In the figure below, C D E is an equilateral triangle. Find ∠ A D E in degrees.
Note: ∠ B E C = 9 5 ∘ , ∠ E C B = 4 5 ∘ , ∠ A B E = 2 5 ∘ , and ∠ B A C = 6 5 ∘ .
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Nice proof! What graphics program are you using to produce the graphic?
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Thanks. It is interesting problem, as always. I use GeoGebra.
Very nice construction of the "triangle congruent to ...". That was a pretty natural point to create esp with the idea of wanting to find congruent triangles as in the inspiration problem, but having the angle between two sides not matching up.
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Calvin, "The Hardest Geometry Problem" is a thing, and the popular version of it involves an isosceles triangle with base angles and couple of crossed lines extending from the base vertices at angles usually in some multiple of 10 degrees. The problem then asks to find some angle in the figure that cannot be found by normal angle chasing. I wonder if such problems can always be resolved by means of congruent and similar triangles. i.e., is there any instance of figures of having angles of integer degrees, with the "answer" also being an integer degrees, that cannot be resolved by means of such congruent and similar triangles? Maybe I'll post another "hardest geometry problem" to see if such a synthetic proof is not forthcoming.
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I don't think that all such (integer, multiple of 5/10) angle questions could be resolved using congruent / similar triangles. Main rationale is because such an approach isn't aggressively promoted as a "try this", but only happens with "notice the lengths / angles are congruent / similar".
That is why with such "construct new point / line" questions, I also try and provide the motivation of why this is the (natural) point to consider, as opposed to the magic of "we pulled this rabbit out of the hat, which was pulled out of another rabbit".
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@Calvin Lin – Right now, it's just a hunch that I have, given 2 triangles with a common base and with interior angles of only integer degrees, should the line through the apexes of the triangles also be at an angle of integer degrees relative to the base (or any other line), then there always exist a synthetic proof for it. I'm still looking for a counterexample. There are literally thousands of such examples, most all of them falling into a relatively small number of synthetic proofs, which differ only in numbers but not in concept.
Here's an easy example: Given two perpendicular lines, and given points A, B, C forming a triangle with all integer degree interior angles, such that points A, B are on the vertical line while point C is on the horizontal line, then there always exist another point D on the horizontal line such that triangle ABD not only has all integer degree interior angels, all lines are at angles with all others by integer degrees. The proof isn't difficult, but triangles of this type are one of the many possible that share a common synthetic proof.
Ingenious method
Really inelegant approach.+1).
excellent. i though it requires some advance geometric theorems. it is shockingly simple and elegant proof.
the main hint for solving this problem is the inspirational problem. when i saw level 5. i thought it requires some advance machinery.
We note that ∠ E B C = 1 8 0 ∘ − ∠ B E C − ∠ E C B = 1 8 0 ∘ − 9 5 ∘ − 4 5 ∘ = 4 0 ∘ . Therefore, ∠ A B C = ∠ A B E + ∠ E B C = 2 5 ∘ + 4 0 ∘ = 6 5 ∘ . This means that ∠ B A C = ∠ A B C = 6 5 ∘ and that A C = B C .
Now, let C D = D E = E C = 1 and A C = B C = a . Considering sine rule in △ E C B :
sin ∠ B E C B C sin 9 5 ∘ a ⟹ a = sin ∠ E C B E C = sin 4 0 ∘ 1 = sin 4 0 ∘ sin 9 5 ∘ . . . ( 1 )
Let ∠ A D E = θ . Then ∠ D A C = 1 8 0 ∘ − ∠ A D C − ∠ D C A = 1 8 0 ∘ − ( θ + 6 0 ∘ ) − 5 5 ∘ = 6 5 ∘ − θ . Considering sine rule in △ A D C :
sin ∠ A D C A C sin ( θ + 6 0 ∘ ) a ⟹ a = sin ∠ D A C C D = sin ( 6 5 ∘ − θ ) 1 = sin ( 6 5 ∘ − θ ) sin ( θ + 6 0 ∘ ) . . . ( 2 )
Then we have:
( 2 ) = ( 1 ) : sin ( 6 5 ∘ − θ ) sin ( θ + 6 0 ∘ ) ⟹ θ = sin 4 0 ∘ sin 9 5 ∘ = sin 4 0 ∘ sin 8 5 ∘ = sin ( 6 5 ∘ − 2 5 ∘ ) sin ( 2 5 ∘ + 6 0 ∘ ) = 2 5 ∘ Note that sin ( 1 8 0 ∘ − x ) = sin x
Same method, I used multiple angles for sin ( θ + 6 0 ∘ ) and sin ( 6 5 ∘ − θ ) , but your shortcut is very neat. Whoops
N o t e t h a t ∠ s E B C = 4 0 o , C D A = 6 0 + α , D A C = 6 5 − α , A B C = 6 5 o , b u t a l s o B A C = 6 5 o . ⟹ A C = B C . W L O G C D = D C = 1 . U s i n g S i n L a w : − Δ E B C , A C = B C = S i n 4 0 S i n 9 5 ∗ 1 . . . . . . . . . . ( 1 ) Δ D A C , A C = S i n ( 6 5 − α ) S i n ( 6 0 + α ) ∗ 1 . . . . . . . . . . ( 2 ) E q u a t i n g ( 1 ) a n d ( 2 ) , e x p a n d i n g S i n ( A + B ) , w e g e t , ( S i n 9 5 ∗ S i n 6 5 − S i n 4 0 ∗ S i n 6 0 ) ∗ C o s α = ( S i n 4 0 ∗ C o s 6 0 + S i n 9 5 ∗ C o s 6 5 ) ∗ S i n α . ∴ R e p l a c i n g v a l u e s o f S i n s , C o s e s a n d A r c T a n , a n d s i m p l i f y i n g α = 2 5 .
