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Using only elementary geometry, determine angle x.

triangle

You may not use trigonometry, such as sines and cosines, the law of sines, the law of cosines, etc.

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  • 2
    $\begingroup$ Sheesh, and I thought the law of cosines was elementary... $\endgroup$ – J. M. is a poor mathematician Sep 12 '11 at 5:03
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    $\begingroup$ The world's hardest easy geometry problem... :) $\endgroup$ – tom Sep 12 '11 at 13:21
  • $\begingroup$ @user236182: You are right. $\endgroup$ – Aryabhata Oct 26 '15 at 23:23
  • $\begingroup$ Recently, it was proved the all problems of Langley's Adventitious Angles could be solved with elementary method. I wrote at wikipedia. $\endgroup$ – Takahiro Waki Oct 2 '16 at 11:49
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adventitious angles

  1. Draw a line $DF$ parallel to $AB$, intersect $BC$ at $F$;
  2. Connect $AF$, intersect $BD$ at $G$;
  3. Connect $CG$.

Now, it's easy to prove that $CE=AG$, and $DF=DG=GF$.

Since $AF=CF$, then $EF=GF$.

Then $EF=DF \Rightarrow \angle FED= \angle FDE$.

While $\angle DFE=\angle ABC=80 ^\circ$, so $\angle DEF=50^\circ$.

From $\angle AEB=30^\circ$, we can get $x=\angle DEA=20^\circ$. [Q.E.D]

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This is known as the problem of "adventitious angles". You'll find many references if you search the web for that phrase.

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Each node of the regular polygon $P_{18}$ sees all other nodes separated with $10^{\circ}$, making this $P_{18}$ a perfect drawing canvas for this problem (left image):

Left image: polygon property. Right image: solution

It is easy to verify that $A$ and $B$ are located on a large diagonal of $P_{18}$.

Now embed the unique, smaller $18$-gon $p_{18}$ inside $P_{18}$, with nodes $A$, $B$ and $C$.

Then we have:

$$AB \overset{\text{similarity}}{\parallel} A’B’\overset{\text{regularity}}{\parallel}A’’B’’$$

Now using the $10^{\circ}$-property, we see that:

$$x = \angle A’’B’’B = 2\times 10^{\circ} = 20^{\circ} \qquad \blacksquare $$

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  • $\begingroup$ I don't even know what it is all about but I just upvoted due to pure beauty... $\endgroup$ – BigbearZzz Oct 3 '16 at 21:38
  • $\begingroup$ I see that A is on the long diagonal shown because the two shorter diagonals that intersect to define A are symmetric with respect to that diagonal. I can even show with a symmetry argument based on three 18-gons in the "beer rings" configuration (so their three pairwise intersections form an equilateral triangle) that the short diagonal CB' goes through B. But how do I know that the 18-gon through C whose 7th vertex clockwise is B goes through A? I.e., there's a whole family of polygons similar to P_18 through C, but why does any go through both A and B? That seems to be the crux here... $\endgroup$ – Glen Whitney Jun 14 '18 at 5:45

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