areas, 4 equal right triangles, 2 circles (2010 Denmark MO- Mohr Contest p1)

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areas, 4 equal right triangles, 2 circles (2010 Denmark MO- Mohr Contest p1)

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#1 h Aug 13, 2020, 10:51 PM

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Four right triangles, each with the sides $1$ and $2$ , are assembled to a figure as shown.
How large a fraction does the area of the small circle make up of that of the big one?

https://1.bp.blogspot.com/-XODK1XKCS0Q/XzXDtcA-xAI/AAAAAAAAMWA/zSLPpf3IcX0rgaRtOxm_F2begnVdUargACLcBGAsYHQ/s0/2010%2BMohr%2Bp1.png

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#2 h Aug 13, 2020, 11:03 PM

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sol

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#3 h Aug 13, 2020, 11:06 PM

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Because the little circle is tangent to the star shape, the radius of the small circle is the altitude of the right triangle from the hypotenuse, which is $\frac{2\sqrt{5}}{5}$ .
The radius of the larger circle is $2$ , or just the long leg of the right triangle.
So, the answer is $(\frac{\frac{2\sqrt{5}}{5}}{2})^2$ , or $\frac{1}{5}$ .

This post has been edited 2 times. Last edited by aie8920, Aug 13, 2020, 11:07 PM

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#4 h Aug 14, 2020, 2:02 AM

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Nice mangekyo.

Anyway the radius of the larger circle is clearly 2. The radius of the smaller circle will be the non-leg height of the right triangle. Which is $\frac{1\times 2}{\sqrt{5}}$ . So the ratio of the areas is $\frac{4}{5\times 4}=\frac{1}{5}$

This post has been edited 1 time. Last edited by natmath, Aug 14, 2020, 2:02 AM

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