TWO OF MY FORMULAS !!!!!!!!!!!!!!!!!!!

Hello, my dear brilliant members. Well, I wanted to give you some simple challenges. These are 2 of my new formulas which are (A)Finding the height of any triangle with sides a,b,c in terms of a, b and c. Obviously, there can be 3 heights but you find for base b. (B) Finding the length of median of any triangle of side a, b, c in terms of a, b and c and the median should be bisecting side b. Well, it is extremely easy and the challenge is that you find these formulas with proper proofs. Don't Google it as I am pretty sure you all can do it. Thank you and everyone comment on this, try and find the formulas yourself and post it in the comments.

Note by Kushagra Sahni
5 years, 3 months ago

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I have gotten both formulae.

(A) hb=2a2b2+2b2c2+2c2a2a4b4c42bh_b = \frac{\sqrt{2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}}}{2b}, where hbh_b denotes the altitude incident on bb

(B) xb=2a2+2c2b22x_b = \frac{\sqrt{2a^{2}+2c^{2}-b^{2}}}{2}, where xbx_b denotes the median incident on bb

Should I post a proof?

Nanayaranaraknas Vahdam - 5 years, 3 months ago

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YOUR MEDIAN ONE IS CORRECT, BUT THE ALTITUDE ONE CAN BE SIMPLIFIED FURTHER. SIMPLIFY IT USING SUITABLE IDENTITIES AND THEN POST A PROOF FOR BOTH

Kushagra Sahni - 5 years, 3 months ago

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AND YES, ALSO TELL ME THAT ARE THESE GOOD DISCOVERIES OR VERY EASY ONES ?

Kushagra Sahni - 5 years, 3 months ago

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COME ON FRIEND, REPLY. AND PLEASE RESHARE THIS NOTE

Kushagra Sahni - 5 years, 3 months ago

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Sorry, been caught up with IIT classes

Nanayaranaraknas Vahdam - 5 years, 3 months ago

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@Nanayaranaraknas Vahdam No problem friend, please reshare this note and simplify your altitude formula and post a proof for both.

Kushagra Sahni - 5 years, 3 months ago

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The altitude formula can be simplified to hb=(a2+b2+c2)24(a2b2+b2c2+c2a2)2bh_b=\frac{\sqrt{(a^{2}+b^{2}+c^{2})^{2}-4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}}{2b}

Nanayaranaraknas Vahdam - 5 years, 3 months ago

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Proofs: (A) Take ABC\triangle ABC with an altitude BDBD of length hh perpendicular to bb. Let DC=xDC=x, and thus, AD=bxAD=b-x. Now, by Pythagoras Theorem, a2x2=c2(bx)2a^{2}-x^{2}=c^{2}-(b-x)^{2} \Rightarrow a2+b2c2=2bxa^{2}+b^{2}-c^{2}=2bx \Rightarrow x=a2+b2c22bx=\frac{a^{2}+b^{2}-c^{2}}{2b}

Also, x2+h2=a2x^{2}+h^{2}=a^{2} \Rightarrow h2=a2+x2h^{2}=a^{2}+x^{2}

Substituting for xx into the first equation, h2=c2(ba2+b2+c22b)2h^{2}=c^{2}-(b-\frac{a^{2}+b^{2}+c^{2}}{2b})^{2}

By simplifying and bringing to the same denominator, we get h=2b2c2+2c2b2+2a2b2a4b4c44b2h=\sqrt{\frac{2b^2c^2 + 2c^2b^2 + 2a^2b^2-a^4-b^4-c^4}{4b^2}} \Rightarrow h=2b2c2+2c2a2+2a2b2a4b4c42bh=\frac{\sqrt{2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4}}{2b}, which can be simplified, as shown in my other comment.

Nanayaranaraknas Vahdam - 5 years, 3 months ago

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It still can be simplified. Apply the identity a^2-b^2

Kushagra Sahni - 5 years, 3 months ago

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And yes. Post a proof for median one as well

Kushagra Sahni - 5 years, 3 months ago

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u suck

ketan bansal - 4 years, 1 month ago

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