Challenging Geometry Puzzle: Problem 1561. 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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See diagram here.

ReplyDeleteExtend BO to intersect AC in E, OD to intersect (O) in F and BH to intersect (O) in G.

We have :

- ΔAOB and ΔBOC isosceles in O ⇒ ∠OAB = ∠ABO = 21° and ∠BCO = ∠CBO = 42°

- ΔAOC isosceles in O and ∠AOC = 2. ∠ABC = 126° ⇒ ∠OAC =∠OCA = (180-∠ABC)/2 = 27°

- ∠BAC = ∠OAB + ∠OAC = 48°

- CH being the altitude from C ⇒ ∠HCA = 90°- ∠BAC = 42°

- In ΔBCE, ∠BCE = ∠BCA = ∠BCO + ∠OAC = 69° and ∠BEC = 180°- ∠BCE - ∠CBO = 69° so ΔBCE is isosceles in B and therefore H, being the altitude from B, also lies on the bisector of ∠CBE

- Hence ∠OBG = ∠OBH = ∠EBH = 21°

- Also, since OD is // to BC, ∠ODA = ∠BCA = 69° = ∠OCA + ∠COD, therefore ∠COD = 42° and ∠BOD = 2.∠BAC + ∠COD = 138°

By construction of F and G, ΔBOF and ΔBOG are both isosceles in O.

In ΔBOF, ∠OBF = (180-∠BOF )/2 = (180-∠BOD )/2 = 21° = ∠OBG = ∠OBH = means that H lies on BF, so that BH and OD actually intersect on (O) in F = G which are one and the same.

Now :

- In ΔGCH, ∠CFH = ∠CFB = ∠CAB = 48° and ∠CHF = 90° - ∠HCA = 48° so ΔGCH is isosceles in C and CH = CF

- In ΔDCF, ∠CDF = ∠ODA = 69° and ∠CFD = 180° - ∠CDF - ∠DCF = 111 - ∠ABF = 69° so that ΔDCF is isosceles in C and CD = CF

So: CH = CD, ΔDCH is isosceles in C and ∠DHC = (180-∠HCD)/2 = (180-∠HCA )/2 = 69°

Let BO extended meet AC at X, BH extended meet AC at U & OD extended at Y, Z be on BC extended, AH extended meet BC at V and let CH extended meet AB at W

ReplyDelete< VAB = 90 - 21 - 42 = 27 and < OAB = OBA = 21 & < OAV = 27 - 21 = 6

< AOC = 2 X < ABC = 126 & so < OCA = OAC = 27 & <VAC = 27 - 6 = 21

Now < BCO = < CBO = 42 & < BCW = 90 - 21 - 42 = 27 & so < OCW = 15

Hence < XOY = < CBO = 42 = < BCO = < COY & OY bisects < XOC

Also < CBU = 90 - 69 = 21 & so BY bisects < CBX

It follows that Y is the excentre of Triangle BOC & so YC bisects < OCZ

So < OCY = 69 & < DCY = 69 - 27 = 42 & since < CDY = < DCB = 69, < DYC = 69

But Triangles HCU & YCU are congruent SAS & So HC = YC = DC

Hence C is the Circumcentre pf Triangle HDY and so < DHY = 1/2 of < DCY = 21

Therefore < CHC = 21 + 48 = 69

Sumith Peiris

Moratuwa

Sri Lanka