problemset1s

# problemset1s - MAT 137Y 2008-09 Winter Session Solutions to...

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Unformatted text preview: MAT 137Y 2008-09 Winter Session, Solutions to Problem Set 1 1 (i) (SHE 1.3 #58) We need to show that a ≤ √ ab , √ ab ≤ a + b 2 , and a + b 2 ≤ b . Since 0 ≤ a ≤ b , we have a 2 ≤ ab , or a ≤ √ ab , proving the first statement. Also, a ≤ b = ⇒ a + b ≤ 2 b = ⇒ a + b 2 ≤ b , proving the third statement. Finally, note that ( a- b ) 2 ≥ = ⇒ a 2- 2 ab + b 2 ≥ = ⇒ a 2 + 2 ab + b 2 ≥ 4 ab = ⇒ ( a + b ) 2 ≥ 4 ab = ⇒ a + b ≥ 2 √ ab = ⇒ a + b 2 ≥ √ ab , which proves the second statement. Therefore a ≤ √ ab ≤ a + b 2 ≤ b . (ii) (SHE 1.4 #62) The three midpoints of the triangle are ( c 2 , ) , ( a + c 2 , b 2 ) , and ( a 2 , b 2 ) . The equations of the three lines are thus y = 2 b 2 a- c x- c 2 , y = b a + c x , y = b a- 2 c ( x- c ) . Equating the last two equations, we get y = b a + c x = b a- 2 c ( x- c ) = ⇒ b a + c- b a- 2 c x = bc 2 c- a = ⇒ x = bc 2 c- a 1 b a + c- b a- 2 c = ⇒ x = bc 2 c- a ( a + c )( a- 2 c ) ab- 2 bc- ab- bc = a + c...
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problemset1s - MAT 137Y 2008-09 Winter Session Solutions to...

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