Challenging Geometry Puzzle: Problem 1550. 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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Let AD & AE extended meet circle ABC (with diameter AC) at P & Q respectively.

ReplyDeleteDraw DM perpendicular to BP, M on BP

BP = CQ (both subtend 20 degrees on the circumference) = y, say

< ACB = < DBP = < DPB = 35 and so,

Triangle BDP is isosceles and hence BM = y/2

Triangles BDM & CEQ are similar (90,35,55) and so

BD/BM = CE/CQ

Therefore BD/(y/2) = 3/y which yields

BD = 3/2

Sumith Peiris

Moratuwa

Sri Lanka

Hi this ready...?

DeleteTrigonometry Solution

ReplyDeleteLet tan35 = t

BD = c.tan20 = c/tan70 = c./(tan(2X35) =c.(1-t^2)/2t........(1)

BE = c.tan35 = ct.......(2)

BC = c.tan55 = c/tan35 = c/t .......(3)

(3) - (2); CE = BC - BE = c.(1/t - t) from (2) & (3)

So c.(1-t^2)/t = 3......(4)

From (1) & (4) BD = 3/2

Sumith Peiris

Moratuwa

Sri Lanka

The trigonometric solution is beautiful because it is simple. The geometric solution is beautiful because it makes ingenious auxiliary constructions.

ReplyDeleteThank you

Delete2nd Geometry Solution - much simpler

ReplyDeleteLet FA be perpendicular to AC, F on CB extended

Let BD = x

Now FD = DA = DC so DE = a - x - 3 and FB = a - 2x

FB = BE = a - 2x = BD + DE = x + (a - 3 - x) = a - 3

Hence 2x = 3 and x = 3/2

Sumith Peiris

Moratuwa

Sri Lanka

Nice solution by creating the right triangle FAC.

ReplyDeleteNotice that AD=CD and, if ADCF is a rhombus, F is the circumcenter of triangle AEC, <FCE=<ADB, FC||=AD, G, projection of F onto CB is midpoint of EC, while triangles CFG and DAB are congruent, so CG=BD, done.

ReplyDelete