I suspect you have that the answer is 13. This is incorrect, as the circles are not tangent to the lines at the same points. To fix this, simply do not give the radius of circle A. I will assume the radius of circle A is unknown and that the circles are tangent to the lines in the same points in the following.

Let the centers of the circles be \(O_A,O_B,O_C,O_D\). Let \(P\) be the intersection point of the tangents, and let circles B and C be tangent at \(M\). Let the radius of circle A be \(r_a\), and similar for circles B,C,D. We will attempt to find \(O_AO_C\).

Note that \(r_b=3\). Since \(A1,A2,A3\) are in arithmetic progression, \(m\angle A2=90^\circ\), and so \(m\angle O_AO_BO_C=90^\circ\).

By symmetry, \(P\) lies on \(O_AO_C\). Since \(\angle O_CMP\cong\angle O_CO_BO_A\) and \(\angle MO_CP\cong\angle O_BO_CO_A\), triangles \(\triangle O_CMP\) and \(\triangle O_CO_BO_A\) are similar by AA similarity. Therefore,
\[\dfrac{r_c}{r_b}=\dfrac{r_c+r_b}{r_b+r_a}\implies r_a=1\]

Then, \(\triangle O_AO_BO_C\) is a right triangle, and
\[O_AO_C=\sqrt{O_AO_B^2+O_BO_C^2}=\sqrt{(r_a+r_b)^2+(r_b+r_c)^2}=\sqrt{4^2+12^2}=4\sqrt{10}\]
–
Daniel Chiu
·
3 years, 8 months ago

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TopNewestI suspect you have that the answer is 13. This is incorrect, as the circles are not tangent to the lines at the same points. To fix this, simply do not give the radius of circle A. I will assume the radius of circle A is unknown and that the circles are tangent to the lines in the same points in the following.

Let the centers of the circles be \(O_A,O_B,O_C,O_D\). Let \(P\) be the intersection point of the tangents, and let circles B and C be tangent at \(M\). Let the radius of circle A be \(r_a\), and similar for circles B,C,D. We will attempt to find \(O_AO_C\).

Note that \(r_b=3\). Since \(A1,A2,A3\) are in arithmetic progression, \(m\angle A2=90^\circ\), and so \(m\angle O_AO_BO_C=90^\circ\).

By symmetry, \(P\) lies on \(O_AO_C\). Since \(\angle O_CMP\cong\angle O_CO_BO_A\) and \(\angle MO_CP\cong\angle O_BO_CO_A\), triangles \(\triangle O_CMP\) and \(\triangle O_CO_BO_A\) are similar by AA similarity. Therefore, \[\dfrac{r_c}{r_b}=\dfrac{r_c+r_b}{r_b+r_a}\implies r_a=1\]

Then, \(\triangle O_AO_BO_C\) is a right triangle, and \[O_AO_C=\sqrt{O_AO_B^2+O_BO_C^2}=\sqrt{(r_a+r_b)^2+(r_b+r_c)^2}=\sqrt{4^2+12^2}=4\sqrt{10}\] – Daniel Chiu · 3 years, 8 months ago

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– Priyansh Sangule · 3 years, 7 months ago

Oh! Thank You for Pointing it out ! :) Nice Solution ;)Log in to reply

Are the adjacent circles supposed to be tangent to the line between them on the same point? – Daniel Liu · 3 years, 8 months ago

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– Priyansh Sangule · 3 years, 8 months ago

Oh yeah .Log in to reply

Calculate the radius of B to be 3 units and A2 would be 90 degrees

Get the distance to be \sqrt{13} + \sqrt{90} == 13.092 – Sanjay Banerji · 3 years, 8 months ago

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– Priyansh Sangule · 3 years, 8 months ago

Well , somethings not rightLog in to reply