Challenging Geometry Puzzle: Problem 1575. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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We give the guidelines of the solution, as there is no space here for a longer and detailed explanation. The interested student may take a paper and a pencil and provide details, specially some routine calculations.
ReplyDeleteTake the point J, as harmonic conjugated of point G with respect points B and C. Let K be the midpoint point of J and G, and draw the circle of diameter JG with center K, and radius KG. This is the Apollonius circle so that all points H of this circle have the property that HB / HC = GB / GC, and HG bissects angle BHC.
This Apollonius' circle intersect the excircle in two points, one is G and the other is symetric of G with respect to KF, the line o centers; call this point H’. To complete the solution we show that angles H’GK = HGK, so that line H’G = HG.
As H’G is perpendicular to KF, we easily check that angles H’GK = KFG. So if we prove that the right triangles KFG and DGE are similar, case SAS, we get the desired result, angles DGE = KFG = H’GK. Indeed, we need to prove KG / FG = ED / DG = S / p (c – b), Where S is the area of triangle ABC, p semiperimeter of ABC, and we suppose AB = c > b = AC.
This ratio will be established by direct calculation of segments involved.
KG = JG / 2 = (JB + GB) / 2. So JB = 2KG – GB. JC = JB + a, a = BC. As J is the harmonic conjugated of G, we have the relation JB / JC = GB / GC = (p – c) / (p – b) (the last equation is consequence of that G is the point of contact of the excircle with side BC).
So, replacing JB by the relation connecting with KG, we get
(2KG – GB)/(2KG – GB + a) = (p – c)/(p – b), solving this KG=(p – b)(p – c)/(c – b).
Now KG / FG = (p – b)(p – c) / ((c – b)FG), multiplying numerator and denominator by p(p – a), we have p(p – a)(p – b)(p – c) / ((c – b)FG(p – a)). By Heron’s formula for the area of a triangle, the numerator is the square of the area of triangle ABC, S2 (square o S), and by the known formula FG(p – a) = S (FG the radius of the excircle), we find KG / FG = S2 / ((c – b).S) = S / (c – b) (1)
For the ratio ED / DG we observe that ED = ha / 2 (h1 = height of triangle ABC with respect to BC) and DG = CG – CD = p – b – bcos LC = p – b – (b(a2 + b2 – c2) / 2ab) = (2a(p – b) – (a2 + b2 – c2)) / 2a = p(c – b) / a, the reader may find this easily.
Then ED / DG = (h1 / 2) / (p(c – b) / a) = (h1.a / 2) / (p(c – b)) = S / (p(c – b)) (2),
(remind that haa / 2 = S).
By (10 and (2), we finally get KG / FG = ED / DG = S / p (c – b) and so angles HGK = DGE = KFG = H’GK, which means H’GE is a straight line as HGE is, then H’ and H belong both to the same line EG and to the same excircle, so H’ = H as was to be proved.
Joaquim Maia
Rio de Janeiro
Brazil