World’s Hardest Easy Geometry Problem

I spent (wasted?) several hours working this problem out years ago but saw it posted recently on Google Plus.  I soon realized that either my solution predated my blog or I never actually posted my solution.

Here’s the problem and my solution is below: (ref: http://thinkzone.wlonk.com/MathFun/Triangle.htm) worlds hardest easy geometry problem

Using only elementary geometry, determine angle x. Provide a step-by-step proof.

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 trigonometry, such as sines and cosines, the law of sines, the law of cosines, etc.

Spoiler Alert: Don’t read any further if you want to give this a shot yourself!

MY SOLUTION:

hardesteasygeomoetrysoln1.   \angle DFG = 60 since it corresponds to \angle FAB (intersection of two parallel lines by another)

2.   \angle FDG = 60 since it corrresponds to \angle ABD

3.  This implies by sum of triangle angles that \angle DGF=60, and hence \triangle DFG is equilateral, giving us | DF| = |DG| = |FG|

4. Also notice that \angle DFE=80 since it corresponds to \angle ABC, which now means that \angle AFC = 140.

5.  Now, consider that \triangle AFE is similar to \triangle CFG since they share the same angle measures.

6.  Also notice that \triangle CGA  is congruent to \triangle AEC since \angle ACG=\angle CAE, \angle ACE= \angle CAG, and they share the side AC.  (SAS)

7.  Now since \triangle CGA = \triangle AEC then |CG| =|AE|, since Corresponding Parts of Congruent Triangles are Congruent (CPCTC).

8.  Now (7) together with (5), means that \triangle AFE is congruent to \triangle CFG.

9.  This means that |EF| = |FG|  (CPCTC).

10.  By (3), this means that |EF| = |DF|, so that \triangle DEF is an isosceles triangle.  So, \angle DEF = \angle EDF = 50 because they must be equal and \angle DEF = 80.

11.  Now, using the fact that \angle AEB = 30 (from the fact the we know \angle BAE = 70 and \angle ABE = 80) we just subtract this from \angle DEF = 50 and we have that \angle DEA = 20.

\square

6 thoughts on “World’s Hardest Easy Geometry Problem

  1. Sorry, but as a Geometry teacher, I wouldn’t give you an A on your “proof”. For one thing, your terminology is off. You use the word “corresponds” in steps 1 and 2 when it appears that you really mean “alternate interior angles”. Corresponding angles are a different animal — you used it correctly in step 4. But main my beef is that you have added points F and G without stating how you got them. It appears that your first step should have been “draw segment DF parallel to segment AB”. You can then draw segment AF, which in turn defines point G, where AF and BD intersect. And you can certainly draw a segment from point A that makes a 60 degree angle with segment AB, but you haven’t shown that segment would necessarily intersect segment BC at your previous point F. Similarly, you can draw the segment that bisects angle C, but you haven’t shown that it necessarily intersects segment BD at point G. I am not saying that your answer is incorrect, just that you haven’t presented a sufficiently rigorous proof.

  2. That’s awesome! I haven’t looked at this in years and your comments ares absolutely spot on. While the proof is sound it is CERTAINLY incomplete as you have indicated. I intuit a great deal too much integer setup. Expect a revision and thanks for your excellent feedback.

  3. Here’s a short, simple method. Use a “unit triangle”, where AB is equal to 1. Alternately, do a few steps where the values of a few sides are calculated in proportion to AB. The AB value will cancel out.

    Calculate Length of DA (Law of Sines)
    Calculate Length of AE (Law of Sines)
    Calculate Length of DE (Pythagorean Theorem)
    Calculate angle x (Law of Sines)

    Angles ADB and AEB are determined using definition of triangles

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