Problem 10 Geometry
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Problem 10: Geometry
HMMT 2008 Guts #21
Problem: Let ABC be a triangle with AB = 5, BC = 4 and AC = 3. Let P and Q be squares inside ABC with disjoint interiors such that they both have one side lying on AB. Also, the two squares each have an edge lying on a common line perpendicular to AB, and P has one vertex on AC and Q has one vertex on BC. Determine the minimum value of the sum of the areas of the two squares.
C
A
P
Q B
Solution: Label A(0, 0) and B(5, 0) and let the intersection of P and AC be D and the intersection of Q and BC be E. Also let P have length p and Q have length q. The slope of line AC is tan ∠CAB = 34 . Then label D
3 4 p, p
. Then label E
7 4p
+ q, q .
The slope of line BC is tan ∠CBA = 34 . Write the equation of line BC as y = − 43 x + 15 4 . 3 7 15 Then q = − 4 4 p + q + 4 . Then 21p + 28q = 60. By the Cauchy-Schwarz inequality, 602 602 122 144 . (p2 + q 2 )(212 + 282 ) ≥ (21p + 28q)2 = 602 . So p2 + q 2 ≥ 2 = = = 21 + 282 352 72 49 Note: We have equality when
p 21
=
q 28 ,
or p =
36 35 , q
=
48 35 .
Solution was written by ch1n353ch3s54a1l and compiled from Art of Problem Solving Forums.
HMMT 2008 Guts #21
Problem: Let ABC be a triangle with AB = 5, BC = 4 and AC = 3. Let P and Q be squares inside ABC with disjoint interiors such that they both have one side lying on AB. Also, the two squares each have an edge lying on a common line perpendicular to AB, and P has one vertex on AC and Q has one vertex on BC. Determine the minimum value of the sum of the areas of the two squares.
C
A
P
Q B
Solution: Label A(0, 0) and B(5, 0) and let the intersection of P and AC be D and the intersection of Q and BC be E. Also let P have length p and Q have length q. The slope of line AC is tan ∠CAB = 34 . Then label D
3 4 p, p
. Then label E
7 4p
+ q, q .
The slope of line BC is tan ∠CBA = 34 . Write the equation of line BC as y = − 43 x + 15 4 . 3 7 15 Then q = − 4 4 p + q + 4 . Then 21p + 28q = 60. By the Cauchy-Schwarz inequality, 602 602 122 144 . (p2 + q 2 )(212 + 282 ) ≥ (21p + 28q)2 = 602 . So p2 + q 2 ≥ 2 = = = 21 + 282 352 72 49 Note: We have equality when
p 21
=
q 28 ,
or p =
36 35 , q
=
48 35 .
Solution was written by ch1n353ch3s54a1l and compiled from Art of Problem Solving Forums.
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