After drawing DF, you could have just used the symmetry of the trapezoid and its diagonals to copy all the angles instead of calculating

I come across this very problem regularly once in every two years and it always takes me forever to solve it, since I don't remember the solution. This year I'm wiser and I will just watch this video without trying it myself.

We could also exploit symmetries to speed up at the start by noticing that since ABC is isosceles, the construction of F can also be defined as a reflection of D along the bisector of angle ACB, so G rests on the symmetry axe, and deduct some distance/angles more easily.

hes making waaay more complicated than it has to be

Shorter: 1.) Triang(AFE) kongruent Triang(CGF) (because Ang(GCF)=Ang(EAF) , Ang(EFG) in common AF=CF) hence EF=GF. 2.) Via equilateral Triang(GDF) we conclude EF=DF 3.) Via isosceles Triang(EDF) we find directly: Ang(FDE)=Ang(DEF)=50° =30°+x ->x=20°

WoW. This is some complicated solution, considering you are already done when you drew the line at 3:30. Due to symmetry angle(DGF)=angle(AGB)=60°. The line CG divides the 20° at C into 2*10° and the 60° at G into 2*30° (again symmetry of the isoceles triangle ABC and the yellow lines). But then triangles CGF and CGD are congruent with CD=CF making GD=GF, and so DGF is isoceles with the top angle being 60°. Therefore the other angles are (180-60)/2=60 and GD=GF=DF. Finally triangles ACE and CBG are congruent with AC=CB, so AE=CG. But then since triangles AEF and CGF are congruent, it follows EF=GF=DF. Also angle(AFB)=angle(ADB)=40°, and so angle(DFE)=180°-60°-40°=80°. This gives angle(DEF)=x+30°=(180°-80°)/2=50° => x=20°.

I find solving these more relaxing than meditation

Solving it analytically: "Wow, that was quite exhaustive, wonder what beautiful and efficient solution will he present". Him: This.

I really like these geometry videos, but I'm very glad that you have slowed down since filming this. You definitely feel rushed in this, while your more recent videos are much more relaxed and calm. So it feels like you have now found your sweet spot, and I'm glad for that :)

8:00 you have a huge mistake in the argument!!! The triangles are in fact congruent, but it is not because of SAS the way you argued!!! The angle of 20 is not between the sides but across the small(!!!!) Side. There is another triangle with these 3 attributes that is not congruent. The angle has to be be across the bigger side. Triangle ABG is isoceles (actually it is equilateral but that is not important) therefore AG=BG and then by SAS the triangles are congruent

I really really love that you don't skip over the trivial steps on your channel, and go through whole solutions thoroughly - thank you! The individual steps aren't difficult, but sometimes figuring out what the next step is is the tricky bit, and you make the logic of the sequence very easy to follow, great communication skills!

Forgot to thank you for your great videos! Well structured and concisely presented. Hope I can find enough time to learn from your number theory lessions... my brain is quite numb in this region ;)

10:17 i dont get it. First of all that's 2 angles and a side so how is it SAS, but also how is CBG involved in triangle CAE in any way?

I saw a bunch of different ways to solve this problem, but I really want to see a method that only uses sine or cosine law without auxiliary lines. Is there a way, Mike?

By the way, i am sorry for only pointing mistakes. I really enjoy your lessons (my favorites are number theory) And i know how much effort you put in your channel. Had to say that too :-)

Wow what a nice solution!

Hey Michael congrats for your first geometry problem keep uploading more Olympiad level geometry problems,and also I requested the 1994 Romanian maths Olympiad number theory question.the question is d1

Starting with the original diagram with no further constructions (beyond the three given lines), reflect the whole thing about the vertical centre line (no need to draw the vertical line). Now there are six internal lines. Then draw the two horizontal construction lines as per the solution. Then start filling in angles in triangles and where parallel lines also permit corresponding angles. Keep on filling... you'll eventually end up at the value for x.

Is it possible to solve this without drawing extra geometry? It seems like it should be since the problem is well posed. I’m trying and its not working out

This is exactly the same math problem that I tried to solve but couldn't. Then I was fifteen, a junior high school student. Forty-four years have passed since then, and I finally got the solution! Thank you very much!

