Challenging Geometry Puzzle: Problem 1574. 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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BI*BBFBD*BD=BJ*BK
ReplyDelete1s Solution:
ReplyDeleteAccording to circles EDFIH and MEDGL, line AED is the radical axis of these circles, and so AE*AD = AH*AI (circle EDFIH) and also AE*AD = AM*AL, then,
AH*AI = AM*AL, so points H, I, L, M are concyclic. Call this circle C1.
Similarly using vértices B and C, we have respectivelly,
H, I, J, K concyclic, call circle C2, and J, K, L, M concyclic, call circle C3.
We want to prove that these three circles are the same, so that H, I, J, K, L, M are concyclic.
If two of the circles C1, C2 or C3 are equal, we are done, because each pair of two circles contain all six points.
Then, suppose by contradiction that the three circles C1, C2 and C3 are different taken two by two.
Vertex “A” of triangle ABC belongs to line AHIB, radical axis of circles C1 and C2, having H, I as common points. Also vertex “A” belongs to line AMLC, radical axis of circles C1 and C3, having M, L as common points. So, vertex “A” has the same power with respect to all three circles C1, C2 and C3. In consequence, vertex “A” shall belong to the radical axis of circles C2 and C3, namely, line BJKC, i.e., side BC of triangle ABC. But this means that vertices “A”, “B” and “C” are colinear and so the triangle ABC is degenerate, certainly contrary to hypothesis.
This contradiction shows that circles C1, C2 and C3 are indeed the same; then points H, I, J, K, L, M are concyclic.
Joaquim Maia
Rio de Janeiro
Brazil
2nd Solution
ReplyDeleteThe preliminaries are the same as the first solution. We repeat here for the sake of completeness.
According to circles EDFIH and MEDGL, line AED is the radical axis of these circles, and so AE*AD = AH*AI (circle EDFIH) and also AE*AD = AM*AL, then,
AH*AI = AM*AL, so points H, I, L, M are concyclic. Call this circle C1.
Similarly using vértices B and C, we have respectivelly,
H, I, J, K concyclic, call circle C2, and J, K, L, M concyclic, call circle C3.
Now take the intersection of circle C1 with side BC of triangle ABC, and call these intersections points J’ and K’. So, H, I, J’, K’, L, M are concyclic (circle C1). We want to prove that J’ and K’ coincide with points J and K, not necessarily in this order.
The power o vertex “B” with respect to circle C1 is BI*BH = BJ’*BK’.
But with respect to circle C2, BI*BH = BJ*BK.
Then BJ’*BK’ = BJ*BK (1).
Similarly we find using power o vertex “C” with respect to circle C3,
CJ’*CK’ = CJ*CK (2).
Using the relations CJ = BC – BJ; CJ’ = BC – BJ’; CK = BC – BK; CK’ = BC – BK’ and substituting into (2) we get,
(BC – BJ’)*(BC – BK’) = (BC – BJ)*(BC – BK); after multiplication,
BC*BC – BC*BK’ – BJ’*BC + BJ’*BK’ = BC*BC – BC*BK – BJ*BC + BJ*BK, cutting common factors, and using relation (1) we finally find,
BJ’ + BK’ = BJ + BK (3).
By (1) and (3), considering BJ’*BK’ = BJ*BK = q and BJ’ + BK’ = BJ + BK = p, we see that both pairs (BJ, BK) and (BJ’, BK’) are solutions of the second degree equation
X**2 – p*X + q = 0.
However, by Algebra, we know that this equation has only two solutions, and we conclude that BJ = BJ’ and BK = BK’, or BJ = BK’ and BK = BJ’. Anyway, the pair of points J, K coincide with J’, K’ in any order. So J, K belong to circle C1; then points H, I, J, K, L, M are concyclic as was to be proved.
Joaquim Maia
Rio de Janeiro
Brazil