Doing geometry with one hand on your back
You may use only elementary geometry, such as the fact that the angles of a triangle add up to 180 degrees and the basic congruent triangle rules (side-angle-side, etc.). You may not use trigonomery, such as sines and cosines, the law of sines, the law of cosines, etc.
The task is to find the angle x in the diagram, but with your hands tied as explained above. Some people has named the problem: The World’s Hardest Easy Geometry Problem.
I borrowed the diagram from http://thinkzone.wlonk.com/MathFun/Triangle.htm.
February 9th, 2010 at 5:45 am
The World’s Hardest Easy Geometry Problem! That’s rather like The World’s Tallest Dwarf.
Angle BAD = 70° + 10° = 80°.
Angle ABE = 60° + 20° = 80°.
Thus angle ACB = 180° - 80° - 80° = 20°.
Angle ADB = 180° - 80° - 60° = 40°.
Angle AEB = 180° - 70° - 80° = 30°.
Let F be on BC so that DF || AB.
Let AF meet BD at G.
Then angle ADF = angle BFD = 180° - 80° = 100° (the angles an intercept makes on the same side of parallels sum to 180°)
Thus angle DFG = angle FDG = 100° - 40° = 60°
Thus angle DGF = 180° - 60° - 60° = 60°, so triangle DFG is equilateral.
Triangle ABC is isosceles with vertex C, which is bisected by CG.
Thus angle ACG = 20°/2 = 10° = angle CAE; and angle CGF = 60°/2 = 30° so angle AGC = 180° - 30° = angle AEC. Thus triangles AGC, CEA are congruent as they have equal angles and have the side AC in common.
Thus AG = CE.
Also AF = BD = CD = CF.
Thus DF = GF = AF- AG = CF - CE = EF, so triangle DEF is isosceles with base DE. Angle DFE = angle ABC (an intercept makes equal angles with parallels) = 80°. Angles DEF and EDF are equal and sum to 180° - 80° = 100°, so angle DEF = 50°.
Thus x = angle DEA = angle DEF - angle AEF = 50° - 30° = 20°.
February 12th, 2010 at 11:20 pm
Thank you for the link, Jan. Thanks to it, I’ve been able to research deeper into the subject of adventitious-angle problems. I’ve tried to compose a few of my own, but I haven’t found any that are solvable but are as hard as this one.
I never liked the point C in the above diagram: the statement of the problem doesn’t need it. Of course you could construct it if you want, but I looked for another proof which didn’t go beyond the bounds of quadrilateral ABCD. I found the following:
Angle BAD = angle BAE + angle EAD = 80°.
Angle EBA = angle EBD + angle DBA = 80°.
Angle AEB = 30° (angles of triangle ABE sum to 180°).
Angle ADB = 40° (angles of triangle ABD sum to 180°).
Let F be on BD so that BF=BA. Angle FBA=60°, so triangle ABF is equilateral.
Extend AF to G on BE. Then triangles ABD and BAG are congruent (side and 2 included angles) because side AB is common, angles BAD and GBA are equal, and angles DBA and BAG are equal. Then angle AGB = angle ADB = 40°. Subtracting the equals AF, BF from the equals AG, BD, FG=FD. Angle GFD=60°, so triangle DFG is equilateral.
A and B are on a circle centred on F, and angle AFB = 60° which is twice angle AEB, so E is also on this circle, so FE=FB. Thus triangle BEF is isosceles on base BE, so angle FEB=20°, so angle EFD=40°, so angle GFE = angle GFD - angle EFD = 20°, so triangle EFG is isosceles on base EF, so EG=FG=DG, so triangle EDG is isosceles on base ED.
Angle EGD = angle EGF - angle DGF = 80°, so angle DEG = (180°-80°)/2 = 50°.
Thus x = angle DEA = angle DEG - angle AEB = 20°.
QEF.
February 13th, 2010 at 6:48 am
Very impressive!