You should continue making videos. So, great they are.

Well... I got a calculable solution that gives you the right answer, but it sure doesn't look pretty. When I started watching the video afterward, I saw him draw 3 new lines and I was like "oh crap, he gets an actual numerical answer, doesn't he?" . Here's what I did: (note: all angles are in degrees AND the spaces are for alignment purposes, sorry if it aligns poorly on your screen) . Step 1. Calculate 3 more angles: . Angle ADB = 180 - 10 - 70 - 60 = 40, Angle AEB = 180 - 70 - 60 - 20 = 30, Angle BDE = 180 - 20 - 30 - x = 130 - x . Step 2. Law of sines on 4 triangles: (I ignore the 3rd equivalence for each because I don't use it) . ADE: AD / sin x = DE / sin 10 BDE: DE / sin 20 = BE / sin(130 - x) ABE: BE / sin 70 = AB / sin 30 ABD: AB / sin 40 = AD / sin 60 . Step 2.5. Multiply both sides and regroup: . AD = (sin x / sin 10) DE DE = (sin 20 / sin(130 - x)) BE BE = (sin 70 / sin 30) AB AB = (sin 40 / sin 60) AD . Step 3. Recursively substitute each equation into the previous equation (e.g. A = 2B, B = 4C => A = 2 * 4 C, except do 3 replacements instead of 1) . AD = (sin x / sin 10) (sin 20 / sin(130 - x)) (sin 70 / sin 30) (sin 40 / sin 60) AD => (sin x / sin 10) (sin 20 / sin(130 - x)) (sin 70 / sin 30) (sin 40 / sin 60) = 1 . Step 4. Multiply both sides and regroup again, then define constant for simplification: . sin(130 - x) = (sin 20 sin 70 sin 40) / (sin 10 sin 30 sin 60) * sin x Let k = (sin 20 sin 70 sin 40) / (sin 10 sin 30 sin 60) => sin(130 - x) = k sin x . Step 5. Expand sin(130 - x) and do some algebra: . sin 130 cos x - cos 130 sin x = k sin x => sin 130 cos x = (cos 130 + k) sin x => cos x / sin x = (cos 130 + k) / sin 130 => cot x = cot 130 + k / sin 130 . Step 6. Isolate x: . arccot(cot x) = arccot(cot 130 + k / sin 130) x = arccot(cot 130 + (sin 20 sin 70 sin 40) / (sin 10 sin 30 sin 60 sin 130)) . Step 7. Note upper/lower bounds of x, because arc trig functions sometimes give you answers outside your bounds: . 180 = Angle CED + Angle AED + Angle AEB = Angle CED + x + 30 => x = 150 - Angle CED => 0 Watch Michael Penn be way more clever than you for 16+ minutes. => Write your entire solution into a KZfaq comment in an attempt to save face. . QED.

We actually had it as a homework in my highschool. There were 2 different solutions in our class. You can use therorem of sines three times and some trygonometric identities

15:50 since triangles ADB and AFB are congruent, angle AFB=angle ADB=40. There is no need for the other angles calculations

Math is so neat when you know what you're doing and everything falls into place. Things like this is just a lot of keep track of while solving.

This puzzle is known as "Langley's Adventitious Angles"; a similar variant of this appeared on the 2014 MATHCOUNTS National Team.

You don't need SAS to show angle BAG is 60 degrees. Trivially angle AFB is 40 degrees. Since triangle FGB must have angles add up to 180 degrees, we have angle FGB is 120 degrees. Therefore angle AGD is also 120 degrees and thus angle DGF = angle AGB = 60 degrees. Since angle GBA was given as 60 degrees, it remains that angle BAG = 60 degrees. Now we have all the other angles necessary to easily find x.

Taking the meaning "look at things from different angles" to new levels

How do we know to compose DF and take a new point G at the intersection?

Generality is not lost even if AB=1. From the sine rule, BD=sin80°/sin40°=2cos40° BE=sin70°/sin30°=2sin70° From the sine rule for △BDE, 2cos40°/sin(x+30°)=2sin70°/sin(130°-x ) ∴cos40°sin(50°+x)=cos20°sin(x+30°) ∴sin(x+90°)=sin(x+50°) ∴(2x+140°)/2=90° ∴x=20°

Dear MP: angle of DEF is obviously same as ABF from similarity. No need for extra work there.

