The answer is 90.

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beautiful solution!

Marià Cano
- 6 years, 6 months ago

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that is great thought! but can you enter the dimension of sides of triangles and if possible explain how you have superimposed sides on each other to make it into a sensible figure. GREAT JOB anyways!!!

abhideep singh
- 6 years, 6 months ago

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The reason I didn't include the dimensions of the triangles is because I didn't want to introduce unnecessary visual clutter. Though having the dimensions on the spot is convenient, they can easily be derived using the grid in the background.

To be honest, this method rarely works. I was just thinking can I do this without calculating since I was pretty lazy on that day. It would be cool if someone could provide a formal proof for whether or not such a figure exists.

Evan Lee
- 6 years, 6 months ago

Very simple way is
Find it through ratio.

1(45°)+2/3(45°)+1/3(45°)=90°

Maseeh Rezazad
- 5 years, 3 months ago

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with 3:1 ratio of the sides for the triangle with the angle c, 3c+c=90 and thus c=45/2. Cud u please explain how did u get c=45/3. Thanks in advance.

Sreyasee Bhattacharjee
- 4 years, 6 months ago

Very elegant!

Davy Ker
- 6 years, 5 months ago

Extraodinary Explanation

Tej Ajet
- 5 years, 6 months ago

This video describe a similar solution without scale triangles https://www.youtube.com/watch?v=m5evLoL0xwg

Victor Duarte da Silva
- 5 years, 6 months ago

i used arctan hahahaha 45 + 27 + 18 = 90 XD

Keiji John Libadisos
- 5 years, 1 month ago

Can you elaborate on the NOTE?

omkar pawar
- 4 years, 11 months ago

I love a good "proof by picture".

Richard Desper
- 4 years, 7 months ago

A truly elegant solution

Peter D Morrison
- 3 years, 1 month ago

I believe that the answer is 94.47122063

Mohamed Yasser
- 5 years, 6 months ago

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Mohamed, you are possibly using sin rather than tan in one of your calculations. Sin = Opp/Hyp and Tan = Opp/Adj legs. And that is why you reach 94.47 and not 90 degrees (I had that mix up before).

Raphael Cavalcanti
- 4 years, 9 months ago

$(i+1)\cdot (i+2) \cdot (i+3) = 10i$ which has angle $\boxed{90}$ .

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Lovely solution.

Saatvik Jain
- 5 years, 4 months ago

i dont understand, can u explain?

Thành Trần
- 5 years, 4 months ago

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You can take those angles like complex numbers', so, if you take the three complex numbers $x$ , $y$ , and $z$ such that $arg(x)=a$ , $arg(y)=b$ and $arg(z)=c$ you will find out that $x=1+i$ , $y=2+i$ and $z=3+i$ . For finding the value of $a+b+c$ we can use that $xyz=rcis(a+b+c)$ where $r$ is the module of $xyz$ . As $xyz=10i$ we have that $a+b+c=\frac{\pi}{2}$ or, in degrees, $a+b+c=90$

Hjalmar Orellana Soto
- 5 years, 4 months ago

Very beautiful solution

Hjalmar Orellana Soto
- 5 years, 4 months ago

On an $xy$ -grid, draw triangle $\Delta ABC$ with coordinates $A(0,3), B(2,2)$ and $C(1,0)$ . Also label the origin $O(0,0)$ and the point $D(3,3)$ .

Then since $AB = BC = \sqrt{5}$ and $AC = \sqrt{10}$ we have that $\Delta ABC$ is a right isosceles triangle, and so $\angle BAC = 45^{\circ}$ , which is also the same value as $a$ .

Now $\angle OAC = c$ and $\angle BAD = b$ . But $\angle OAC + \angle BAC + \angle BAD = 90^{\circ}$ , so $a + b + c = \boxed{90^{\circ}}$ .

Comment: Note that this can be written as

$\arctan(1) + \arctan(\frac{1}{2}) + \arctan(\frac{1}{3}) = \frac{\pi}{2}$ .

Now for $x \gt 0$ we have that $\arctan(\frac{1}{x}) = \frac{\pi}{2} - \arctan(x)$ , (proof below).

So upon substitution we have that

$(\frac{\pi}{2} - \arctan(1)) + (\frac{\pi}{2} - \arctan(2)) + (\frac{\pi}{2} - \arctan(3)) = \frac{\pi}{2}$

$\Longrightarrow \arctan(1) + \arctan(2) + \arctan(3) = \pi$ ,

which is kind of cool. :)

Proof: Let $f(x) = \arctan(x) + \arctan(\frac{1}{x})$ for $x \gt 0$ .

Then $f'(x) = \frac{1}{1 + x^{2}} + (\frac{1}{1 + \frac{1}{x^{2}}})*(-\frac{1}{x^{2}}) = \frac{1}{1 + x^{2}} - \frac{1}{1 + x^{2}} = 0$ .

Thus $f(x)$ must equal some constant for $x \gt 0$ . So plug in any positive value for $x$ to find out what this constant is. We have $f(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$ , and so $\arctan(\frac{1}{x}) = \frac{\pi}{2} - \arctan(x)$ for $x \gt 0$ .

