How To Solve The Hardest Easy Geometry Problem
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- Опубликовано: 3 сен 2016
- 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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A small correction: at 5:04 I meant to say BG = BF. We wil shortly prove BG = GF.
Pls why do we have to draw a line while solving this question rather than solving it as it is
@@onyebuchiukeoma1551 because that's how you *can* solve it...
@Vinz Vinz It isn't assumed. Its because triangle BFG is equilateral (all angles are 60 deg, therefore all sides of that triangle are equal) :)
plz someone check my comment lmao its somewhere in the newest
@Vinz Vinz I was thinking the same thing - how do we know they're equal?! I didn't understand that statement.
And how does one come to think of drawing that magical initial isosceles triangle in the first place?
the answer is you either don't because that specific method only works in this exact scenario or you learn big people maths and try to form a general solution for whatever angles EBF, CBE, BCF & ECF are.
you are gonna need some trigonometry!
the only reason it works in this case is cause the entire question is done in a giant isosceles :P
You don't need trig, plain old geometry works fine, just know 180° is a triangle, 360° is a square, and start solving.
Hahahah that's what I was embarrassed to ask...
this isnt the easiest way
Experienced Geometry Nerds.
did it in 5 seconds , I just had to grab my protractor
Lmao
Lmao
Lmao
Got a quick draw holster for your protractor do ya?
Lmao
How did no one not notice that there is ABCEFG but no D?
thats not the point at all
Lol doesnt matter
Point D is the intersection of BE & CF, duh! :P
Because this question is a girl.
You'll get the D later
Damn, you really made this unnecessarily difficult.
True he did make it more difficult, theres other easier ways to figure it out
What is the easiest way?
@@atikshagarwal5147 FC and BE make an X and when theres an x theres 2 same angles that are opposite to each other.
@@x-x2060 Using that and exterior angle theorem is how I got it.
@@davefickess7973 could you elaborate please?
This is the first time I can answer the problem in your videos, you are awesome, things like this are really motivational, thanx 4 ur videos
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.
Dude I swear this is the only problem I did by myself. Not too hard, but he made it sound kinda complicated. I just did some basic stuff.
I've do these types of questions on a daily basis and seem to have no problem whatsoever.
@@armaansharma8349 it's a class 9 question so I solved it, I'm too in class 9
I am preparing for Olympiad and I also did it myself
*Math video exists*
Indian dudes: c'mon let's brag here.
I really enjoyed this video. Its in some way a perfect presentation and solution all in one
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.
This problem was not so hard.But you made it look really complicated
I assume u have passed Jee Adv by now🤣
It is easy to talk
Try to solve it
he is over complicated it its easy, I solved this question in 3 mins when I was jus twelve years old :|
edit: now that I'm older, I realize how idiotic and braggy that sounds XD
bundle maarte reh... tere jaise feku bahut dekhe hain... exam mein saari hawa nikal jaati hai...
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
What's funnier is when people say,
"All you need to know is that the sum of all triangle's angles equal to 180° while quadrangle's equal to 360°"
"Verticle angles are equal to each other, so you get that left = right = 70°"
but there are LITERALLY obvious common facts that people wouldn't feel the need to bring up.
How ABOUT you tell us how to solve it by doing your so called "oNLy cOunTiNg, nO dRAwiNg" method.
I've really really wanted to see that for like 4 years already, or since the video was posted.
The problem with the "merely counting" method is that it would always end up getting 180°=180° as the final count and result. You'd always get stuck on that. You wouldn't get anywhere or even move forward.
if those "obvious common facts" were the mere needed things to know of, then such question wouldn't have come out in the first place.
Stop this hipocrisy and bs once and for all.
Once again, you either solved this trigonometrically OR just wanted attention and reactions by commenting another "yOu pOinTleSsLy mAdE iT hArD" due to what the majority of the comment section have been saying.
i was so focused in this question that i didn't even realise that i am watching the solution on your channel..
Really excellent sir.Thankyou for sharing such a good problem with us.
