Very confusing triangle

Geometry Level 3

Two sides of a triangle are 5 and 12 cm long, and the measure of the third side is an integer (in cm.). If the triangle must also be obtuse, then how many such triangles are possible?


The answer is 6.

Biswajit Barik by Biswajit Barik , 20, India

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2 solutions

There are 6 possible cases for the third sides..... AND THE REQUIRED CASES ARE: 8 , 9 , 10 , 14 , 15 , 16 The only thing which we have to notice are: (1) The sum of any two sides of a triangle is greater than the third side. So, let the length of the third side be /(a/) then
(i) 5 + /(a/) > 12 => /(a/) > 7 (ii) 5 + 12 > a => /(a/) < 17 .'. 7 < a < 17

(2) For any obtuse angled triangle : x^2 + y^2 < z^2 where z > x and y

(i) When a < 12 then a must be 8 , 9 , 10 and11 Now 5^2 + a^2 < 12^2 This criteria is satisfied by 8 , 9 , 10 Similarly when a>12 the other three values are 14 , 15 , 16

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How can you say that there are only 6 case as the third side remain in7<x<17

Biswajit Barik - 4 years, 5 months ago

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Well, (i think) you are too lazy to read the second part of the solution. It is clearly mentioning the reason. Please do read the second :

Vishwash Kumar ΓΞΩ - 4 years, 5 months ago

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Well it has been too long to see someone on Brilliant from Mumbai. Nice to see you. ::)

Vishwash Kumar ΓΞΩ - 4 years, 5 months ago

Oh yeah that's what I am saying

Biswajit Barik - 4 years, 5 months ago

Thanks for feedback rohit I am a totally new person on brilliant and from three month I can't found a way to post my problems but after Calvin sir, S effort I post my first problem but I was very sour of it as the question is from a good book whose publication name is Pearson I also gave the question to many teachers but they did not give a satisfied solution and I am from Orissa and I live in Mumbai with my brother it is very nice to see such tender students solving such good problems those whom 10 class students found it difficult keep it up😀

Biswajit Barik - 4 years, 5 months ago

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THANKS FOR YOUR KIND WORDS

Vishwash Kumar ΓΞΩ - 4 years, 5 months ago

I will post some good problems from major maths Olympiad in near future and this will help you

Biswajit Barik - 4 years, 5 months ago

@Biswajit Barik Well I love olympiads and so I would love your that effort. I have seen pearson book that is An IIT FOUNDATION SERIESes.

Vishwash Kumar ΓΞΩ - 4 years, 5 months ago

The condition for obtuse angle is, cos of angle opposite the largest side must be <0. Cos Rule.
(1) If the third side is largest , it must be < sum of the other two sides=5+12=17. So possible lengths are,..13, 14, 15, 16 .
Third side 13 is a right angle triangle. But 14, 15, 16 have Cos(0pposite 13)<0..
(2) If 12 is the largest side, sum of the other two sides > 12. So possible lengths are 11, 10, 9, 8.
But for 11, Cos(0pposite 12)=(5^2+11^2-12^2) /' ...>0. But 10, 9, 8 have Cos(0pposite 12)<0..
So SIX triangles are possible.


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