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For some positive integers p, there is a quadrilateral ABCD with positive integer side lengths, perimeter p, right angles at B and C, AB = 2, and CD = AD. How many different values of p < 2015 are possible?
(A) 30
(B) 31
(C) 61
(D) 62
(E) 63
Are You Up For the Challenge: 700 Level Questions
(A) 30
(B) 31
(C) 61
(D) 62
(E) 63
Are You Up For the Challenge: 700 Level Questions
29 Jun 2022, 19:31
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For some positive integers p, there is a quadrilateral ABCD with positive integer side lengths, perimeter p, right angles at B and C, AB = 2, and CD = AD. How many different values of p < 2015 are possible?
(A) 30
(B) 31
(C) 61
(D) 62
(E) 63
(A) 30
(B) 31
(C) 61
(D) 62
(E) 63
Either there is a better way, or there is a better way for you to use your time studying for the GMAT!!
Add a point Z along CD such that AZ is perpendicular to CD. Make AD = CD = x. Make AZ = BC =y. AB = CZ = 2. DZ = x-2.
\((x-2)^2+y^2=x^2\)
\(x^2-4x+4+y^2=x^2\)
\(y^2+4 = 4x\)
\(x=\frac{y^2+4}{4}\)
AB+BC+CD+AD < 2015
2+y+x+x < 2015
2x+y+2 < 2015
2x+y < 2013
Plugging in \(x=\frac{y^2+4}{4}\):
\(\frac{y^2+4}{2}+y<2013\)
\(y^2+4+2y<4026\)
\(y^2+2y+4<4026\)
\(y^2+2y+1<4023\)
\((y+1)^2<4023\)
\((y+1)<\sqrt{4023}\)
Since y is an integer, y+1 <= 63.
y <= 62
We know from the equation above in red that y has to be even, and y has to be positive. There are 31 possible values for y, which means 31 possible corresponding values for x and 31 corresponding values for p.
Answer choice B.
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