(Note that for $x \lt 0$ we would have $f(x) = -\frac{\pi}{2}$ , so the formula for $x \lt 0$ would be $\arctan(\frac{1}{x}) = -\frac{\pi}{2} - \arctan(x)$ .)

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Sir can u please send me your facebook id so that i can add you to my friends list.

Venky Venkatesh
- 6 years, 6 months ago

This is the real solution. Thanks

Burak Cagatay
- 5 years, 6 months ago

tan -1 x + tan -1 y= tan -1 (x+ y)/(1-xy)

Now tan a = 1, tan b = 1/2 and tan c = 1/3

Now a + b +c = tan -1 ( 1) + tan-1 (1/2) + tan -1(1/3) =45 + tan-1 (( 1/2 +1/3)/(1-(1/2)x(1/3))) = 45 + tan-1(1)= 45+45= 90 degrees

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im almost there i solve it almost like that and get 89

Motchi Leuterio
- 6 years, 5 months ago

I support it . :)

Abdul Hai
- 6 years, 6 months ago

very nice soloution

Mohamed Abdelhady
- 6 years, 6 months ago

how did you get the tan values? Now tan a = 1, tan b = 1/2 and tan c = 1/3

hegde aahana
- 3 years, 12 months ago

Solve in the same manner. Straight forward . Nice one Nidhin

Bhargav Upadhyay
- 6 years, 4 months ago

tan (b+c) = (tan (b) + tan (c)) / (1-tan (b)*tan(c))

= (1/2 + 1/3) / (1 - (1/2)*(1/3))

= (5/6) / (1 - 1/6)

= (5/6)/(5/6)

= 1

Hence, (b+c) = 45 and (a+b+c) = 90

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I like this solution. Haha, but I use calculator

Kyaw Zaw
- 6 years, 6 months ago

Exactly what I was looking for, instead of inverse trigonometric ratios! Thank you!

Ram Padmanabhan
- 5 years, 8 months ago

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That's not actually the case. Using trigonometry, we have that

$b = \arctan(\frac{1}{2}) = 26.565^{\circ}$ and $c = \arctan(\frac{1}{3}) = 18.435^{\circ}$ ,

each to $3$ decimal places. These also add to $45^{\circ}$ , but individually are different than the values you give.

P.S.. Welcome to Brilliant. Sorry I had to start out by disagreeing with you. :(

Brian Charlesworth
- 6 years, 6 months ago

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without calculataor

Jaydeep Goyal
- 6 years, 6 months ago

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I know that. Check out my solution; it requires nothing more than the Pythagorean Theorem. I was simply pointing out to Veena that her solution method was flawed in that the angle values she quoted for b and c were not correct, even though they happened to add to 45 degrees.

Brian Charlesworth
- 6 years, 6 months ago

This solution is completely false , because you say that tan(b)=sin(b) ,which cannot be true in right-angled triangle.

Kristian Vasilev
- 6 years, 6 months ago

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yes tan(b)=sin(b) can true when b is very very small. It means when b tends to zero only then the relation can be true. But it is not true for this case as the angles are not tends to zero.

Trishit Chandra
- 6 years, 6 months ago

This problem was actually on one Numberphile video (The Three Square Geometry Problem).You can watch it if you like.

Abdur Rehman Zahid
- 6 years, 6 months ago

Why are b and c 30 and 15?

Omkar Kulkarni
- 6 years, 6 months ago

how can you know angle b and c measures 30 and 15 respectively

mika servi
- 6 years, 6 months ago

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read the above post and you need a bit of trigonometrical ratios and it's knowledge

Nisarg Patel
- 6 years, 6 months ago

According to your solution arctan(1/2)=arctan(1/root3) and finally 1/2= 1/root3. But this is not true.

Trishit Chandra
- 6 years, 6 months ago

Nice solution mam.

Siraj Mustafa
- 6 years, 6 months ago

thts wrong logic ( even i didnt get the amnswer to this question )

Rohan Naik
- 6 years, 6 months ago

no,you are wrong. According to you b=30 and if we see in the figure tan(b)=1/2 but tan(30)=1/(3^1/2)

akhil kumar
- 6 years, 6 months ago

a=45 tatz fyn/......hw come yu say b= 30???& c= 15???? can yu please explain??

Kaartik Krishnan
- 6 years, 5 months ago

How b=30 and c=15

ABHISHEK RAJAKUMAR DHANASHETTI
- 6 years, 6 months ago

The solution is based on trigonometry

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arctan(1)+arctan(1/2)+arctan(1/3) = 90d

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It is not true that $b = 30 ^ \circ$ and $c = 15^\circ$ . In fact, it should be slightly surprising that the sum is indeed $90^\circ$ .

I have seen this type of question first time andI solved this by only watching the question

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2x3 Solution

Since scaling, translating, reflecting, and rotating do not change the triangles' angles, we can reorganize them like so. The sum of the angles becomes trivial.

Note: I scaled the right triangles with angle $a, b,$ and $c$ by $2, \sqrt{2},$ and $1$ respectively.