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
8 years, 4 months ago

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6 votes

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Comments

The ratio is 1/41/4. Simply rotate the smaller triangle so that its corners lie at the midpoints of the sides of the larger triangle.

Francis Gerard Magtibay - 8 years, 4 months ago

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i like yr answer

RoMa Refaat - 8 years, 4 months ago

Refer here for the figure. Let OO be the center of the circle and the radius of the circle be rr. The equilateral triangle ΔABC\Delta ABC is inscribed inside the circle and the circle is inscribed inside the equilateral triangle ΔDEF\Delta DEF. Let AA′AA' be perpendicular to BCBC and DD′DD' be perpendicular to EFEF. Since ΔABC\Delta ABC and ΔDEF\Delta DEF are equilateral triangle, AA′AA' and DD′DD' are both medians to the respective sides from respective vertices and both pass through OO, which is also the centroid. Recall that the centroid divides the medians in the ratio 2:12:1. Hence, we get that AOAA′=DODD′=23  ⟹  AA′=3r2;DD′=3r\dfrac{AO}{AA'} = \dfrac{DO}{DD'} = \dfrac23 \implies AA' = \dfrac{3r}2; DD' = 3r Hence, Area of ΔABCArea of ΔDEF=12⋅AA′⋅BC12⋅DD′⋅EF=AA′DD′⋅BCEF=AA′DD′⋅A′CD′F=(AA′DD′)2=14\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 - 8 years, 4 months ago

I think it's 14\frac{1}{4}...

or maybe I have made wrong in calculations...

Galih Pradananta - 8 years, 4 months ago

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

Alan Liang - 8 years, 4 months ago

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Don't you know that r=Asr=\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=abc4Ar=\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 - 8 years, 4 months ago

i think the ratio is 1/4

nasir afroze - 8 years, 4 months ago

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

QI HUAN TAN - 8 years, 4 months ago

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

=> ratio of areas = 1 : 4

Vinay Sipani - 8 years, 4 months ago

\frac{1}{4}

Rushikesh Jogdand - 8 years, 4 months ago

It's 1/4.

Jandy Diaz - 8 years, 4 months ago

1/4

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

1/4

Harsa Mitra - 8 years, 4 months ago

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 - 8 years, 4 months ago

14\frac{1}{4}

Zi Song Yeoh - 8 years, 4 months ago

according to me it should be 1/4

MANAN CHHEDA - 8 years, 3 months ago

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

Kumail Raza - 8 years, 4 months ago

1/5

Arjel alvarez - 8 years, 4 months ago

1:3

ved prakash - 8 years, 4 months ago

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Fail.

Tim Ye - 8 years, 4 months ago
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