Imagine it
Two congruent squares A B C D and P Q R S are positioned such that they share a common area defined by Δ P Q B . The ratio of the area of Δ P Q B to the area of polygon A Q R S P C D is 2 2 3 .
If the side length of both squares is s , then the perimeter of polygon A Q R S P C D is n m s , where m and n are coprime positive integers . Find m + n .
The answer is 33.
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1 solution
an easy one
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I missed it since I did not convert values in terms of s.
. [...] indicates area of ...
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I do not see a place for an alternate solution, so I will treat it as a comment Let QB = x, and BP = y, so the area of the triangle is (1/2)xy The area of the polygon is the area of the yellow section + area of the blue section = 2(s^2 - (1/2)xy) = 2s^2 -xy. The ratio of areas is 3/22 = (1/2)xy/(2s^2 - xy), , xy/(4s^2 - 2xy) = 3/22, !2s^2 - 6xy = 22xy, 3s^2 = 7xy. From the triangle, x^2 + y^2 = s^2 .Then 3x^2 + 3y^2 = 7xy, and 3x^2 + 6xy + 3y^2 = 13xy, so 3(x + y)^2 = 13xy, and x + y = sqrt((13/3)xy). The polygon perimeter is AQ + QR +RS +SP +PC +CD +DA = (s - x) +s + s + s + s - y + s + s = 7s -(x + y) = 7s- sqrt((13/3)xy) .But xy =3s^2/7. Substituting, the polygon perimeter is 7s -s*sqrt(13/7) = [7 - sqrt(13/7)]s. The difference is how we define the perimeter of the polygon. Ed Gray
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last word: 28/5 = 5.6, 7 - sqrt(13/7) = 5.637... Ed Gray
I don't believe that the perimeter of the polygon is basically the perimeter of the two squares minus the perimeter of triangle PQB. The line segment QP could never have contributed to the perimeters of the two polygons; therefore, I do not think the answer is correct. I used a bit more algebra and got an answer of s*sqrt(7 -sqrt(13/7)), But the algebra was nifty. Ed Gray
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I mis-worded my original dissent, but I think I am still correct. That line segment never entered into the perimeter of the defined polygon,so should not be subtracted out; my apologies Ed Gray
Relevant wiki: Length and Area - Composite Figures - Intermediate
Let G be the area of triangle P Q B . Given that its ratio to the total polygon area is 2 2 3 , we can solve for the area of Δ P Q B . By inspection, we will know that the total area of the two squares is 2 s 2 − G .
To find G ,
2 s 2 − G G = 2 2 3
which we get G = 2 5 6 s 2
Now, we know that P Q B is a right triangle with legs a and b both less than s and hypotenuse s . From here we know G = 2 a b . We can now derive these equations:
a 2 + b 2 = s 2
a b = 2 5 1 2 s 2
from which we can find ( a + b ) 2
( a + b ) 2 = a 2 + b 2 + 2 a b
= s 2 + 2 5 2 4 s 2
Giving us ( a + b ) = 5 7 s .
The perimeter of the polygon is basically the perimeter of the two squares minus the perimeter of Δ P Q B . That is
p = 8 s − ( a + b + s )
p = 7 s − ( a + b )
p = 7 s − 5 7 s
p = 5 2 8 s
And there we have it. 28+5= 33