Ochoas Problem Enlightened

By E. A. J. Garca.

Recall Daos Eyeballing a Ball problem. In this case the segments IJ, GH

are not necessarily congruent to those segments corresponding to the

Praying Eyes Theorem. So, one would ask: when segments IJ, GH are

equal to those of the Praying Eyes Theorem? Well, Ochoa has answered the

question by misunderstanding Daos problem.

Ochoas image #1

In Ochoas image at CTK-Facebook, he takes ABD as a triangle being

circle centred at C its incircle, which is different to Daos configuration.

Ochoa also stated that MN equals IJ = GH.

Now I will prove that IJ = GH = MN, but I will change the notations.

Consider two circles (A), (C). From A draw a tangent line to (C) in I. From

C draw a tangent line to (A) in J. Let E be the intersection of AI and CJ.

Draw a circle centred at E and tangent to AC in F. Now, from A draw a

tangent line to (E) in G and from C draw another tangent line to (E) in H.

Let K be the intersection of AG, CH.

Let CT IO. TIC = 180 90 JIA = 90 JIA

TCI = 90 (90 JIA) = JIA = JCA.

ECF = HCE and CP = CQ CE PQ.

It follows easily that triangles COI and CPQ are congruent and

therefore OI = PQ.

Now I will prove that G, H are on segment IJ.

Let ACK be a triangle. Let E be the incenter of triangle ACK. Let FGH be

the intouch triangle of ACK. Let bisector AE intersect circle (CHEF) in I.

Then it turns out that G, H, I are collinear.

Proof:

AKC = 180 KAC KCA

KGH = (180 AKC) / 2

KGH = KAC / 2 + KCA / 2

AGH = 180 KAC / 2 KCA / 2

AEC = 180 KAC / 2 KCA / 2

Hence, AGH = AEC

Also, EIH = ACE. So, AIG is a triangle and is similar to AEC. Which

means that H is on GI and therefore, G, H, I are collinear.

Analogously, G, H, J are collinear and, therefore, G, H are on IJ.

MJ = GH [Praying Eyes Theorem wrt circles (A), (E)]

IO = GH [Praying Eyes Theorem wrt circles (C), (E)]

MJ = IO [Praying Eyes Theorem wrt circles (A), (C)]

Summarizing:

MJ = IO = GH = PQ = RS.

So now, in order to understand Ochoas image associated to the

archimedean circles consider the case when semicircles (A), (C) are tangent

so we get an arbelos. As MJ = IO = GH = PQ = RS and

MJ = IO = 2RaRc/(Ra + Rc), then, we will have a new triplet of

archimedean circles.

Ochoas image #2