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.
daniel buttons I did the same thing :P
If you add up the opposite angle by this method and the results they don't add 180 degrees
@@juanmatias87 opposite angles don't provide 180 degree. Linear pair angles ( angles on a straight line) gives 180 degree and that's right in this question.
Yes me too and my x was 40 degrees because of that.
As vpn x is 60
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.
Dude thank you I was having a headache about this and you made it ten times more enjoyable
ruclips.net/video/JzzEqECodn8/видео.html
i can explain it to u only using angle sum property and without any construction 🙂🙂
@@deepakmanwaniappdev bro please tell me
@@Crazy77772
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
huh? I'm not sure how everyone in the comments don't realise that you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use ANY isosceles triangle logic at ALL.
I solved this in a very different way. After attempting to label and solve for every angle in the diagram (and finding nothing useful) I decided I needed a shift in perspective. So I noticed that FE could be rotated counterclockwise (about F) to form the line CFE and triangle BCF. With this new shift in perspective, I was able to use the given 60 degrees and my previously solved angle CFA (and vertical angles) to find that x was 30 by x=180-20-130. I'm not sure this would work in all cases and when he mentioned isosceles triangles I was sure I got it wrong. But, surprisingly, I got the right answer.
I remember struggling with the infamous "Butterfly Problem".
5:04 you mean BG and BF
exactly. He should fix it in an annotation
You're right, thanks for letting me know. Sorry for the mistake, I have added an annotation and included the correction in the video description.
He ment BigFriendlyGiant BFG
Presh, I think you got x wrong. This is unless crossing lines don't create equal angles on opposite sides. Maybe I am wrong, but, in the triangle with angles 50º and 60º, the remaining angle is 70º, and so is the one on the opposite side . But the problem with your solution comes when you add this angle to F, which is 70º and x which is 30º. It doesn't add up to 180. it adds up to 170º
Almost. The angle you found is related to the line "C-F". In the final drawing, when he finds 70º, the line he uses is "F-G".
i started watching 1 vid. now im addicted to his voice! omg
+1
wow that really impress me a lot! excellent job! well done!
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.
Nice one. 10/10
TEEHEE!
I thought you were serious until the end 😩
I knew it its like my mother always said u gotta look for that D.
did you take into account the GBF, gay best friend triangle?
Ok I had no idea you were allowed to add in random lines! How on earth are your supposed to know to do that?!
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).
you can figure out the method in a few seconds. you don't have to draw anything at all! this question doesn't need to use any isosceles triangle logic at all. regardless, cool thought
I have seen many videos of yours. But this one is classic
I stared at the thumbnail for 5 minutes and sold it. then click on the video.
What?
solved*
good work......"Will Hunting"
same I didn't keep track of how long though I think it was 2-3
Tory Berry SAME! Although I got 50... I came so close... although I haven't taken any geometry. But I do know the three angels = 180 and what he said in the video. Maybe I should have checked my work.
nice problem IMO I tried solving it as 4 equations and 4 unknowns, but the equations I used were no good
me too. there are dependent so no way to solve :(
I tried that as well and I got a solution like x = x or something, which is obviously a completely meaningless answer.
+Laurelindo well, you are right, the value of x is equal to the value of x
Yeah, but it's still meaningless when trying to find th- *realizes you are Captain Obvious* Oh....
I did the same !
Everyone: Talking about how to solve it
Me: “Hey, that 20° angle looks the same as x, probably is 20°”
Hello it is really tricky to resolve this question. Why I felt a bit wired on this question:
1. What can trigger the thought of the line BG?
2. What if the question change for example if the angle b and angle c change from 80 to 85 degrees?
I see the solution here is for special case , won’t work generically.
Am I right? Please advice
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
I kinda feel like this is giving the solution backwards.
I would find it way more logical to construct point G starting from the assumption that an Isosceles traingle GFE would help us if we could find the value of angle FGE.
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
Is it possible to solve this problem without using properties of side lengths at all? I was trying to do angle substitutions and systems of equations and getting nowhere, I forgot about isosceles triangles...