In Brazil we call it "Triângulo Russo" (Russian Triangle), despite we know it has never appeared in any Russian Olympiads. It is a fun and well-known problem, but always interesting to see the solution

I got this after a heap of ugly sine and cosine rules putting the base of the triangle equalling X (which later cancels in the cosine and sine rules). This was after I discovered that 4 equations in the 4 unknown angles is completely powerless (which was a very strange feeling). Why does it happen that 4 equations in 4 unknown angles yields nothing. Is it just because of this particular angle set up or would it happen generally.

5:48 if triangle ADB is congruent to triangle AFB then angle BAF (which is BAG) is congruent to angle ABD

15:56 you may use angle ABC = angle DFE by corresponding angle

Great problem. I'd like to see more geometry problems here.

I found a way to solve this using only trig and without any auxiliary lines. We will just use Law of Sines (LoS) and the double angle formula. Let AB = 1, as this won't change the angle we desire and it makes the problem more easy. Note that angle ADB = 40, so using LoS on triangle ADB tells us 1/sin(40) = DB/sin(80) -> DB = sin(80)/sin(40) = 2cos(40) by the double angle formula. Now, note that angle AEB = 30. Using LoS on triangle AEB tells us 1/sin(30) = EB/sin(70) -> EB = 2sin(70) (I used sin(30) = 1/2 here). So we know now that DB = 2cos(40) and EB = 2sin(70). Note finally that since angle AEB = 30, angle EDB = 130-x. Using LoS on triangle DEB tells us DB/sin(DEB) = EB/sin(EDB) -> 2cos(40)/sin(30+x) = 2sin(70)/sin(130-x) -> cos(40)sin(130-x) = sin(70)sin(30+x) -> sin(50)sin(130-x) = sin(70)sin(30+x). We proceed by guess and check. Since there is a sin(50) on the left, we guess sin(30+x) must equal sin(50), so x = 20. Plugging this into our final equation, we see sin(50)sin(130-20) = sin(50)sin(70) = sin(70)sin(30+20), so 20 is our solution. We can analyze graphs of both sides of the equation to confirm that this is the only solution.

I don't understand the congruency of CE = AG. For lines to be congruent, isn't it necessary that their length be equal?

You can use a helpful Lemma for this problem

8:07 does not work. For SAS to apply, the angle *must* be the one defined by the two sides. The thesis is still vslid by simmetry though.

After drawing CG we immediately have ∠GCB = ∠FAE = 10°, CF = AF, and ∠CFA = ∠AFC. So ▲CFG ≌ ▲AFE due to ASA. Then we have EF = GF = DF, from which the result follows.

Wow this problem was on Mind Your Descriptions and then on TecMath and now here. Why do all these youtubers love this problem so much??

Unfortunate. The missing part is WHY? Why I start with F and G. Where do I aim? I know this would make the lesson longer, but immensely more useful. When teaching math, the hardest thing is always to explain WHY you go certain direction. Do you remember proofs in calculus, where you start with something like pi*delta/2 to get in the end to pure epsilon? Because the author refined the proof to be "elegant"? The same thing here. P.S. I understand the motivation to use "simple" tools. But "a line intersecting two parallel lines" would be shorter and may be more obvious, still being very basic and simple.. I would expect Michael to explain, what restrictions he used on the tools and why. P.P.S. Thank you for the lessons. They are generally great and inspiring.

Just tried it from MindYourDecision, and indeed, it was hard though very simple to solve. It broke my heart that I couldn't solve it 😅

No need of construction, congruency, similarity. Just by the fact that sum of angles of triangle is 180, you get 4 variable, 4 equation. Just the very elementary computation solves it.

Could be solved more simply by using equivalence of opposing angles where DB crosses AE (F say) and the sum of angles around D and E, and in CDE and DEF being 180. Introducing angles w = BDE, y = CDE and z = DEC, and writing in all other angles that can be calculated directly, it boils down to four equations in four variables. You don't need that it's an isisceles triangle.

