If ( x , y , z ) be the lengths of perpendiculars from any interior point P of a tri

migongoniwt

migongoniwt

Answered question

2022-06-15

If ( x , y , z) be the lengths of perpendiculars from any interior point P of a triangle A B C on sides B C, C A and A B respectively then find the minimum value of :
x 2 + y 2 + z 2
The sides of triangle being a , b , c.

I thought of using Lagrange's method of multipliers but am not able to find another function in terms of x , y , z and a , b , cAny help will be appreciated. Thanks.

Answer & Explanation

Esteban Johnson

Esteban Johnson

Beginner2022-06-16Added 15 answers

Here is a solution without the method of Lagrange multipliers. Note that a x + b y + c z = 2 S where S is the area of the triangle. Therefore, by Cauchy-Schwarz inequality we have
4 S 2 a 2 + b 2 + c 2 = ( a x + b y + c z ) 2 a 2 + b 2 + c 2 x 2 + y 2 + z 2 .
The equality holds when x : y : z = a : b : c. The corresponding point is the radical center of the three Apollonian circles associated with the triangle.
Quintin Stafford

Quintin Stafford

Beginner2022-06-17Added 4 answers

As in x 2 + y 2 + z 2 all are squares , sum of squares is minimum when all terms (for which squares are taken) are equal. So the point should be the orthocentre of triangle with equal altitudes . if altitude be 'h'. distance from orthocentre to base is h/2 , then minimum is 3 ( h 2 ) / 4. height can be expresses in terms of area which is in terms of sides . so a relation is obtained as you wanted .

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