I don't think so. When you try relying only on angle properties there are four angles that seem impossible to work out, one being x itself. Knowing any one of them will then lead you to the other three but you just can't get it without doing other things.
I did. Weirdly enough I come to x=50. Might sound stupid but I solved every angle without creating new lines and every single angle is coherent. I'm still trying to prove myself wrong.
I posted a photo of my solution. I hope you can see if I'm wrong
Mauricio Huicochea Toledo Your solution cannot be disproven from angle tracing alone. Indeed your system of equations does not have a unique solution. If x + w = 140 and y + w = 160 then subtracting gives y - x = 20 so y = x + 20. Then substituting this in z + y = 130 gives z + x + 20 = 130 or z + x = 110. I deduced your last equation from the others, so it isn't independent, and you really have four variables and three equations. So the system has infinitely many solutions and the error must be how you deduced that x can only equal 50.
I did not used a system of equations, I tried but it looped. It didn't worked.
Let the point of intersection of BE and CF be O then ,
In triangle BOC , angle BOC = 180° - (60°+50°) = 70°
Now angle EOF = angle BOC = 70° since they are vertically opposite angles
And according to your solution , in triangle EOF , angle EOF = 180° - (30°+70°) = 80°
So , why are we having two different values for angle EOF ?
Is there any mistake ?
In his solution about here 4:56 He makes another triangle and that triangle passes through the top angle which means when he said that the top angle was 70 he was referring to this triangle 6:28 (GFE) not CFE triangle
Which means the top angle in CFE was actually 80 (70+10)and not just 70.(The 10 was used to create the triangle in 4:56)
proof: 80(CFE) +30(x or BEF) +70(FOE) = 180 checks out
thank you very much for this knowledge
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
I tried it aswell, it is actually possible, by just adding up the angles you're missing 2 angles that can be replaced by y/Z then you can form an equation with both of these and solve for Z and y then you have all the angles and can get to x, I wrote that down somewhere earlier don't want to do it again tho
@@Skimbleshanks73 It's not possible. There's not enough information to solve those equations. Go on and actually try solving those equations and you'll understand.
@@demifiend9 It's actually possible by applying sine law, but the calculation is a little bit messy and compound angle formula is needed for solving the trigonometric equation. The method in the video is much more beautiful
@@Skimbleshanks73 Yes, you can generate two equations in two unknowns but the equations do not produce a unique solution. No matter how you try solving the two equations you only end up with platitudes like 0 = 0 or y = y. If you try expressing your set of linear equations in matrix/vector notation, you will find that the matrix has a determinant of zero and so cannot be inverted.
@@demifiend9 x I think can have many values. I might be wrong as well
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.
Yes, that's almost exactly what I did, but I used 1 instead of a. But this video was a serious letdown, because I was hoping to learn how to do this *kind* of problem. However, the presented solution wouldn't work if the lines got shifted by a few degrees, whereas the law of sines strategy would still do the job. It felt like watching a video about breaking into a computer and being told "When prompted for the password, just write the correct password, and you're in! Super easy!"
@@davidhoracek6758 I also used 1 and did the same thing lol
@@davidhoracek6758 without knowing the value of sin (20 degrees)?
Well, it was very usefull if you say that these angles are a particular case.
I lost a lot of time by creating the general solution for any angles, but you came out with a speific solution for only these angles.
Good problem and a great solution ...
plz keep posting such questions👍😊
Ans is 60 degree
Interesting question. just have a question, when you sketched BG, how did you know that it will create an angle of 20°?. Was it just an assumption?
I believe that was a mistake on his part. What he should have said was he needed Line BG to make an Isosceles triangle BGC. The 20 degrees was not a priority value (I thought he just copied the existing 20 degrees too), but just a result of making an isoscles triangle BGC.
Thanks for the video!!
