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Geometry thing... Again?

I would've posted this on the other thread but it got closed so I'm creating another one. I hope there's nothing wrong with doing this. And by the way, this isn't my homework or anything, I'm just practicing some geometry, but I end up getting stuck at some problems which I only have the answer but not the solution.

This time I'll be more clear so no one wastes their time trying with wrong information.

In a isosceles triangle, AB=AC, and the angle Â=80. In BC, we choose a point D such that <DAB=30, and in AB, we choose E such that <BCE=20. Find the value of <DEC.

Image

To anyone who give it a try, thank you :)
Now for the moment of truth to see if I suck at math as much as I think. I probably overcomplicated it by using too many steps; I found the answer though (unless I was way off).



tl;dr angle DEC = 40

Also I just realized that I didn't need step #5 at all so ignore that
according to my shit calculation which is so bad and lazy I don't even wanna show it. I believe angle DEC is 50°
Pat, I think you made a mistake between steps 7 and 8. In step 7 you found BEC = 110° but in step 8 you treat DEA as 110° instead ?_?

I'm completely clueless, but I'll leave this here in case I was getting somewhere with drawing random lines idk
I have no idea how to find anything about ED and its angles. Must involve some obscure properties again.
It's easily the best thing I've done
So why the empty numb?
Pat, you never found <ADE. You can't just assume it's 40°.
The rest of your logic is okay but that one flaw brings down everything.

I'm still struggling on the problem, and so is the rest of the JUMP1/2 team LOL. Waiting for worldpeace again *sigh*
As people pointed out, you didn't really find ADE, Pat, so that's not a correct solution :( Thanks anyway!

lolyoshi: it isn't 50, unfortunately. I don't want to tell the result (or at least just yet) so people don't focus their effort on making that specific value appear there.

Koopster: thanks for helping again mate. I actually tried to do the same exact thing you did, but couldn't reach anywhere as well. Now, seeing that, I think I'll probably post one of my tries.

ft029: hey man, thanks for the help again! And also for asking the JUMP1/2 team to help as well :D I guess you guys like these challenges as much as me. If worldpeace nails this one again, he's a true god, dayum. By the way, if it's possible, could you ask him if he learned his techniques in a book or something I could study as well? I mean, since most here are from different countries, we might learn stuff in very different ways. For instance, apparently russians have some nice geometry problems which I also struggle to solve, and that also reflects in their mathematics olympiads problems.
You're all a bunch of nerds. I guess you can keep using this thread to pose more questions if people somehow have fun with this?? I closed the other one cause I figured you were looking for a solution for homework or whatever, I'll leave it open if it's just a thing you wanna keep doing.


smh @ u nerds
Your layout has been removed.
I'm too lazy to go through the entire problem, but here's my stab at the method based on what I remember from geometry years ago:

I was immediately reminded of a little thing called Langley's Adventitious Angles, which involves drawing a few additional isosceles triangles inside of the main one. You can use the fact that two sides + two angles are the same in each to work your way into the unknown angle in this problem.

I distinctly remember my high school teacher saying that problems like this are used in admissions tests at MIT/Harvard/etc - they use simple concepts, but involve thinking outside the box.
(Not solution, but I'm getting really, really close)
There are TONS of unmarked isosceles triangles here. Like, TONS. Then use your knowledge of angles and congruent sides. I can taste victory, but I still need to finish it off.

One of the isosceles triangles I found was very difficult to find and involved cyclic quadrilateral stuff, so I think I might have done it in a really roundabout way. But who cares.


Fun fact: I've been thinking about this problem for about 2 hours straight. Wtf am I doing with my life.
S.N.N.: Yep, that's a classic. Here we know this problem as the "Russian Triangle Problem", but seeing this article, I don't know why we really call it like that lol.

If you take the image in the first post and rotate it to the right, you'll get something really close to that triangle. In one of my attempts I tried to find a solution taking the same steps as you would take to solve this Langley Triangle, but I couldn't reach anywhere because of the bigger angles that doesn't let you insert a bunch of isosceles/equilateral triangles where I needed and the fact that, when you rotate the triangle, you don't have the sides with same length where you need. There's a (20,80,80) triangle, the CDP (P being the intersection of AD with CE), which is the exact Langley Triangle, but I don't see how it could help since the angle we want is in the opposite side :/

EDIT: ft020, what if I tell you I've been thinking about this problem for 4 days straight xD
wow I really do suck balls at math
Yeah this problem is pretty ridiculous, which is funny because initially I thought it was easier than the first one LOL. Anyway, after several hours of the rest of us trying to crack it worldpeace finally came to the rescue and solved it. I won't post the solution here though so everyone else can try.

e: ok since trs cheated and use law of sines i'll post the better proof worldpeace found (in spoilers if you dont want to see it yet)



1. Draw a line AP that divides <DAC into 30° and 20°, then let M be a point on line AP such that AE = AM and conclude:

- Triangle AEM is equilateral and <AEM=60°.

2. Since (<ECA*2) = 30*2 = <EMA and M is on the perpendicular bisector of line AE, conclude:

- M is the circumcenter of triangle AEC
- EM = MC
- DE = DM too, because line AD perpendicularly bisects line EM
- DA = DC
- △CDM and △ADE are congruent

3. Let x = <MDC, conclude:

- x = <MDC = <EDA (because of congruence) = <MDA (again, AD is the bisector)
- <ADC = 80° = 2*x
- x = 40°
- <AED = 180°-30°-40° = 110°.

