[Guest]

In an acute angled triangle ABC of sides a, b and c, an incircle is drawn.

3 tangents are drawn to this circle such that they are parellel to the sides of the triangle ABC.

Each of these 3 tangents forms 3 smaller triangles inside the original triangle ABC.

Incircles are drawn inside each of these 3 smaller triangles.

What is the total area covered by the 4 circles?

I read this problem somewhere and don't remember the exact wording of the problem, so in case you need clarification, pls ask.

PS: I dont know the answer and what I've worked out may not be correct... so we'll have to go by looking at your solution logic to see if it is right. So, if you post your answer, pls post the method/reasoning also.

[Guest]

I am sorry. I did not read the question correctly

Edited September 3, 2009 by jerbil

[Guest]

the radius of the large circle ® can be described as

r = c • ( (sin(A/2) • sin(B/2) ) / cos(C/2) )

therefore

area = (c • ( (sin(A/2) • sin(B/2) ) / cos(C/2) ) )² • pi

the sum of the area of the smaller circles will be equal to one-third of the area of the large circle

the total area = (4/3) • c • ( (sin(A/2) • sin(B/2) ) / cos(C/2) ) )² • pi

[Guest]

[Guest]

the radius of the large circle ® can be described as

r = c • ( (sin(A/2) • sin(B/2) ) / cos(C/2) )

therefore

area = (c • ( (sin(A/2) • sin(B/2) ) / cos(C/2) ) )² • pi

the sum of the area of the smaller circles will be equal to one-third of the area of the large circle

the total area = (4/3) • c • ( (sin(A/2) • sin(B/2) ) / cos(C/2) ) )² • pi

Are you sure your formula for incircle raius is correct?

Even if it is correct, assuming that the area of smaller circles will be 1/3 of the bigger is definitely not correct. That would be the case only for an equilateral triangle not for a general triangle.

[Guest]

the formula for the area of the circle is correct, but not the only possible formula

as for the smaller circles, each is a ratio of the larger one determined by the length of the segment from the shared corner through the opposing tangent to the same segment minus the diameter of the large circle

[Guest]

as for the smaller circles, each is a ratio of the larger one determined by the length of the segment made by the angle bisector to the same segment minus the diameter of the large circle

[plainglazed]

The radius of circle inscribed in a triangle is: sqrt{(s-a)(s-b)(s-c)/s} where s = 1/2 the perimeter of the triangle or (a+b+c)/2

So the area of the large circle is simply pi{(s-a)(s-b)(s-c)/s}.

If we know the three legs of the smaller triangles we can calculate the corresponding area of the smaller inscribed circles the same way.

Let's break down the sides of the large triangle as follows:

a = a_B + a_A + a_C where:

a_B is the segment from B to the point of intersection of the tangent line parallel to b and segment BC

a_C is the segment from C to the point of intersection of the tangent line parallel to c and segment BC

a_A is the segment between the two

b = b_C + b_B + b_A where:

b_C is the segment from C to the point of intersection of the tangent line parallel to c and segment AC

b_A is the segment from A to the point of intersection of the tangent line parallel to a and segment AC

b_B is the segment between the two

c = c_A + c_C + c_B where:

c_A is the segment from A to the point of intersection of the tangent line parallel to a and segment BA

c_B is the segment from B to the point of intersection of the tangent line parallel to b and segment BA

c_C is the segment between the two

now the three small triangles have sides labelled as follows:

c_A,b_A,{a-a_B-a_C} (small triangle at A)

c_B,a_B,{b-b_A-b_c} (small triangle at B)

a_C,b_C,{c-c_A-c_B} (small triangle at C)

now for the lengths of the legs:

a_B = {a-2r/sin©}

a_C = {a-2r/sin(B)}

b_A = {b-2r/sin©}

b_C = {b-2r/sin(A)}

c_B = {c-2r/sin(A)}

c_A = {c-2r/sin(B)}

Plug and reduce...

This is so symmetrical it seems like it would reduce quite nicely. Too much for me tho...

(Sorry for the cumbersome labeling description but no time or skills to work up a drawing)

[plainglazed]

...sketch than a drawing but like I said, limited skills. And as for the time, didn't really need lunch anyway.

[Guest]

...sketch than a drawing but like I said, limited skills. And as for the time, didn't really need lunch anyway.

Dunno why all the apologies. From the start I thought this was merely a semi-perimeter problem, where the three subsidiary circles were similar to the original triangle.

[Guest]

For a triangle with sides a, b, c it's area is:

A = \sqrt [s(s-a)(s-b)(s-c)]

where s = 1/2(a + b + c)

thus the radius of the inner circle is:

r = A/s,

the other three circles have the radius:

r_c = r(h_c - 2r)/h_c, this is the one opposite c, where h_c = A/(1/2c) = 2A/c,

r_b = r(h_b - 2r)/h_b, this is the one opposite b, where h_b = A/(1/2b) = 2A/b,

r_a = r(h_a - 2r)/h_a, this is the one opposite a, where h_a = A/(1/2a) = 2A/a.