تمرين جميل رائع. شرح واضح مرتب. رسم واضح .شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
This can be solved using the sine rule and cosine rule without adding any additional lines but it takes quite a bit longer
Thats the proper way to go.. If the angles mentioned in this diagram are any different, this method of using isoceless triangles to find angles wont work (cuz, we are supposed to get BGF as an equilateral triangle)
@@drainedzombie2508
✌️☺️STEP-BY-STEP UNIQUE METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
It's interesting how so many people are claiming to only need "vertical angle theorem" to solve the entire problem and end up with a cyclic argument or wrong algebra. people should really check first before posting their solution ;)
I used it, and I see no way to get x or the opposite angle in that way! Doesn't make sense. Seriously, you can't use that way and people have to realize that. There aren't any vertically opposite lines to x!
construct a circle of radius BE...
draw a 180° line on BC and mark the intersection of the line and the circle as G making a new line CG
Now since angle B equals 80°
in triangle BEG
angle B = 120..
Since BE = BG (radius of the circle) we can conclude that the new triangle BEG is an isosceles triangle and since the interior angle of a triangle is 180°.. In triangle BGE .. angle beg = angle bge
therefore 2x + 120 = 180
x = 60/2
x = 30°
@@WOEEW wow. My brain is too fuzzy to analyse that right now, but I do see sense in it. I'll come back later when I'm sane, haha
@@harinirajesh3838 you don't have to complicate it like the original video
@@BerkayCeylan if you contract a circle from radius BE , you will end up will only a circle of radius be from the point b and e
Great. I just learned a new to incorporate into solving other problems. Thank You
Thanks, I always wanted to know the value of x.
how do we know bg = bf? you said it does, but idk how you knew that.
also, how do we know ge = gf?
+Andrew Howe in an isosceles triangle, the sides opposite equal angles have equal length. he said this at the start of the video
NecroLord I know what an isosceles traingle is, but how do we know that it makes an isosceles triangle?
∠CBF=80
∠BCF=50, therefore ∠BFC=180-80-50=50
from this follows that BC=BF
since BC=BG was already shown we can now conclude that BG=BF
i guess newlines get removed for some reason :$
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 😊
Very well done. I am ashamed to not be able to accomplish the same. Respect.
@@ARDAYILMAZ72 - don't be ashamed. Sometimes, a problem turns out to be difficult for us, because we don't see what many others have seen. Look at their solution and learn.
. . . BTW I would wait with the Respect for ranjan, until they have shown their solution.
I saw only the triangles ABE and CFB. With those isosceles triangles everything can be calculated. There is also the triangle CBvertex, from which you can calculate complementary angles. No way you have to trace any further triangles.
Pro! Thank you.
i was so lostt when you started drawing those additional triangles
i can explain it to u only using angle sum property and without any construction 🙂🙂(price 30 Indian rupees
@@deepakmanwaniappdev
I've explained (to all) using symmetrical angle constructions inside the isosceles triangle. I've explained this FOR FREE.
@@deepakmanwaniappdev
it's impossible, fraudster 😆
For the purposes of this example, we'll call the unlabeled Vertex G. I'll simply go through the values for the easily identified angles, and go from there.
BGC is 70°. BAC is 20°. From that, we can identify that FGE is also 70°. As a result, both CGE and BGF are 110°. CEG is a 40° angle, leading GEA at 140°
Next, I'll look at quadrilateral AEGF, with it currently known angles of 70°, 20°, and 140°. The angles of a quadrilateral add up to 360°, so the final angle, GFA, is 130°.
I point this out because GFE and EFA must add up to that 130°. Likewise, GEF (in the diagram as "x") and FEA must add up to 140°. We also know that AEF and AFE must add up to 160°, and GFE and GEF must add up to 110°.
After trial and error, there is only 1 set of angles that fit those requirements:
AEF: 100°
AFE: 60°
GFE: 70°
GEF: 40° (This was the angle identified as "x", in the diagram.)