4. Finally we also know that angle AEC = 70°, so the answer is 40°.
It's really strange when you go "fuck geometry" and instead turn to trigonometry to get your answer. But here we are.

Anyway, let's assume the two sides that are equal in length are each 1 unit long. The actual length doesn't matter, but we don't want to be dealing with variables any more than necessary.

While you can figure out most of the angles in the figure below, these are the only ones we care about:


(Click it!)

In order to determine the value of <DEC using trigonometry, we need to know the lengths of the sides labeled in green, blue, and purple. We'll start with the blue side.

You can use the law of sines to determine the length of each side of a triangle, given that you know the measurements of all the angles and the length of at least one side. Given that we want to find the length of line CE and we know the length of line AC, we can use triangle ACE to calculate. We know all of the angle measurements and the length of one side (1 unit), so we can plug in all the numbers to solve for the length of the blue line.

Code
x/sin(80°) = 1/sin(70°)

multiply both sides by sin(80°) to isolate x

x = sin(80°)/sin(70°) ≈ 1.048011


Similarly, you can use triangle ACD to find the green line.

Code
x/sin(50°) = 1/sin(80°)

multiply both sides by sin(50°) to isolate x

x = sin(50°)/sin(80°) ≈ 0.777862




At this point I didn't remember exactly how to continue and turned to WolframAlpha. We know the length of line CE and the length of line CD, as well as the angle <ECD (20°). Using this widget, we can plug in these three values and the length of the purple side is calculated for us! The first answer it gives you is 0.4139, but if you scroll down it gives a more precise 0.413892 (this is why I rounded the other lengths to 6 digits, by the way).



Now that we have the length of all three sides of triangle CDE, we can solve for <DEC using the law of sines again.

Code
sin(x)/0.777862 = sin(20°)/0.413892

Multiply both sides by 0.777862.

sin(x) = [0.777862*sin(20°)]/0.413892

Divide both sides by sin(). 
Note that x/sin() is the 
inverse of sin, or arcsin(x).

x = arcsin([[0.777862*sin(20°)]/0.413892) ≈ 40°




Our answer is not *exactly* 40° because we rounded the lengths of the three colored sides, but it would have been exactly 40° if we hadn't rounded.


Noivern: I really appreciate your solution, which is right indeed. It's just that we can't really calculate the sine of these angles without a calculator, so I wouldn't say it's a totally fine solution if this was a question in a test (which indeed it is). Maybe you can use some sum/product formulas to find the sine of these angles in terms of known angles, but I can see it would be super tedious to do.

LHB: He solved it again only with constructions? How does he even do that ._. I wonder how hard these problems are for him.

Anyway, thanks everyone, I really appreciate your help :)

One of my teachers showed me how to solve it. I'll post it as well, because it's different from both worldpeace's and Noivern's solution.

Image

- Prolong CE until you can draw DT=DC (Sorry, forgot to include T in the image, but I think you'll understand). The construction implies that <DTC=20 as well. Note that <CDT=140, because the other two angles are 20, but we know that <CDA=80, so <ADT=60
- Note that the triangle ADC is isosceles (two angles of 50), so AD=DC=DT.
- Trace TA and close the triangle. Note that it's an isosceles triangle with an angle of 60, so it's actually equilateral. This means that <BAT=30 as well.
- <BAT=30 implies that AB is a bisector, but since the triangle is equilateral, it's also a perpendicular bisector, so DE=DT, and also TDE=20.
- Finally, the angle we want is 20+20 (external angle theorem)



Anyway........... I really don't know how much you guys are enjoying this (probably aren't at all xD), but this is another one I couldn't solve too. I guess I suck at geometry, oh well. If you guys want to give it a try:

The triangle is rectangle in A, and AB=CD. <BAD=x, <DBC=2x and <DCB=x. Find the value of x.

Image
Originally posted by undefinied3
The triangle is rectangle in A, and AB=CD. <BAD=x, <DBC=2x and <DCB=x. Find the value of x.

Image


Did you manage to solve this? I gave it a try and I can't really figure out what to do next. Here is what I did:
- Extend the line AD until K, where BK = AB.
- Draw the segment CK.
- Triangle CND is congruent to triangle KNB (AAS).
- So BDCK is an isosceles trapezoid, because the diagonals of a trapezoid are congruent.

Image

I think I'm almost there, but I really couldn't figure out what I'm missing.
Yep, I solved this one later, sorry for not posting the solution earlier.

There's one really simple solution using trigonometry but I don't like those, so I'll give the geometric one.

Image

Reflect the triangle to the left. Consider the point D', reflection of D. The segment DD' is also AB since AD'DB is parallelogram, and that closes an equilateral triangle. AC is the altitude of that triangle, so it's also a bisector, resulting in <ACD=30.
Now, take triangle ADB and "add" on top of CD. This construction adds the green line (the one which connects to a point which I forgot to name. let's call it P), which has the same size as DB, so BDP is isosceles, and <DPB=2x. This makes CPB isosceles and, as a consequence, <PDC=x as well, but since that triangle is congruent to ADB, <DBA=x.
Now we have a triangle with angles 90, 30+x, 3x.
90+30+x+3x=180 -> 4x=60 -> x=15


I'm not sure how many of you are still interested in that geometry stuff, but here's another one just to always keep one challenge here,

Find x in terms of theta, given that AB+BC=CD
Image