Tiggsy

Clear the fractions

xy = x + y .................(1)

xz = 2x + 2z .............(2)

yz = y + z .................(3)

Subtract eq(1) from eq(3),

yz - yx = z - x,

y(z - x)  - (z - x) = 0,

(z - x)(y - 1) = 0.

Implies that z = x, or y = 1.

If y = 1 then, in eq(1), x = x + 1, so reject that.

If z = x, then in eq(2), z^2 = 4z,

z(z - 4) = 0,

z = 0 or z = 4.

If z = 0 then x = y = 0, so reject that.

Finally, z = 4, x = 4, y = 4/3.

\(a^{3} + 3ab^{2}=679, \\ 3a^{2}b+b^{3}=615.\)

Subtract the second equation from the first,

\(a^{3}-3a^{2}b+3ab^{2}-b^{3}=64, \\ (a-b)^{3}=64, \\ a-b = 4.\)

The general term of the series can be taken to be 1/(n*(n + 3)) where n proceeds by 3 for each term, n = 1, 4, 7, 19, ... , 94, 97.

(It could also be taken to be 1/((3n - 2)(3n + 1)) where n proceeds in steps of 1 from 1 to 33, which was the choice of #2.)

Expanding 1/(n*(n + 3) ) into the partial fractions fractions 1/(3n) - 1/(3(n + 3)),

n = 1  : 1/(1*4)     = 1/3 - 1/12,

n = 4  : 1/(4*7)     = 1/12 - 1/21,

n = 7  : 1/(7*10)   = 1/21 - 1/30,

n = 10: 1/(10*13) = 1/30 - 1/39,

.............

.............

n = 94 : 1/(94*97)   = 1/282 - 1/291,

n = 97 : 1/(97*100) = 1/291 - 1/300.

Adding all of the equations,

1/(1*4) + 1/(4*7) +1/(7*10) + ... + 1/(97*100) = 1/3 - 1/300 = 100/300 - 1/300 = 99/300 = 33/100.

Suppose that the area of the triangle is A, then (half base times height)

A = AB*CF/2, so AB = 2*A/CF,

A = BC*AD/2, so BC = 2*A/AD = 2*A/12,

A = CA*/BE/2, so CA = 2*A/BE = 2*A/15.

For the triangle to exist,

AB + BC > CA, so 1/CF + 1/12 > 1/15,

BC + CA > AB, so 1/12 + 1/15 > 1/CF,

CA + AB > BC, so 1/15 + 1/CF > 1/12.

The third of these conditions leads to CF < 60 meaning that the largest integer value for CF is 59.

One of the other two conditions gives a minimum for CF.

Take a look at the answer above, #1.

There's actually a numerical mistake on the third line, but if you correct that it will get you to the result that you are looking for.

At the completed square version of the equation, if you substitute in the y co-ordinate of the centre you can calculate the extreme values for x, (the x co-ordinates of the vertices for that direction of the axes).

Similarly if you substitute in the x co-ordinate of the centre you can calculate the extreme values for y.

It turns out that the major axis is parallel with the y-axis so it's the distance between the extreme values of y that you need.

The equation will have either

1, no real roots, meaning that the line doesn't intersect with the parabola,

2, two real roots, meaning that the line intersects the parabola at two (different) points, or

3, two equal roots, meaning that the line contacts the parabola at a single point. In that case the line will be a tangent to the parabola.

Answered yesterday.

There's a mistake in the answer given above, the slope of the parabola should be 2ax + b, not 2a + b.

That leads to incorrect values for a and b.

Here's a solution not involving calculus.

To find out where the straight line and the parabola intersect equate the two expressions for y,

\(-11x-7=ax^{2}+bx+1,\\ \text{so} \\ ax^{2}+(b+11)x+8=0.\)

We need this to have equal roots at x = 2.

That gets you the two equations

\(-(b+11)/2a=2 \\ \text{and} \\ (b+11)^{2}-32a=0,\)

which can be solved simultaneously for a and b.

(a = 2, b = -19)

ABC is a right-angled triangle, the right angle being at C.

The altitude from C onto AB meets AB at H.

AH = x, HB = y, and CH = h.

\((b+1)(8b^{2}+1)=2023,\\ 8b^{3}+8b^{2}+b-2022=0,\\b=6,\\b+1=7,\\8b^{2}+1=289,\\(x-7)(x-289)=x^{2}-296x+2023. \)