Thank you Presh, I got it wrong first but now I understand
Draw a line through E parallel to CB and another through B parallel to AC. The two intersect in, say, G forming a parallelogram CBGE. Let P be the point on BE that completes the equilateral triangle CBP. Then CP=CB=BF (equilateral triangle) and CE=BG (parallelogram). ∠ECP=80°-60°=20° and also ∠GBF=∠BAC=20°; so that, by SAS, ΔECP=ΔGBF. It follows that ∠BGF=∠CEB(=180°-60°-80°)=40°. And, since ∠BGE=80° (parallelogram), GF is the bisector of angle BGE. On the other hand, BF is the bisector of angle EBG. The point F is, therefore, the incenter of ΔEBG, the intersection of its angle bisectors. So FE is the bisector of ∠ BEG (=∠CBE=60°) and ∠BEF=∠X= 30°.
What about a general solution ? Suppose you have the same problem, but not with specific angles given.
What would x be if the given angles were say a,b,c,d ?
I guess you can do that with analytical geometry and a lot of effort
create a circle from the line BE as a radius
Well its not important that the angle x will be findable in every case..
@@srpenguinbr
I have the general solution to this. The solution method (of symmetrical angle constructions inside the isosceles triangle) in the links at the end can be used in general case...
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
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.
Andrew Stoll how did u get the 80?
Yes, i do it in head in 2 minutes using this method.
110, in truth
Wheres the parallel line?
@@digaddog6099 Draw the perpendiculars from points C and E,after that you can do this exercise in a much easier way than he did in the video.
Thanks for all
How to find coordinates
Polar spherical coordinates?
Space coordinates?
Well , i saw if we can see the folded triangle we could save it quickly. Think BC as the base and E is the vertex whichs is folded on BF .
The interior angles are 50 100
(60+20.2 = 100 20 is folded so when we open the triangle it makes 40 ) and x . So x is 30. If i m wrong please tell me *-*
4:55 "we'll consider this triangle BFG"
Doomguy approves
I didnt thought about drawing new lines. Time has passed since elementary chool
wow, I did geometry in high school. isn't memorization of theorems too strange and boring for tiny kids? well... i guess high schoolers have their own issues
Where did you get 20 degrees? There's only one angle there that would make your construction work, and you picked that exact angle. How could I reproduce that number without being given it?
The proof is flawed. You don't really need the 20° angle, you just need to construct BG so that CGB=80°.
Locate G on AC such that BG = BC. Easily constructible now, still a bit of deus ex machina.
The initial assumption of a 20deg triangle is just a guess, or, in other words one of many other guesses that could be made at some start. The whole exercise-as it is presented-is not an instructive procedure to get to the bottom of such a task, but an instruction for gambling. SIN and COS rules have to be applied to avoid waste of time.
how GF = GE? the angles that acrrosed by line GF and BE are 80 and 100 degrees
you didn't mention that triangle BGE formed as isosceles.. so BG=GE, since BG=GF, therefore GF=GE.
You made the solution lot more complicated than it needed to be
He was doing it using really basic maths, hence the title "hardest easy geometry problem", using more complicated trig would be going against the point of the problem.
yh I did it in my head
How do you go about solving these questions? I never know where to begin
ruclips.net/video/JzzEqECodn8/видео.html
Hey!
Can you please provide proof for BG=GF?
BTW I like all your videos. please don't stop posting videos like these.
thanks!
Regards,
Math lover.
If in an isosceles triangle one of its angles 60, the triangle is equilateral.
"Did you figure out this problem?"
Fucking no.
Is the point lable D being skipped a clue?
Assign it, rather obviously, to the central intersection.
ruclips.net/video/JzzEqECodn8/видео.html
Yes, add point D below A, such that BCD is 90°
You can then prove BC is parallel to EF, that x=60, and that Prash is wrong.
You can complete the rest of the triangles by adding to 180° to check the answer (Prash’s x= 30 fails this test)
This really helped a lot. My teacher told us to come here to get the solution and it worked
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 so its imposible for x to be 30 and the other angle being 70.
Wow !!! I'm impressed!!
It was pretty easy tbh just dividing 180 and 360 and with some plus and minus i could conclude this
Could you describe the steps you took?
