8

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Radius of $A$ is $2$, Radius of $B$ is $1$, Radius of $C$ is $4$, Radius of $X$ is $3$. Find the radius of $D$

Could someone please help with this question? :)

EDIT: How I tried to solve it - I tried using similar triangles and other things but to be honest I really don't know how to do this question. This led me to getting 5 as the (incorrect) answer so I'm still not sure what to do really..

However, I found that the tangents of the radii are parallel, if that helps anyone!

Update again: I thought about using the cosine law and setting up two triangles: CX and C's perpendicular radius and CA and C's perpendicular radius

Yet another update: I thought about maybe trying this

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Hiraphor
  • 179
  • Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Mar 26 '15 at 07:49
  • I can prove the centers of the outer circles are concyclic, but I haven't found how/if it helps. – Peter Woolfitt Mar 26 '15 at 08:57
  • Writing $a$, $b$, $c$, $d$ for the radii of $\bigcirc A$, $\bigcirc B$, $\bigcirc C$, $\bigcirc D$, it happens that $ac = bd$. (The radius of $\bigcirc X$ is irrelevant!) I got this result by a tedious coordinate argument; I'll seek an elegant one later. – Blue Mar 26 '15 at 10:48
  • @Blue sweet! ${}{}{}$ – Peter Woolfitt Mar 26 '15 at 10:55

1 Answers1

1

Sorry for this bad translation but it's originally in japanese, hopefully you can understand!

The contact of the circle X and QR, E, the contact of the circle D and QR, I and F. (Following contacts, please look at the figure) and the inscribed circle X, I explained in Sobase~tsuen D. And inscribed circle, from the nature of the near contact circle, (inscribed circle X of radius) × (Sobase~tsuen D of radius) = QE × QF, ※ supplemental reference (QE = RF = a, QF = RE = d, You radius x beside contact circle.) Thus, ad = 3x = ... ① Similarly, inscribed circle X, to think in Sobase~tsuen C, and you have (PH = PG = b,) ab = 3 × 4 ... ② inscribed circle B, and think in Sobase~tsuen X, (and with SG = SK = c,) BC = 1 × 3 ... ③ inscribed circle A, to think in Sobase~tsuen X, cd = 2 × 3 ... ④ than ① × ③ = ② × ④, x = 8 Therefore, the radius of the circle D is, 8 (A radius) × (C radius of) = (radius of B) × (radius of D) to be, I understand. Supplemental ※ proof, △ XQE∽ △ QDF than (∠XQD = 90 °, more, ∠XQE = ∠QDF) QE: DF = XE: QF than, DF × XE = QE × QF

Hiraphor
  • 179