Nice circle geometry question. (1 Viewer)

[seanieg89]

Let ABCD be a cyclic quadrilateral and let X be the point of intersection of the tangents at A and C. (We assume that AC is not a diameter of the circumcircle of ABCD.)

Prove that X lies on BD if and only if AB.CD=AD.BC.

 

[jyu]

Let ABCD be a cyclic quadrilateral and let X be the point of intersection of the tangents at A and C. (We assume that AC is not a diameter of the circumcircle of ABCD.)

Prove that X lies on BD if and only if AB.CD=AD.BC.

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Reasonable proof?

 

[seanieg89]

Pretty good. Not entirely happy with the "work backwards to complete the proof" though...I think some justification is needed for why e=a+b, f=pi-b+c is the only possible solution.

 

[johnpap]

If BDX collinear, then by power of a point, XD.XB = XA^2 = XC^2, so XDA and XDC are similar to XAB and XAC respectively (Since you have side ratios and the included angle and stuff...). Then AD/AB = XA/XB = XC/XB = DC/BC, so AD/AB = DC/BC, so AB.CD = AD.BC.
I'm not entirely sure what you're allowed to do in the HSC though for the reverse direction though. I mean normally you'd just say, if AB.CD = AD.BC, let D' be the point where BX intersects the circle (Assuming the diagram has B further from X than D but it doesn't really matter if you have to permute points anyway), then by the first part of the question, AB.CD' = AD'.BC, so AD'/CD' = AD/CD, which implies D' = D because for a moving point Z on the arc AC, the value AZ/ZD is continuous and increasing over the range [0,infinity+) as Z tracks from A to C, so f(Z) = AZ/ZC is an injective function on all the points on that arc so you know D' = D when AD'/CD' = AD/CD (alternatively if you want to be complicated about it the Apollonius circle with the appropriate ratio with respect to AC can only go through the circumcircle of ABCD once on the arc AC - since it goes through some internal ratio divisor of AC and the corresponding external ratio divisor - which again gives uniqueness), but I'm not sure how well that proof would be received in the actual HSC...

 

[seanieg89]

Hey yep, the circle of Apollonius (or line if the fixed ratio is 1:1) is how I proved D=D'. That injectivity argument is clever . Probably wouldn't be received well in the HSC but then again, most HSC geometry questions are much easier.

 

[largarithmic]

If BDX collinear, then by power of a point, XD.XB = XA^2 = XC^2, so XDA and XDC are similar to XAB and XAC respectively (Since you have side ratios and the included angle and stuff...). Then AD/AB = XA/XB = XC/XB = DC/BC, so AD/AB = DC/BC, so AB.CD = AD.BC.
I'm not entirely sure what you're allowed to do in the HSC though for the reverse direction though. I mean normally you'd just say, if AB.CD = AD.BC, let D' be the point where BX intersects the circle (Assuming the diagram has B further from X than D but it doesn't really matter if you have to permute points anyway), then by the first part of the question, AB.CD' = AD'.BC, so AD'/CD' = AD/CD, which implies D' = D because for a moving point Z on the arc AC, the value AZ/ZD is continuous and increasing over the range [0,infinity+) as Z tracks from A to C, so f(Z) = AZ/ZC is an injective function on all the points on that arc so you know D' = D when AD'/CD' = AD/CD (alternatively if you want to be complicated about it the Apollonius circle with the appropriate ratio with respect to AC can only go through the circumcircle of ABCD once on the arc AC - since it goes through some internal ratio divisor of AC and the corresponding external ratio divisor - which again gives uniqueness), but I'm not sure how well that proof would be received in the actual HSC...

HEY JOHN GUESS WHO

You can also do a really neat trick using Ptolemy's theorem (wikipedia it, its neat and elementary, has come up in HSC before):

Suppose there exists D, D' distinct on the same arc AC such that AD/CD = AD'/CD', i.e. AD.CD' = AD'.CD . Also ACDD' (or in the other order) is cyclic, so you must have either AD.CD' = AD'.CD + AC.DD' or AD'.CD = AD.CD' + AC.DD' depending on the order of A,C,D,D' around the circle. In either case, you obtain AC.DD' = 0, which clearly requires D = D'.

edit: you can also do this with cross ratio but john you should be solving stuff using normal geometry