# World's hardest easy geometry problem

livinonaprayer
have you encountered it and/or solved it? what did you think?

JonnyG
I got infinitely many solutions...anyone else get that too?

Hornbein
I got infinitely many solutions...anyone else get that too?

I figured all the easy angles, then was left with three equations and four unknowns. So I manfully gave up. I'm too dumb to do it.

livinonaprayer
I figured all the easy angles, then was left with three equations and four unknowns. So I manfully gave up. I'm too dumb to do it.
nooo don't give up! it took me 5 hours to find the solution. and you don't need complicated equations, you're only allowed to use simple geometry.

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livinonaprayer
I got infinitely many solutions...anyone else get that too?
i think there's only one solution? i got only one solution and after going over it multiple times i even checked online to confirm that my way of getting there was correct. remember, you're only supposed to use basic geometry, no trig or some sort of super advanced math.

zoobyshoe
nooo don't give up! it took me 5 hours to find the solution. and you don't need complicated equations, you're only allowed to use simple geometry.
I spent about an hour on it yesterday, and an hour today, paying careful attention to looking for a solution by means of the specific elementary geometry principles they list at the bottom. I have no idea where two parallel lines being cut by a third fits in, nor where the equilateral triangle might come into play. But, they wouldn't have presented them if they weren't relevant, so I figure once I realize in what way they can be employed here, the solution should be immanent.

livinonaprayer
I spent about an hour on it yesterday, and an hour today, paying careful attention to looking for a solution by means of the specific elementary geometry principles they list at the bottom. I have no idea where two parallel lines being cut by a third fits in, nor where the equilateral triangle might come into play. But, they wouldn't have presented them if they weren't relevant, so I figure once I realize in what way they can be employed here, the solution should be immanent.
they are important, just like the rest of the information given.
you will have to add more lines inside the triangle to reach the solution

zoobyshoe
I'm still too stoic to check out the hint. However, I will ask you, was the hint necessary in your case?

livinonaprayer
I'm still too stoic to check out the hint. However, I will ask you, was the hint necessary in your case?
nope, but that's because i always solve geometry problems like that. the hint isn't an actual step closer to the solution, it kinda points to to a 'technique' you MUST use to find it (you can't solve it otherwise, it's kinda like trying to write a computer program without knowing part of the code language or writing style, impossible). i would check out the hint if you're not familiar with high school geometry.

zoobyshoe
nope, but that's because i always solve geometry problems like that. the hint isn't an actual step closer to the solution, it kinda points to to a 'technique' you MUST use to find it (you can't solve it otherwise, it's kinda like trying to write a computer program without knowing part of the code language or writing style, impossible). i would check out the hint if you're not familiar with high school geometry.
OK, I checked out the hint and it didn't help at all, since I'd already been doing that.

livinonaprayer
OK, I checked out the hint and it didn't help at all, since I'd already been doing that.
ohh... so keep trying! try looking at the problem in different ways, exploring new directions. see what the already existing information has to offer. make a list of moves that you've already figured out that might be useful.

Staff Emeritus
Homework Helper
Pretty obvious if you use trig. My point of view is: any proof is a good proof, so I don't bother finding it without trig haha

zoobyshoe
Pretty obvious if you use trig. My point of view is: any proof is a good proof, so I don't bother finding it without trig haha
Oh, sheez, you know better than that, micro.

livinonaprayer
it's a brain teaser with rules! using trig is forbidden.

Homework Helper
Gold Member
I just solved it! Finally! (geometrically, without trigonometric functions) My goodness, it was not without effort though. There were too many false starts to recount. I honestly went through about half a notebook of paper on this problem. The effort as a whole did not take merely hours, but rather days of my life. (Of course, most of that was a nest of dead ends and trudging down the wrong paths.) [Edit: Although in retrospect, the solution isn't very difficult at all. The really difficult part is figuring out what to look for, such that you can form a solution -- once you know the solution, it seems almost easy. It's getting to that point that is certainly not (at least is was certainly not for me).]

Henceforth, the Christmas season of 2015 shall forever be known to me as "The Christmas of geometry" or perhaps better, "The Christmas where collinsmark spend several days with his nose burried in a self-made, eternally expanding, degenerate* army of occultish pictographs littered with mysterious hieroglyphs, while he was oblivious to all else around him, even in the presence of family, friends, coffee shop patrons and pub-goers."

I say this because looking back at some of my diagrams, which by some accounts might have qualities of Christmas symbols (there's a shape of Christmas tree and are some some stars around in there somewhere), it really looks more like a guy with a horned goat-head -- or perhaps like something straight out the Necronomicon.   Which makes me wonder how many times throughout history some poor sap was accused of witchcraft and burned at the stake, when he or she might have just been doing a geometry problem.

Anyway, I think I am prepared to post a solution (within spoiler tags of course), after organizing my notes a bit, if @livinonaprayer (the OP) requests it. I won't post it unless requested though, since it sounds like the OP might prefer people to work on this independently* without too many hints or somebody else's solution.

If you are reading this and working on the problem on your own, don't become too dependent* on a single combination of shapes. And if you find yourself frustrated, don't become a mad degenerate*. But rather try to stay congruent* with an open minded outlook on all the shapes and shapes within shapes.

