True or false? Any triangle we can h_{A}^{2}+h_{B}^{2} geq 2h_{C}^{2}, where h_A, h_B, h_C are lengths of the heights from vertices A, B and C.

themobius6s

themobius6s

Answered question

2022-09-27

True or false? Any triangle we can h A 2 + h B 2 2 h C 2 , where h A , h B , h C are lengths of the heights from vertices A, B and C
Let A B C any triangle and the lengths of the heights from vertices A, B and C are, respectively, h A , h B , h C . Then how should we prove that, h A 2 + h B 2 2 h C 2 , h C 2 + h A 2 2 h B 2 a n d h B 2 + h C 2 2 h A 2

Answer & Explanation

Joel Reese

Joel Reese

Beginner2022-09-28Added 17 answers

Step 1
We may use following inequality:
h A + h b + h c < a + b + c = 2 p           ( 1 )
Also:
h A + h B + h C > p
Let's suppose:
h A + h B + h C p + p 2 = 3 p 2
Squaring both sides we get:
h A 2 + h B 2 = h C 2 + 9 4 p 2 3 p h C 2 h A h B
So we must show that:
t = 9 4 p 2 3 p h C 2 h A h B h C 2
Step 2
If h A = h B = ϵ then p h C and we have:
t = 9 4 × 4 h C 2 3 h C 2 ϵ = 6 h C 2 ϵ
So there is possibility to have a triangle in which:
h A 2 + h B 2 h C 2
But there is no guarantee for this relation to be symmetric. for example Consider Right angle triangle with sides 3, 4, 5 where:
h B = 4, h C = 3 and h A = 2.4 we have:
h A 2 + h B 2 = 2.4 2 + 4 2 = 21.76 > 2 × 3 2 = 18
h B 2 + h C 2 = 9 + 16 = 25 > 2 × 2.4 2 = 11.54
But: h A 2 + h C 2 = 14.76 < 2 × 4 2 = 32
jhenezhubby01ff

jhenezhubby01ff

Beginner2022-09-29Added 1 answers

Explanation:
Consider a very narrow pointy isosceles triangle, with one angle close to 0 and the others nearly 90 .

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