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Circle, Circle, Triangle, Circle! (Posted on 2004-09-08) Difficulty: 4 of 5
Consider a circle O with a diameter AB, shown here in green. Draw a second circle (red) with diameter AC, such that C is on AB. Draw an isosceles triangle with base CB and third vertex D on circle O. Draw a third circle (X), tangent to the first three figures. Prove that the line from C to the center of circle X is perpendicular to AB.

No Solution Yet Submitted by SilverKnight    
Rating: 3.5556 (9 votes)

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Solution Solution | Comment 4 of 10 |

Thoroughly enjoyed this problem.  Here's my solution:

Let w1 denote the red circle and w2 the green circle.  Let r be the radius w1 and let R be the radius w2.  Let x=2r(R-r)/(R+r).  Draw a circle w3 externally tangent to w1 and internally tangent to w2 with radius x.  Note that such a circle exists since x<R-r. 

Let P, Q, and S be the centers of w1, w2, and w3, respectively.  Let a=angle(PQS).  Applying the law of cosines to triangle PQS, we have

PS^2 = PQ^2 + QS^2 - 2(PQ)(QS)cos(PQS)

or

(r+x)^2 = (R-r)^2 + (R-x)^2 - 2(R-r)(R-x)cos(a)

or, after some algebraic manipulation,

cos(a)=(R^2-Rr-Rx-rx)/((R-r)(R-x)).

Now note that

QC/QS = (2r-R)/(R-x),

which after some algebraic manipulation (making use of the fact that x=2r(R-r)/(R+r)), equals

-(R^2-Rr-Rx-rx)/((R-r)(R-x)),

or cos(180-a)=cos(180-PQS)=cos(CQS).  This implies that triangle CQS is a right triangle with right angle at C.  In other words, CS is perpendicular to AB.  It would suffice to show that w3=X. 

Now draw a line through C tangent to w3, "away" from w1.  Let this line meet w2 at T and w3 at U.  Drop a perpendicular from T to a point M on AB.  Since CS and TM are parallel, we have that angle(SCT)=angle(CTM).  Thus the right triangles CUS and TMC are similar, implying that

CU/US = TM/MC                  (Equation 1)

We can compute CU from the Pythagorean theorem on triangle CUS:

CU = sqrt(CS^2-SU^2).       (Equation 2)

To get the value CS^2, we again need the Pythagorean theorem on triangle PCS:

CS^2 = PS^2 - CP^2
         = (r+x)^2 - r^2
         = x^2 + 2rx,

so plugging back into Equation 2,

CU = sqrt(x^2+2rx - x^2)
     = sqrt(2rx).

Finally, plugging back into Equation 1,

sqrt(2rx)/x = TM/MC,

or letting MC = y,

TM = y*sqrt(2r/x).

Since TM is the altitude from the hypotenuse of right triangle ATB, we have the relation

TM^2 = AM*MB,

or

2ry^2/x = (2R-y-2r)(y+2r).

We can solve this equation for y.  When we do so (using the relation x=2r(R-r)/(R+r)), we find:

y = R-r.

What does this mean?  It means that M is the midpoint of CB, since we now have

BM = BA - AM
     = BA - AC - CM
     = 2R - 2r - y
     = R-r
     = y
     = MC.

Thus, T coincides with the point D as described in the problem.  It follows that w3 coincides with X, which completes the proof.


  Posted by David Shin on 2004-09-09 16:09:12
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