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An isosceles triangle of area 3072 contains 3 non-overlapping circles, each of diameter 30. What is its base?

Find the length of the base of an isosceles triangle if has area 3072 and it contains 3 non-overlapping circles each with diameter 30.
@Elio D: good that you drew a picture. The answer I got is different from yours. Perhaps you should check again.

The smallest equilateral triangle which is tangent to 3 tangent circles inside of it has a side a=30(1+sqrt(3))~81.961524 and a height h=15(3+sqrt(3))~70.980762.
Since 3072=3*2^10 ==> a=85 1/3 and h=72, b=sqrt((a/2)^2+h^2).

I will post the drawing with more details tomorrow.

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2 Responses to “An isosceles triangle of area 3072 contains 3 non-overlapping circles, each of diameter 30. What is its base?”

  1. From my drawing, I SUSPECT, that the diameter of the top circle, added to the diameter of the two bottom ones is 2/3 of the way up the altitude of the triangle. I say I suspect since I haven’t had enough time to prove it. IF that is the case, then the altitude is 90, and so

    3072 = (1/2)B(90) which means the base B = 3072/45 = 68.2 and 2/3. I have NO PROOF of this.
    References :

  2. The smallest equilateral triangle which is tangent to 3 tangent circles inside of it has a side a=30(1+sqrt(3))~81.961524 and a height h=15(3+sqrt(3))~70.980762.
    Since 3072=3*2^10 ==> a=85 1/3 and h=72, b=sqrt((a/2)^2+h^2).

    I will post the drawing with more details tomorrow.
    References :

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