Question

HELP MEDAL!!!!!!!!!!! Find all values of A such that the curves y = Ax^2 + Ax + 1/24 and x = Ay^2 + Ay + 1/24 are tangent to each other.

Answer 1

@campbell_st @cwrw238 @.Sam. @Nurali @robo @jhonyy9 @SolomonZelman @shamil98 @eliassaab @karatechopper @Kainui @april115 @Isaiah.Feynman @tester97 @fifi @Gatorgirl @Mr.ClayLordMath @AccidentalAiChan @mathmale @Whiteboy1949 @ilyautumn @brando1996 @luka1998 @INeedHelpPlease? @waterpologoalie @ComeAlongP0nd @lawls @JessicaBBY @emmigrace222 @tatumrocks @chandowd @ChasingMaggie @help@helpineedit @ShadowLegendX @ChasingMaggie @chandowd @Awesome781 @ocelot40 @benjamin2123 @christopher_miller420 @drenl @akk @Maddy17793 @farabor @sunnyymarie @RanandoOnley @gummibea

Answer 2

Putnam 2007 question A1

Answer 3

@wolfe8 half-of-the-earth tag xDDDD

Answer 4

y = Ax^2 + Ax + 1/24
x = Ay^2 + Ay + 1/24

tangent to each means
x=y'
y=x'
first case
y'=x
2Ax+A=Ay^2+Ay+1/24
2Ax+A=A(Ax^2 + Ax + 1/24 )^2+A( Ax^2 + Ax + 1/24)+1/24 >>> eq1

x'=y
2Ay+A=Ax^2 + Ax + 1/24
2A(Ax^2 + Ax + 1/24 )+A=Ax^2 + Ax + 1/24>>>eq1

Answer 5

Curves are tangent to each other at some point implies they share the same tangent at that point. The slope of both curves must be the same at that point. Let that point be (p,q).

y = Ax^2 + Ax + 1/24
y' = 2Ax + A. At (p,q), y' = 2Ap + A = A(2p + 1)

x = Ay^2 + Ay + 1/24
1 = 2Ayy' + Ay' = Ay'(2y + 1)
y' = 1 / A(2y+1). At (p,q), y' = 1 / A(2q + 1)

Both slopes must be the same at (p,q). Therefore,
A(2p + 1) = 1 / A(2q + 1) or
A^2(2p+1)(2q+1) = 1 ----- (1)

The point (p,q) must also be on both curves and must satisfy both equations.
q = Ap^2 + Ap + 1/24 ------ (2)
p = Aq^2 + Aq + 1/24 ------ (3)

Three simultaneous equations and three unknowns.
Solve for p,q, and A.
(We are interested only in A)

A = 2/3, p = 1/4, q = 1/4
A = 3/2, p = -1/6, q = -1/6

Answer 6

There are two more A values: (13 - sqrt(601))/12 and (13 + sqrt(601))/12

Answer: A = 2/3, 3/2, (13 - sqrt(601))/12 and (13 + sqrt(601))/12