Challenging Geometry Puzzle: Problem 1603. 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 P and Q be the midpoints of CD and AC, respectively.
ReplyDeleteTriangle PQF is an equilateral triangle, and triangle CPF is an isosceles triangle.
Since angle DCF=15°, GD=tan15°=2-√3
Therefore, AG=1-(2-√3)=√3 -1
From CF⊥FD, point F is on a semicircle of radius 1/2 centered on point P. Therefore, △PQF is an equilateral triangle with ∠FQP=60°, and ∠PQE=30°, so ∠PFE=15°.
DeleteDid you assume that H,F,D are collinear? If so why and how? Only if H,F,D are collinear can you say that CF is perpendicular to CF
DeleteShall thank you to clarify
I see, that's right.
DeleteI'm not sure if this is a geometric proof, but if you take point G' on AB where GD=AG', then rotating -90° around point Q moves △CDG to △DAG', and since they are congruent, points H, F, and D are on the same line, and DF⊥FC.
Actually, my initial idea was to take a coordinate plane with Q as the origin,
and set ∠PQE=θ, C(1/2, 1/2), E(1/2 cosθ, 1/2 sinθ), F(1/2 sinθ, -1/2 cosθ), and from the fact that points C, E, and F are on the same line, I derived θ=30°.
The result may be correct. refering to lines 2 and 3 of your comment .Please show more details that
ReplyDelete1.Triangle PQF is an equilateral triangle,
2Triangle CPF is an isosceles triangle.
3. angle DCF=15°
Let the center of circle is O, it's radius is r=1/2. OC is equal to √2r. If we draw a perpendicular OJ to EF from O, then OJ=r/√2 (as EFHK is square and OJE is right isosceles triangle), Triangle OJC is right triangle with OC=2OJ, Triangle OJC must be 30-60-90 triangle, hence Angle OCJ = Angle ACG=30 degrees.
ReplyDelete