Michael, I suspect that by having a parallel to DE passing through A lots of angles can be inferred leading to an easier argument... Havent tried though .. Nice stuff 😀

This is a general problem taught to many Olympiad aspirants , I myself was taught the same in 9th grade however the solution to it was a bit easier

For a simple exercise in discovering the angles of triangles, this is bizarrely complex. 70+60+50 50+&c....

I have a book (in polish language) for students interested into mathematical competitions/olympiads/etc., that have a chapter "one problem, n solutions" (don't remember n, but it is >=6) and shows n solutions for this problem. All of them quite long.

I came across this problem a few years ago. It looked easy but was deceptively hard. It took me quite a while to figure it out on my own. I did make some constructionsut I don't recall having to construct DF and AG. By the way . . . Some funny choice of words here by Michael. While we do say congruent shapes, such as congruent triangles, we don't say 'congruent line segments' or 'congruent angles'. We'd rather say it the more usual way - EQUAL line segments or EQUAL angles. Furthermore we never make actual measurements when solving maths problems. Measuring is unacceptable in plane geometry. It is a strict taboo. So it has always been perplexing me why they say something like 'angle measure' or 'measure of a line segment' . . . I think the appropriate terms are 'SIZE' for angles and 'LENGTH' for line segments.

Quite rare to see a geometry video from you, sir. I think you should definitely keep more of this kind coming!

You can skip 90% of these steps. Like we know triangle ABC and triangle DFC are similar because DF is parallel to AB, so obviously angle DFC and angle ABC are congruent

I remember solving this a while back in a much cleaner way

i attempted doing this purely with basic arithmetic/algebra based on the given geometry, no drawing any extra lines, and eventually gave up. can someone please put me at ease and confirm that it is in fact impossible to do it that way?

I don't see how CE is congruent with AG, though. Even a blind man can see that they're not equal

Found a still faster method by noting 1.) that the point O of intersection of AE with CG is the center of the circumscribed circle of Triang(ABC). 2.) Then constructing a rotated copy of Triang(ABC) called Triang(A´B´C´) within the circumscribed circle such that the top Point C´of the new triangle is identical to Point A, O=O´ and C=B´. Since in this new rotated triangle we can 3.) find a point G´(corresp. to point G) which is equal to point E since points G´OA are kollinear points. 4.) Hence we can find the corresponding equilateral triangles Triang(DFG)=Triang(FD´G´) to be tilted by Ang(DFD´)=20° and 5.) we find the isosceles triangle Triang(FDG´) (note G´=E)!. Since Ang(FDG´)=Ang(FG´D), Ang(DFG´)=60°+20°=80° and since FG´=FD we obtain Ang(FG´D)=(180°-80°)/2=50°=Ang[A(G´=E)B]+x=30°+x ->x=20° QED

It could be shorter demonstration ¨!

Alternatively, apply the law of sines, AD/DE=sin(x)/sin10, DB/AD=sin60/sin80, DE/DB=sin20/sin(x+30); We have, sin(x)*sin60*sin20=sin10*sin80*sin(x+30), tan(x)=sqrt(3)sin10/(4sin20*cos10-3sin10)=sqrt(3)sin20/(4sin20(1+cos20)-3sin20) =sqrt(3)sin20/(2sin(60-20)+sin20)=tan20 Thus, x=20.

I tried making every triangle separate and start filling in sides and angles assuming AB was given and hoping it would cancel out, well... it did cancel out but I have a monstrous equation with multiple sin(10), sin(40), sin(80) and some are in the denominator and are under a sqrt... im gonna assume I've just discovered a new identity for 20 instead of not being able to simplify well

YEAH, please professor, more classical euclidean geometry

Now do world's easiest "hard" problem

Because DF is parallel to AB, wouldn't the angle DFC (where DF intersects BC) be the same as angle ABC? Wouldn't that be easier than worrying about the angles in triangle GBF?

Well is 8:00 the triangles ACG and BCG are congruent by SAS where AG = BG not CG=CG , i guess that's a silly little mistake there 😅 Edit - ooh again , at 10:16 i guess you've made another mistake .. angle CBG is no where related to triangle CAE

Probability video next please

8:30 how that comes? There should be two sides and angle BETWEEN.

Now please show us the solution for the hardest 'hardest' problem

I've seen this decades ago!

