# 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
6 years, 2 months ago

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I have gotten both formulae.

(A) $h_b = \frac{\sqrt{2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}}}{2b}$, where $h_b$ denotes the altitude incident on $b$

(B) $x_b = \frac{\sqrt{2a^{2}+2c^{2}-b^{2}}}{2}$, where $x_b$ denotes the median incident on $b$

Should I post a proof?

- 6 years, 2 months ago

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

- 6 years, 2 months ago

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

- 6 years, 2 months ago

- 6 years, 2 months ago

Sorry, been caught up with IIT classes

- 6 years, 2 months ago

- 6 years, 2 months ago

The altitude formula can be simplified to $h_b=\frac{\sqrt{(a^{2}+b^{2}+c^{2})^{2}-4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}}{2b}$

- 6 years, 1 month ago

Proofs: (A) Take $\triangle ABC$ with an altitude $BD$ of length $h$ perpendicular to $b$. Let $DC=x$, and thus, $AD=b-x$. Now, by Pythagoras Theorem, $a^{2}-x^{2}=c^{2}-(b-x)^{2}$ $\Rightarrow$ $a^{2}+b^{2}-c^{2}=2bx$ $\Rightarrow$ $x=\frac{a^{2}+b^{2}-c^{2}}{2b}$

Also, $x^{2}+h^{2}=a^{2}$ $\Rightarrow$ $h^{2}=a^{2}+x^{2}$

Substituting for $x$ into the first equation, $h^{2}=c^{2}-(b-\frac{a^{2}+b^{2}+c^{2}}{2b})^{2}$

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

- 6 years, 1 month ago

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

- 6 years, 1 month ago

And yes. Post a proof for median one as well

- 6 years, 1 month ago

u suck

- 5 years ago