Let P be an arbitrary point inside a plane triangle rABC, PE, PF, PG be perpendiculars dropped from P to the sides AC, BC, AB respectively, r₁, r₂, r₃ be the lengths of these perpendiculars, and p be the perimeter of  rEFG.
Prove that  (r₁ + r₂ )cos(C/2) + (r₁ + r₃ )cos(A/2) + (r₂ + r₃  )cos(B/2) £ p.

Solution

ÐC + ÐCEF + ÐEFC = p
ÐCEF + ÐFEP = p/2 and ÐEFC + ÐPFE = p/2
so ÐC + ÐCEF + ÐEFC = ÐCEF + ÐFEP + ÐEFC + ÐPFE so ÐC = ÐFEP + ÐPFE
so ÐC + ÐFPE = p  so ÐFPE = p - ÐC

cos(p - C ) = cos C
Law of Cosines:
(EF)² = r₁² + r₂² - 2r₁ r₂ cos(p - C) = r₁² - 2 r₁ r₂ + r₂² + 2 r₁ r₂ - 2r₁ r₂ cos C =
(r₁ - r₂)² + 2r₁ r₂ (1 + cos C)     
since cos²x = ½ + ½ cos 2x  (half-angle formula)
= (r₁ - r₂)² + 2r₁ r₂ (2 cos² (C/2)) = (r₁ - r₂)² + 4r₁ r₂ cos² (C/2)
³ (r₁ - r₂)² cos² (C/2) + 4 r₁ r₂ cos² (C/2)
      since cos²x £ 1
= (r₁ +  r₂)² cos² (C/2)
So, EF ³ (r₁ +  r₂) cos (C/2)

Use similar arguments for FG and EG and add them.

    We received solutions from Aaron Albert and Shawn Foster for Find K. Shawn's solution was correct while Aaron's was close but this is not horseshoes. So to all of us mathematical types it is fairly obvious that Shawn is the winner of the non-drawing for the pizza. Congratulations!

Find k such that f(x) = k|x - 2| is a probability density function on [a, b].

Solution

= k((-½ x² + 2x)_a² + (½ x² - 2x)₂^b =
k(-2 + 4 + ½ a² - 2a + ½ b² -2b - 2 + 4) =
4 + ½ a² + ½ b² - 2a - 2b = 1
k = 1/(4 + ½ a² + ½ b² - 2a - 2b)

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