+John Kerpan can't really explain but here is my sketch
I didnt use G at all to get the solution
***** No its not... you can just use several equations and substitution to solve it. Just looking at it I almost immediately knew it was 30. I needed to add 0 lines to solve it. Show anyone that does expert level Sudoku puzzles every day this and ask them to solve it. and odds are they will go to trial and error and solve it pretty fast.
That being said.... the trial and error method while effective with the problem is usually something you do because you do not know or at least are unsure.
You can solve this several ways, however the way they EXPECT you to solve it is this way. Others either are much more time consuming or are brute forcing the answer.
I think another answer could be 50° .is it right? If yes or no please let me know.
Awesome...keep it up.
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.
Maxi E s
wow, thats one way of solving this but it is super complicated
Cartesian coordinates are sometimes ugly, but often work to solve the problem. Reliably like a steamroller flattening the problem.
I did something similar, and needed excel for the calculations, but also got to x = 30°
Analytic brain. I like that. I would habe done it if the angles had abit better numbers(like I don't habe a point with irrational coordinates)
because i realised i needed to take the ratios of lengths into consideration.
i just drew the triangle and measured 30 degrees.
i also did alot of angle working out but it was redundant so yeah...
Johny Broxy u spelt triangle wrong
@@zenzex4166 you spelt "you" wrong...
How can you make sure the
Awesome solution..
I just guessed 30° because is looks similar to the other 30°angle
I think that's how he got his 20° perfect angle to start with anyway. hehe. it's 40!
no. for triangle BCA he just added angle B (20+60) + angle C (30+50) = 160 180-160=20 Not too fucking hard if you went to 6th grade.
"did you figure out this problem?" boy you know i didn't
Neither did Prash, his answer is wrong!
x=60
After I found every angle possible from the information given, (I added point D at the intersection where the 110 and 70 degree angles in the center are), I was left with EFA--FEB=20 and just happened to notice that if x=30 then some of the triangles would be similar and all the math fit so I took that as a good enough 🤷♂
Its Very good and thanks.
How does one even come up with this solution? This seems so created out of thin air that I am forced to believe he tried every single approach there is to solving the problem until he found a solution...
So far this video hasn't taught me anything since I am just learning a wierd solution to a wierd problem by heart
He didn't come up with the solution. This problem is known for having to draw auxiliary lines in order to solve it. Using the common theorems alone would not lead to the answer. Yep, it is indeed a weird solution for a weird problem.
+Paolo Patron you can solve this without drawing anything about 100 seconds of deduction the problem isn't all the complicated if you understand straight lines are 180 and that intersecting lines will be 360 you can deduce all you need to know to find x
Explain how then.
Joshua Woodford That's the thing. You can find all the other angles using supplementary and complementary values using interior and exterior angles but you won't be able to find the necessary angles in order to solve for x.
A "hundred seconds of deduction" is more than enough time to realize that the problem is impossible without auxiliary angles if you actually think.
+MadCodex I'll try .. for explanation purpose I'll call the intersecting lines in the middle R you'll have to figure the angle I'm speaking of.... to start the first angle of r can be determined by the fact bcr has to equal 180 and bc are know, r equals 70 .. since straight lines half to equal 180 angle r1 has to be 110 to complete the straight line same holds true for the two angles opposite have to be 70 and 110 ... now we can determine triangle bfr1 since 2 angles are known and r1 which means f in that instance is 50 also the angle opposite in triangle fae has to be 50 which means the remaining angle f in triangle fer2 can be determined which is 80 , that gives us angle f 80 and r2 70 means x has to be the remaining 30.. hope it helps hard to explain through text
"we are going to draw angle BG"
😂😂
how do we know the construction part
please make a video
Not only did I not get the answer, but memories, horrible memories, of my past (and in particular 11th grade after-school geometry lessons so I wouldn't fail math that year) came back like it was 1999 all over again! *cue the Britney Spears music!*
Draw a circle through B , F , and E and G as a central point
then x = (1/2) (60)=30
What?
