A geometric inequality

Geometry Level 3

A B C ABC is a triangle with area 1 u 2 u^2 . Is M M the product of the sum of the height`s and the sum of the sides of A B C ABC ; its say, M = ( h 1 + h 2 + h 3 ) × ( L 1 + L 2 + L 3 ) M=(h_1+h_2+h_3)\times(L_1+L_2+L_3)

Is M > 12 M>12 ?

Yes No

Alejandro Castillo by Alejandro Castillo , 23, Mexico

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Given that h i L i 2 = 1 \frac{h_iL_i}{2}=1 then h i L i = 2 h_iL_i=2 . As M = ( h 1 + h 2 + h 3 ) × ( L 1 + L 2 + L 3 ) = h 1 L 1 + h 2 L 2 + h 3 L 3 ⏟ 2 + 2 + 2 = 6 + h 1 ( L 2 + L 3 ) + h 2 ( L 1 + L 3 ) + h 3 ( L 1 + L 2 ) M=(h_1+h_2+h_3)\times(L_1+L_2+L_3)=\underbrace { h_{ 1 }L_{ 1 }+h_{ 2 }L_{ 2 }+h_{ 3 }L_{ 3 } }_{ 2+2+2=6 }+h_1(L_2+L_3)+h_2(L_1+L_3)+h_3(L_1+L_2) . Then, for the triangular inequality we can ensure that L 1 + L 2 > L 3 , L 2 + L 3 > L 1 L_1+L_2>L_3, L_2+L_3>L_1 and L 1 + L 3 > L 2 L_1+L_3>L_2 therefore we can ensure that h 1 ( L 2 + L 3 ) + h 2 ( L 1 + L 3 ) + h 3 ( L 1 + L 2 ) > h 1 L 1 + h 2 L 2 + h 3 L 3 > 6 h_1(L_2+L_3)+h_2(L_1+L_3)+h_3(L_1+L_2)>h_1L_1+h_2L_2+h_3L_3>6 then M > 6 + 6 > 12 M>6+6>12

2 Helpful 0 Interesting 1 Brilliant 0 Confused

A nice solution to a nice question,+1! Did it the same way!:)

Rishabh Tiwari - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...