks like a homework... $\endgroup$ – CiaPan Aug 12 '20 at 14:47
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    $\begingroup$ Duplicate of math.stackexchange.com/q/1766545/369453 :D $\endgroup$ – Marius Aug 12 '20 at 15:10
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    $\begingroup$ @GarethMcCaughan Thanks for pointing that out, I've fixed it. Also, good catch by Marius! I didn't know that this had been asked before. It's purely coincidential :) $\endgroup$ – Aniruddha Deb Aug 12 '20 at 15:16

The answer is:



if you draw radii where the circles are tangent to the triangles with tangent points marked as $D$ (smaller circle) and $E$ (larger circle; this forms right angles), you get the following similar triangles: $\triangle FDA \sim \triangle FEB \sim \triangle FXG$. $A$ is the center of the small circle and $B$ is the center of the large one. $X$ is the midpoint of $HG$.

From this, we get this proportion:

$\frac{FA}{FB} = \frac{DA}{EB}$ OR $\frac{FA}{FA+3} = \frac{1}{2}$ because of the radii lengths.



Using Pythagorean Theorem:

$FD = \sqrt8$

Using our similar triangle relations from the beginning, we get:

$XG = 2\sqrt2$

Finally, we use the area of a triangle formula to get:


  • $\begingroup$ Sorry, I got tired of showing my work... $\endgroup$ – Voldemort's Wrath Aug 12 '20 at 14:52
  • $\begingroup$ I'm sure this is a good answer, but I got kind of lost when you started drawing new things. If you had a picture, that would help greatly. $\endgroup$ – Chipster Aug 12 '20 at 22:38
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    $\begingroup$ @Chipster - No, you're right... I'll add that in when I find time! $\endgroup$ – Voldemort's Wrath Aug 13 '20 at 11:57
  • $\begingroup$ A picture would help, but even with no picture, please could you clarify which point is $D$ and which is $E$? $\endgroup$ – Rosie F Aug 14 '20 at 8:57
  • $\begingroup$ @RosieF -- $D$ is for the smaller circle and $E$ is for the larger one. $\endgroup$ – Voldemort's Wrath Aug 14 '20 at 15:48

Let $h$ denote the height of the triangle from F onto GH. Then, by Pythagoras' theorem, $$FG^2=h^2+(GH/2)^2,$$ and by triangles' similarity $$h:(GH/2) = (FG-GH/2):2$$ and $$(h-4):h = 1:2.$$

similar triangles

Can you continue from this?

From the last we get $h-4=h/2$ which resolves to $h=8.$ Plug it to previous two to get

$$\begin{cases}FG^2=64+(GH/2)^2\\8:(GH/2) = (FG-GH/2):2\end{cases}$$ The last equation is equivalent to $$16:(GH/2) = (FG-GH/2)$$


$$FG = (GH/2)+16:(GH/2)$$ and $$FG^2 = (GH/2)^2 + 32 + 256:(GH/2)^2$$

Compare it to the first equation: $$64+(GH/2)^2 = (GH/2)^2 + 32 + 256:(GH/2)^2$$ $$64 = 32 + 256:(GH/2)^2$$ $$32 = 256:(GH/2)^2$$ $$(GH/2)^2 = 256:32 = 8$$ $$GH/2 = \sqrt 8 = 2\sqrt 2$$

Finally the area sought is

$$S_{\triangle FGH} = h\cdot GH/2 = 16\sqrt 2.$$

  • $\begingroup$ See my answer; I think I got it?? $\endgroup$ – Voldemort's Wrath Aug 12 '20 at 14:50
  • $\begingroup$ @Voldemort'sWrath Yes, you did. I didn't show my solution at first, because the question looks like a homework (my comment), so I wanted the author to show some effort first. Now I uncommented the hiden part with the same result as yours. $\endgroup$ – CiaPan Aug 13 '20 at 7:39

Let me chip in with a streamlined answer.

$8\sqrt 8$

All the smaller circle is telling us is that the height of the triangle must be $8$, so let's jot that down and from here pretend the small circle never existed.

The points where the circle (there is only one circle!) touches the triangle divide the long sides into two segments of lengths $y>x$ and the base into two equal segments of length $x$. Let us now compute the area $A$ of the triangle from base and height $A=8x$, from incircle radius and circumference $A=2(2x+y)$ and from Heron's formula $A^2=x^2y(2x+y)$. Comparing the first two gives $2x=y$, together with the last we get $x=\sqrt 8$ and $A=8\sqrt 8$.

  • $\begingroup$ Please explain where A = 2(2x + y) came from. Thanks. $\endgroup$ – asg Aug 13 '20 at 8:56
  • $\begingroup$ @asg as it says "from incircle radius and circumference". $2x+y$ is the semicircumference often written $s$ of the triangle. $\endgroup$ – Paul Panzer Aug 13 '20 at 9:21

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