That was the longest treatment of this puzzle. I found the method did not stop often enough to explain why you were figuring out some aspects which were obvious (for example CG was an angle bisector by virtue of how G came about so many congruences were surplus, or CD was equal to BD from the outset as they were equal angles from a common side, calculating DFE when it was defined by virtue of DF parallel to AB, etc.). Essentially, this is a three step problem you over complicated.

@10:40 isn't that supposed to be ASA?

To calculate DFG just use DF//AB and corresponding angle is ok

Please solve Swedish Olympiad 1978 Q2 Loved your video Man!😍😍

Is x=20 degrees a unique solution?

11:20 ok how did we get CE=AG from that tho?

I remember seeing this problem on MindYourDecision

Using symmetries would make it look a lot easier.

Interesting. More geometry Olympiads problems would be cool.

I solved this problem with another way 20 years ago, when I was student. My way was more complicated than this.

9:57 that doesn't help in any way. You need to show it's equal to

In some lessons you need to be able to follow and understand the teacher very well. In each of the lessons presented to the majority of students in the social network, students can gain knowledge in the areas of study. On this explanatory activity

Angle DAE seems wrongly written as 10 degree?

Shouldn't it be 10 without the calculations. Angle A and E in triangle GAZ and angle DEZ are the same Z is between AE

Amazing

Awesome! !😉☺☺☺😍😍

Your last step could have been much shorter. As soon as you found out DEF was isosceles, you could have found angle DFE by parallel angle thm with ABF (since AB || DE), so you didn't need to do those extra angle solves.

There must be a super-cunning and easier way yet to be discovered to solve this!

L DFE is 80 because DF is parallel to AB. No need to calculate any further.

You don't need to draw any extra lines to solve this. If the intersection of AE and DB is labeled F, AFB = 180-70-60 which is 50 and is the same as angel DFE. That tells you AFD and BFE =130. From there fill in the remaining loose angles until you are left with CDE CED FDE and FED. They can be put into 4 equations CDE + CED = 160, CED + DEF = 150, CDE + FDE = 140, DEF +FDE = 130. 4 equations, 4 variables, solve. It's only a hard Geometry problem if you, ya know, try to use Geometry and not Trig and Algebra.

10:23 huge mistake

SAS ..... Angle has to be inbetween 2 sides ..... ???? Please clear doubt

Twelve solutions to this problem: www.cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml

There's another of solving this particular problem. Geometry problems can obviously be solved in numerous ways.

It can help in RAY OPTICKS! agree?

Why is everyone telling him what he is doing wrong and how it can be done more efficiently. It doesn't matter how you get to answer as long as you get it that's a win in my book.

for some reason Supertramps you take the long way home comes to mind

exactly as advertised

Honestly, once the value of EDF was found, the remaining solution got too complicated. Since DF is parallel to AB, triangle ACB is congruent with DCF. So angle DFC would be 80, like at the base. No need to calculate DFG and BFG.

Hello professor; on my side, I did not use your method which is fast and very [too?] clever (therefore without interest for the pupils), but a more direct method, certainly longer, but not artificial, and especially which works some be the values ​​of the angles! Thanks to very specific data on the angles, we obtain an equilateral DGF triangle; which makes it easy to determine x, which has the value [still too special] of 20 °; in a different case, you would be blocked with your method ... In summary, your exercise with the data announced, is feasible BUT very difficult for students 13-14 years old [who have not yet learned trigonometry], but feasible by students 16-17 years old, who studied trigonometry , and who know the classical formulas and the resolutions of "standard" trigonometric equations. Anyway, thank you for this exercise, which allowed me to discover a very fun simplification of writing: E = 2cos30 ° / (1 + 4 sin70 °), and to be the subject of a small exercise in trigonometry for students aged 16-17. Greetings from Paris. professor essef, in mathematics (active for over a year on KZfaq and Wikipedia, in astronomy & astrophysics). (my like has the number 421 (lol)).

For those interested, using trigonometry: -math.stackexchange.com/questions/6942, forth solution in the Seyed's answer.

Prove this one: The lengths of two of the internal angle bisectors of a triangle are equal; show that the triangle is isosceles.