Much simpler solving with supplementary angles.
please explain!
@@cs8833
They can only comment, they can't explain.
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
@@cs8833
✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
Very interesting problems.
Great Mathematics.... Keep it up..
*Math Teacher*:Ok solve this problem
*Me*: *2 minutes later*, 30
*Math teacher*: How did you get this where is your working out?
*Me*: I did it in my head 😑.
Ur sypposes to sgow qorking out
SAMEEEEE
@@div_yam2304 rip spelling
Peni
What a familiar situation!!!
I've forgotten a lot the stuff I was taught about geometry around 40 years ago. So I just used what I remembered. The fact that all the angles of a Triangle add up to 180°, lines that cross add up to 360°, angles coming off a straight line add to 180° and the angles of a Quadrilateral add to 360°. Then I went looking for all of the above shapes and situations.
Call the point where the lines BE & FC cross G
We know ∠ABC = 80° & ∠ACB = 80°
∴ ∠ BAC = 20°
We also know that ∠ BGC = 70°
Now consider Δ BGC We know ∠ BGC is 70°
Now consider Δ BCE We know ∠ BEC = 40°
Now consider Δ ACF We know ∠ AFC = 130°
Now consider Δ BCF We Know ∠ BFC = 50°
Now consider Δ ABE We Know ∠ AEB = 140°
Now consider Δ CEG We Know ∠ CGE = 110°
Now consider Δ BFG We Know ∠ BGF = 110°
Because we know 3 of the angles at the crossing point of the line at point G and they add to 360°
∴ ∠ EGF = 70°
Then I got stuck. I couldn't make any more of the shapes I remembered about.
I Knew I'd got 110° to share between the angles BFE and CEF in the Quadrilateral BCEF And 160° in the Triangle AEF
So I looked at the solution.
You cheating bastard :) You add extra lines and angles - Also I'd completely forgotten about the properties of Isosceles triangles :)
Thanks though. It was nice to drag some things I was taught so long ago out of the back of my head.
😂 I said the same thing about isosceles triangles. That got me!
What software did you use to teach here?
Excellent !
Other way: Use "trigonometric ceva" for triangle EBC and for point F outside of the triangle.
a 10 year old in Singapore, my country, can do this. and I'm actually astonished people are using trigonometry when you don't even need to mention the word isosceles at all. quite funny nonetheless
I think it is 40 degrees because in triangle CBO we have 60 +50=110
Then180-110=70
Since triangle is issocless according vertical opposite angles one of its angle is 70 degrees since it is a isosceles triangle it becomes 70+70+x=180 then x will be 40 degrees....I think it is my opinion so if it's wrong sorry....
Yes I tried it repeatedly and got 40 degrees somehow
@@kazishahjalal6852
Here's the most logical and easiest way...
[ Just in short -
I've solved it using ONLY BASIC GEOMETRY. But the image and explanation is little complicated so can't explain everything in comment here. Kindly share your mail so that I can share the solution attachment.
Please note that I'm IB Math teacher (and not any imposter or cheat). So do trust me and share the mail id when interested. ]
All dear friends, here's to tell you all that after 2-3 days of rigorous toiling and thinking over the Hardest Easy problem, I finally managed to solve this one on 26th November 2021 in about 3-4 hours using ONLY BASIC GEOMETRY.
However, since youtube rules don't allow to share the image in comments, the best way to share the solution to all interested friends would be by mailing them.
So kindly write your email id in this comment box and I'll share you the solution. I'm sure you all will enjoy this solution.
NOTE :
I tried this problem yesterday but wasn't able to crack it. Finally in the morning, after observing for about an hour, finally managed to relate the info obtained using several constructions done in the intricate solution diagram. I actually had tears of joy after realizing I'd solved this, and wanted to cry out bit loud. ☺
I'll share both the easiest solution and the intricate solution as an attachment.