I am inviting proofs for this one.
@Michael Mendrin Nice question :) @Calvin Lin Thanks for sharing link to Figma , I found it super easy and useful to draw although I use geometer's sketchpad normally for such geometry related questions.
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Pandya, long time no see! How's life treating you these days? And have you heard again from any of the old Y!A regulars? Brian Charlesworth and Pi Han Goh are still active here.
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Sorry for delayed reply, I started my own consultancy start-up so was super tied up in setting up everything and yes, I have had interaction with Pi han Goh over here but not with Brian so far. Will try to catch up with him as well. I am happy to learn more people from our Y!A era are part of this ecosystem. In Y!A Duke, Arms A and couple of other old members are still active. I still miss those old days when Dragan K, Quadrillerator, Zo Maar, Vasek, Falzoon, Zeta, JB, Pneuma, torquestomp, Rita, late Mr. Madhukar and others were active on Y!A. Those were real fun days :(
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@Vikram Pandya – and gianlino too. Yeah, we had a lot of great conversations going, and we went all out on really tough problems, even if sometimes nobody could solve them. Then Y!A for corporate reasons decided to make cliques much more difficult to maintain. And so over time we lost touch with each other, and people went their separate ways.
Still, it's good to hear that you've got a start-up consultancy going, I wish you all the best. I'll wait for the next super-hard problem from you. I feel like posting one myself about right now.
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@Michael Mendrin – True, I have messaged all the old contacts through Y!A email, mostly Duke and Dragan K will join Brilliant soon :) BW nice new question on Sierpinski Carpet.
sir, if you have called it a level 3 problem i could have solved it. when i saw it is a level 5 problem. i was doing complicated things like reflecting things etc etc. the inspirational problem is the hint.
It is not clear to me which angle is 45. Can you clarify? I'm guessing ∠ E C B .
We can update the image and place in sectors to make angle labelling more obvious.
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It's exactly like the other problem that inspired this one. Yes, ∠ E C B . I've added a note to make that clear.
For the sake of a more consistent graphic standard, I can include the arcs.
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Is this image good?
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@Calvin Lin – Can you get me the link to the graphics program you guys are using? It is very good.
I'm all for consistent graphics standards.
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@Michael Mendrin – We use Figma , with the following colors
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@Calvin Lin – Hey, but just how many people are working at Figma in SF? But now that I know Brilliant is using this as a kind of a standard, I'll have a closer look at it. Mathematica is great in many ways, but it sucks at graphical design.
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∠ A C E = 5 calculated for quadrilateral A C E B .
Triangles B C E and A C D have two sides of the same length: B C = A C , E C = C D . Their corresponding angles are different though: ∠ B C E = 4 5 , ∠ A C D = 6 0 − 5 = 5 5 . We can easily create a triangle congruent to △ A C D by the following construction.
Let F denote point of intersection of B E and A C . ∠ B F C = 9 0 . Let G denote reflection of E about A C . ∠ F C E = ∠ F C G = 5 . Triangles B G C and A C D are congruent because ∠ B C G = ∠ A C D = 5 5 , therefore ∠ A D C = ∠ B G C = 8 5 . ∠ A D E = ∠ A D C − 6 0 = 2 5 .