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Circle inscribed in a triangle with triangle inscribed circle...

An equilateral triangle is inscribed in a circle, which (the circle) is also inscribed in another equilateral triangle. what is the ratio of the areas of the inner triangle and outside triangle?

Note by Alan Liang
4 years, 8 months ago

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The ratio is \(1/4\). Simply rotate the smaller triangle so that its corners lie at the midpoints of the sides of the larger triangle.

Francis Gerard Magtibay - 4 years, 8 months ago

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i like yr answer

Roma Refaat - 4 years, 8 months ago

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Refer here for the figure. Let \(O\) be the center of the circle and the radius of the circle be \(r\). The equilateral triangle \(\Delta ABC\) is inscribed inside the circle and the circle is inscribed inside the equilateral triangle \(\Delta DEF\). Let \(AA'\) be perpendicular to \(BC\) and \(DD'\) be perpendicular to \(EF\). Since \(\Delta ABC\) and \(\Delta DEF\) are equilateral triangle, \(AA'\) and \(DD'\) are both medians to the respective sides from respective vertices and both pass through \(O\), which is also the centroid. Recall that the centroid divides the medians in the ratio \(2:1\). Hence, we get that \[\dfrac{AO}{AA'} = \dfrac{DO}{DD'} = \dfrac23 \implies AA' = \dfrac{3r}2; DD' = 3r\] Hence, \[\dfrac{\text{Area of }\Delta ABC}{\text{Area of }\Delta DEF} = \dfrac{\dfrac12 \cdot AA' \cdot BC}{\dfrac12 \cdot DD' \cdot EF} = \dfrac{AA'}{DD'} \cdot \dfrac{BC}{EF} = \dfrac{AA'}{DD'} \cdot \dfrac{A'C}{D'F} = \left(\dfrac{AA'}{DD'} \right)^2 = \dfrac14\]

Marvis Narasakibma - 4 years, 8 months ago

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i think the ratio is 1/4

Nasir Afroze - 4 years, 8 months ago

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I think it's \(\frac{1}{4}\)...

or maybe I have made wrong in calculations...

Galih Pradananta - 4 years, 8 months ago

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Thanks, but could you give me steps on how you did it?

Alan Liang - 4 years, 8 months ago

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Don't you know that \(r=\frac{A}{s}\) if the circle is inscribed in a triangle. With r=radius of circle, A=area of triangle and s=a half of triangle's perimeter... \(r=\frac{abc}{4A}\) if the circle is circumcircle with a,b,c=side of the triangle. Or I'll give the link for you? Sorry I can't draw the picture for you.

Galih Pradananta - 4 years, 8 months ago

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according to me it should be 1/4

Manan Chheda - 4 years, 7 months ago

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\(\frac{1}{4}\)

Zi Song Yeoh - 4 years, 8 months ago

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it can be done by rotation, rotate the inner triangle by 60 degrees with centroid fixed, so now the inner triangle has its vertices on the midpoints of the smaller triangle!! so 1/4 is tthe answer.

Rahul Bhagtani - 4 years, 8 months ago

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1/4

Harsa Mitra - 4 years, 8 months ago

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1/4

M. Ibrahim Aftab Khan - 4 years, 8 months ago

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It's 1/4.

Jandy Diaz - 4 years, 8 months ago

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\frac{1}{4}

Rushikesh Jogdand - 4 years, 8 months ago

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Two triangles are similar, altitudes are 3R and 3R/2 => ratio of altitudes = 1 : 2

=> ratio of areas = 1 : 4

Vinay Sipani - 4 years, 8 months ago

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Rotate the inner triangle so that it is the medial triangle of the outside triangle.

Qi Huan Tan - 4 years, 8 months ago

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Yeah.... I'll go with 1:4

Kumail Raza - 4 years, 8 months ago

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1/5

Arjel Alvarez - 4 years, 8 months ago

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1:3

Ved Prakash - 4 years, 8 months ago

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Fail.

Tim Ye - 4 years, 8 months ago

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