Otherwise check if these work-
1. (Solution written again for easier understanding)
drive.google.com/file/d/1LjXF6eXwtipno5LLYbQmM9cptYZ_cC81/view?usp=drivesdk
2. (The actual solution I'd tried to figure out x)
drive.google.com/file/d/1Lkf3NTl1xY0QneofGMlTldunIAr93ZRS/view?usp=drivesdk
Thank you and love you all.
😘😘😘😊😊😊
@@kazishahjalal6852
✌️☺️STEP-BY-STEP UNIQUE BASIC GEOMETRY METHOD THE WAY ALMOST NO ONE HAS SOLVED TILL NOW☺️
Here's how it goes.
Shortest construction :
Through E, draw line parallel to BC to meet AB at H, HE || BC, and hence using Thales theorem in isosceles triangle ABC, we have HB = EC. Triangles HBC & ECB are congruent by SAS Test, so BHC = 40° (because we can find BEC = 40° as we know the other two angles of triangle BEC) and then BCH = 60°. Let CH meet BE at K & FE at G. Join KF.
Solution :
Isosceles triangle BCF has BC = BF, and triangle BKC is equilateral, so....
BC = CK = BK = BF
In triangle BKF, BK = BF
So, BKF = BFK = 80°
Using linear pair addition on line CKH at K,
FKH
=180° - (60° + 80°) = 40°
So,
FKH = FHK (or BHC)
= 40°
This means,
FK = FH
And HKE = BKC = 60° (vertically opposite angles)
Also,
EHK (or EHC) = BCH &
HEK (or HEB) = CBE,
all are 60° each (reason is because they form alternate pair angles with
HE || BC and angles CBE & BCH are already known)
This means,
Triangle HKE is equilateral....
So, EK = EH
Lastly.....
Both triangles FKH & EKH share the same base KH, and both are atleast isosceles....
So, the line joining their opposite vertices F & E, will be the perpendicular bisector to common base KH, and it'll also be the bisector to angles HEK & HFK, using isosceles triangle property....
Hence,
Angle KEF
= x
= Half of angle HEK
= Half of 60°
= 30°
BINGO !!!!! ✌️👍☺️
Note :
Solving this problem 2 years back was a struggling experience of 18 odd hours in two different days, but the joy of working through the solution was as sweet as the childbirth to a mother. Now I don't even have my noted solution of earlier, still I entirely framed it through again fairly easily.
Cheers to all. ✌️😘☺️
Thank you sir🙏
Is there a way to solve it without additional lines?
You just posted the same problem.. with.. 20° and 10°
Like just now..
Couldn't figure it out the "simple" way so I used Law of Sines and Law of Cosines instead. I began by assuming BC=10 so as to employ the Law of Sines. The length of BC can be any value as it would simply scale the whole figure and have no effect on the angle measures. Using the Law of Sines, I worked my way around the figure, using previous results to gain new results. First I found CE, then BE, then the segment from B to the interior intersection. Then the segment from C to the interior intersection. Then CF. Then some segment subtraction to find the segments from F and E to the interior intersection. Now we are down to the smallest triangle with angle x. I used the Law of cosines to solve for the missing side. Finally, used the Law of Sines once more to find the angle of 30 degrees for x.
Please check MY SOLUTIONS: ruclips.net/video/0Jrjp2AMFIg/видео.html
the easiest way
They should teach the Law of Sines and the Law of Cosines early on in math, like in Geometry. I first learned it in PreCalc and it’s relatively straightforward.
I uSeD TrIgOnOmEtRiC cEvA's tHeOrEm
@@kingklaus2115 It definitely can be understood by someone in a geometry class. Using the SOH CAH TOA a clever student might accidentally come up with it themself. When I took geometry it was a part of the curriculum.
@@bartholomewhalliburton9854 That’s good to hear. It’s not even a difficult concept and like you said a student could easily come across it on accident. All geometry classes should add it to the curriculum early on.
Tried many ways and this one worked
Just have doubt. In 5:57 , in the triangle BFG angleB+angleF+angleG=180
angleG=70 which shows that BG is not equal to GF
I just think that this question is a typo