*(These terms were not used accidentally. )

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• livinonaprayer and Silicon Waffle
zoobyshoe
Congratulations collinsmark, but please don't post any more hints! Why don't you just PM livinonaprayer with your proof.

• Silicon Waffle and collinsmark
livinonaprayer
@collinsmark congrats!1 the satisfaction after solving it is amazing. regarding the solution, i think if it were posted here it might tempt people to peek so you can indeed pm me if you want. the whole solution is out there on the internet anyways so if someone is really desperate they could find it if they wanted to.

• collinsmark and Silicon Waffle
vin300
I have created a solution in pictures but seems to be impossible to upload. If the central intersection is called O, then various angles are obtained as follows: C=20, AOB=70, DOE=70, DOA=110, ADO=50, EOB=110, OEB=40, CDO=130, CEO=140. Thus far, almost every angle except x and its complement have been derived. To find x, the only possible way seems to be exactly find every elemental length. Use sine rule. AC=CB = 100. By sine rule, AB=34.729, from AB, DB=44.6475, from AB again, EB = 46.79 AB= AD, From EB, CE=53.21. Use cosine rule to find DE, as CD=100-AD= 65.271 DE found to be 23.756. Use sine rule once again with DE to obtain angle EDC nearly 50 degrees. We know the rest already, hence x= 30 degrees

vin300

Homework Helper
Gold Member
I have created a solution in pictures but seems to be impossible to upload. If the central intersection is called O, then various angles are obtained as follows: C=20, AOB=70, DOE=70, DOA=110, ADO=50, EOB=110, OEB=40, CDO=130, CEO=140. Thus far, almost every angle except x and its complement have been derived. To find x, the only possible way seems to be exactly find every elemental length. Use sine rule. AC=CB = 100. By sine rule, AB=34.729, from AB, DB=44.6475, from AB again, EB = 46.79 AB= AD, From EB, CE=53.21. Use cosine rule to find DE, as CD=100-AD= 65.271 DE found to be 23.756. Use sine rule once again with DE to obtain angle EDC nearly 50 degrees. We know the rest already, hence x= 30 degrees

And by the way, we're not allowed to use trigonometric functions or rules as indicated in the link containing the problem, "You may not use trigonometry, such as sines and cosines, the law of sines, the law of cosines, etc."

[Edit: Although to be completely forthcoming, there was a point where I found $x = 30^\circ$, but that was before I realized that I made a mistake. After fixing the mistake and rethinking the problem, I came up with a different answer. Like I mentioned, I went down many, many wrong paths before finally reaching a good solution. By the way, are you referring to the first problem in the link? The link has two problems, the "hardest" and "second hardest." I haven't attempted the second one yet.]

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livinonaprayer
I have created a solution in pictures but seems to be impossible to upload. If the central intersection is called O, then various angles are obtained as follows: C=20, AOB=70, DOE=70, DOA=110, ADO=50, EOB=110, OEB=40, CDO=130, CEO=140. Thus far, almost every angle except x and its complement have been derived. To find x, the only possible way seems to be exactly find every elemental length. Use sine rule. AC=CB = 100. By sine rule, AB=34.729, from AB, DB=44.6475, from AB again, EB = 46.79 AB= AD, From EB, CE=53.21. Use cosine rule to find DE, as CD=100-AD= 65.271 DE found to be 23.756. Use sine rule once again with DE to obtain angle EDC nearly 50 degrees. We know the rest already, hence x= 30 degrees
are you talking about the second one? i haven't tried to solve it yet and I'm too lazy to look up high school trig right now so... as for problem 1 (the one i was talking about), x isn't 30 degrees. for both problems trig isn't allowed and you can definitely find the solution by just using simple geometry. (and please post this under a spoiler warning next time). keep trying!

zoobyshoe
According to the site where the problem is posted:

Here is everything you need to know to solve the above problems.

Lines and Angles

• When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees.
• When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.
Triangles

• The sum of the interior angles of a triangle is 180 degrees.
• An isosceles triangle has two equal sides and the two angles opposite those sides are equal.
• An equilateral triangle has all sides equal and all angles equal to 60 degrees.
• A right triangle has one angle equal to 90 degrees.
• Two triangles are called similar if they have the same angles (same shape).
• Two triangles are called congruent if they have the same angles and the same sides (same shape and size).
• SAS (Side-Angle-Side): Two triangles are congruent if two pairs of sides of the two triangles are equal, and the included angles are equal.
• SSS (Side-Side-Side): Two triangles are congruent if three pairs of sides of the two triangles are equal.
• ASA (Angle-Side-Angle): Two triangles are congruent if two pairs of angles of the two triangles are equal, and the included sides are equal.
• AAS (Angle-Angle-Side): Two triangles are congruent if two pairs of angles of the two triangles are equal, and a pair of corresponding non-included sides are equal.
• AA (Angle-Angle): Two triangles are similar if two pairs of angles of the two triangles are equal.
Notice in his commentary he mentions hours of frustration and even the possibility of being driven insane for those who tackle this. But, apparently, angle x can be found with just the above, very elementary, geometric principles.

vin300
The answer was posted for the second hardest problem in #3 with base angles 30, 50, 60, 20. calcs seem to have made the result slightly erroneous, because the angle is not 30 degrees, but 25. Another way must be found out.

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