How To Solve The Hardest Easy Geometry Problem

In the figure, what is the value of angle x? This problem is known as Langley's Adventitious Angles. It is also known as the hardest easy geometry problem because it can be solved by elementary methods but it is notoriously difficult to work out. Can you figure it out? The video presents a solution to this tricky geometry problem.

*At 5:04 I misspoke. I meant BG = BF. We wil shortly prove BG = GF.

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Math with Bad Drawings Solution

World's hardest easy geometry problem

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[MindYourDecisions]

A small correction: at 5:04 I meant to say BG = BF. We wil shortly prove BG = GF.

[leif]

And how does one come to think of drawing that magical initial isosceles triangle in the first place?

[redlt]

did it in 5 seconds, I just had to grab my protractor

[rcnayak_]

I am proving it in another way, but may not be elementary geometry, but perhaps easier as drawing a line of BG in the first place may not occur to mind quickly. Instead of drawing line BG, draw a line EG parallel to BC. Now since BC is parallel to EG, the angle GEA is 80 degree, so that the angle BEG is 60 degree (the angle CEB is already 40 degree). Now look at the two triangles BFE and FGE, considering BF and FG as their bases and have a common height (perpendicular measure), the ratio of their area are in the ratio of their bases, that is area of the triangle BFE/ area of the triangle FEG = BF/FG ). Again the area of the triangle BFE can also be expressed as 1/2*BE*EF* Sin BEF (similar as 1/2*a*b*SinC where a and b are the sides and C is their included angle). Similarly area of the triangle FGE is 1/2*EF*EG* Sin FEG. Therefore, the ratio of the two triangles becomes, BE *Sin BEF/EG *Sin FEG (EF eliminated) which is equal to BF/FG (proved earlier). That is, BE*Sin BEF/EG *Sin FEG = BF/FG. Rearranging, BE/BF = EG/GF *(Sin FEG/Sin BEF). We also know from angle bisector theorem that (see the drawing), if BE/BF = EG/GF, then angle BEF must be equal to angle FEG. Comparing the above two, we find that these relations will hold true only when Sin FEG/Sin BEF =1 (considering BE/BF and EG/GF not equal to 1). In other words, angle FEG = angle BEF. But angle BEG is 60 (we have already proved in the beginning). Therefore angle BEF, x (as we named) and angle FEG are each, 60/2 = 30 degree. This proof utilizes area of triangle theorems.

[maximedition]

How did no one not notice that there is ABCEFG but no D?

[hessylaguna]

solved it! Before watching:
this is how its done;
look at Triangle BAE. this triangle is in a relationship with a different triangle so it spells "bae". We must find this triangle's bae by using the boyfriend line (line BF). We see that triangle BFE and BAE share a common fetish known as "x" this is reasonable proof as to why they are together.
But what is their fetish? That is what the question is asking. lets look at other clues. notice how there is no D in the picture. That is because the D is in the middle (lol). The intersecting piece must be "d" and the angle which contains the fetish is named FED. This is not a coincidence, they have a food fetish which only about 30% of triangles have! if you were to put this into math talk then "x = 30 degrees"
and that is how you solve this one.

[duncansmol]

Damn, you really made this unnecessarily difficult.

[Tacticaviator]

Damn, after watching like 10 of these videos I finally solved this one myself, I'm so proud of myself for solving something that a Chinese 5 year old would do in seconds.

[hookerWithATool]

I solved it using parallel segments you can draw a segment which is parallel to EB from A, then you can connect the segment to point F which would give you two congruent triangles by SAS. The rest is pretty simple actually.

[somapal]

This problem was not so hard.But you made it look really complicated

[maxie]

Beautiful trick!

I had to determine coordinates of E and F in cartesian. Assuming the coordinate of B is (0, 0) and coordinate of C is (1, 0), I created 4 function for several lines:
1. BE line is y = x(tan 60)
2. CE line is y = tan 80 - x(tan 80). From BE and CE equation, I obtained the coordinate of E --> (0.766, 1.32683)
3. BF line is y = x(tan 80)
4. CF line is y = tan 50 - x(tan 50). From BF and CF equation, I obtained the coordinate of F --> (0.17365, 0.985)

And then I calculated the slope of FE line, m = 0.5771
I put Z between E and C so that the slope of FZ line is 0. The angle EFZ would be arctan m. So EFZ = 30 deg.
Knowing that FZ line parallel to BC line, the angle EZF = ECB = 80 deg.
x = 180 - EFZ - EZF - BEC = 180 - 30 - 80 - 40 = 30 deg.

It's ugly, but it works.

[andrewstoll]

I've seen a few other mention using opposite interior angles equal to (2) 70° and (2) 80° angles.
I think you made it harder than it needed to be.

[ranjan]

I saw 2 isosceles triangles, labelled the sides and used sine rule and derived a relation between sinx and the other angles. Then it's just how good u are at trigo manipulation 😊

[BlackFiresong]

I figured it out! Your method was far easier and more elegant than what I ended up doing, though. I basically called the length of the leftmost side "a" and calculated various other side lengths in terms of "a" via the sine rule, and then applied the cosine rule at the very end to yield a loooong arcsin expression for the value of x which I then plugged into Wolfram Alpha to get a result of 30 degrees.

[spiderjerusalem]

Hillarious that people here have been complaining about how the OP's attempt made it hard, but when they try to elaborate it 𝗴𝗲𝗼𝗺𝗲𝘁𝗿𝗶𝗰𝗮𝗹𝗹𝘆, they either somehow suddenly know the measure of one of the unsolvable angles or merely straight ahead solve it trigonometrically

The problem with the "oNLy cOuNtiNg, nO dRaWiNg" method is that you would always end up getting "both-sides-equal-equation" such as 180°=180°, x=x, etc, as the final count and result.

You'd always get stuck on that. You wouldn't get anywhere or even move forward.

if we merely had to know that sum of all triangle's angles equal to 180° and square's equal to 360° and vertical angles equal to each other,
then such question wouldn't have come out in the first place.

I've doubted these comments for a long time. They either solved it trigonometrically without having read the title
or commented another "yOu pOinTlEsSly mAdE iT hARd" merely for attention and reactions due to what the majority of the comment section have been saying

[llaffer]

I tried this using only (A + B + C = 180). I then added unknowns to an additional 3 angles and created a system of four equations and four unknowns but they didn't have a solution as one variable would always get cancelled out leaving an untrue statement (130 = -20, for example).

[TheOoorrrr]

I see people keep saying "I solved it in a different way, it's an easy question"... Ok, so share the proof instead of bragging about it

[danielbuttons]

What If you used triangle between angle 50 and 60 degrees then that angle will be 70 degrees and since opposite angles are equal the other angle is also 70 degrees.

[juliuss]

This can be solved using the sine rule and cosine rule without adding any additional lines but it takes quite a bit longer

[dalevillanca]

i started watching 1 vid. now im addicted to his voice! omg