Graphing Quadratics and Inequalities: Star problems

Let the center be $(h,k)$.

Equal squared distances to $(0,0)$ and $(6,0)$ give $h^2=(h-6)^2$, so $h=3$.

Comparing $(0,0)$ with $(0,8)$ similarly gives $k=4$.

The radius is $\sqrt{3^2+4^2}=5$, so the circle is

Put $u=x-3$ and $v=y-4$.

Then $u^2+v^2=25$ and $x+y=7+u+v$.

Since $(u-v)^2\ge0$, we have

Hence

Equality in the square inequality requires $u=v$.

To attain the upper bound their common value must be positive:

This pair lies on the circle and yields the stated point and value.

The equality conditions also prove uniqueness.

Every nonvertical line through $(0,-1)$ has the form $y=mx-1$ for a real number $m$.

A point is on both graphs precisely when $x^2-4x+3=mx-1$, or

The coefficient of $x^2$ never vanishes, so the usual discriminant test applies for every slope.

There is exactly one distinct real solution if and only if

Thus $m+4=4$ or $m+4=-4$, giving $m=0$ or $m=-8$.

For $m=0$ the equation is $(x-2)^2=0$ and the common point is $(2,-1)$.

For $m=-8$ it is $(x+2)^2=0$ and the common point is $(-2,15)$.

Both points satisfy their original line and parabola equations.

Since every permitted line was represented by $m$, no other line can qualify.

The nonvertical condition matters: the vertical line $x=0$ would also have exactly one meeting point.

Write

The axis is $x=5/2$, so the two inputs $x$ and $5-x$ always give the same height.

No integer is fixed by this pairing.

There are only finitely many points below the axis because the square eventually becomes larger than any fixed number.

They therefore occur in pairs, proving part (b).

For exactly four qualifying integers, the nearest four integers to $5/2$ must qualify and all others must fail.

These are $1,2,3,4$.

The height at $1$ and $4$ is $k-4$, so they qualify exactly when $k<4$.

The next pair, $0$ and $5$, has height $k$, so it must satisfy $k\ge0$.

If $0\le k<4$, the inner pair $2,3$ has height $k-6<0$, while every integer outside $1,2,3,4$ is at least as far from the axis as $0$ or $5$ and has nonnegative height.

Thus the range is sufficient as well as necessary.

At $k=0$ the outer pair lies on the axis and does not count; at $k=4$ the pair $1,4$ no longer counts.

Let the upper vertices have horizontal coordinates $u<v$.

Their heights agree, so $12-u^2=12-v^2$, giving $u^2=v^2$.

Since $u\ne v$, we must have $u=-v$.

Write $v=t>0$.

Positive height requires

Thus every allowed rectangle, including those not initially described as centered, has width $2t$ and height $12-t^2$.

Its area is

At $t=2$ this is $32$.

To prove no other value is larger, calculate and factor the difference:

For the entire allowed interval $0<t<\sqrt{12}$, the factor $t+4$ is positive and the square is nonnegative.

Hence $A\le32$, with equality only when $t=2$.

The resulting height is $8$, yielding exactly the rectangle listed.

This proof optimizes a cubic expression using factoring; a graphing estimate alone would not establish the global maximum.

A perpendicular to $3x+4y=t$ through the origin has points $(3s,4s)$.

Its intersection with the line satisfies $25s=t$, so the foot is $M=(3t/25,4t/25)$ and

A perpendicular from a circle center bisects a chord: the two right triangles have the same hypotenuse and common perpendicular leg, so their other legs are equal.

The circle radius is $5$ and half the desired chord is $3$.

Pythagoras gives $OM^2+3^2=5^2$, hence $t^2/25=16$ and $t=\pm20$.

This condition was necessary for every such chord.

A vector of length $3$ parallel to the line is $(12/5,-9/5)$, because $(4,-3)$ has length $5$.

For $t=20$, the midpoint is $(12/5,16/5)$; adding and subtracting that vector gives $(24/5,7/5)$ and $(0,5)$.

Negating these points gives the endpoints for $t=-20$.

The listed points satisfy both original equations and have distance $6$, so both candidates work.

The vertical difference is

Its vertex has input $a+1$.

We need $D(x)\ge0$ throughout $[0,4]$, so the location of this vertex determines the relevant minimum.

If $a<-1$, the vertex lies left of the interval, and the square in $D(x)=(x-a-1)^2+2-a-a^2$ is smallest at $x=0$.

We need $D(0)=a+3\ge0$, giving

If $-1\le a\le3$, the vertex lies inside.

Its value is

This is nonnegative exactly for $-2\le a\le1$; intersecting with the current case gives

If $a>3$, the minimum is at $x=4$, where

There are no solutions in this case.

Combining the cases gives $[-3,1]$ and proves sufficiency.

Endpoint checks alone give $a\ge-3$ and $a\le11/7$.

For example, $a=3/2$ passes both checks, since $D(0)=9/2$ and $D(4)=1/2$, but has $D(5/2)=-7/4$, so it fails inside the window.

Substitute $y=x^2-2$ and set $u=x^2$.

The circle equation becomes $u+(u-2)^2=r^2$, or

If $r<\sqrt7/2$, the right side is negative, so there are no points.

At $r=\sqrt7/2$, the single value $u=3/2$ is positive and gives two points $(\pm\sqrt{3/2},-1/2)$.

For larger radii the two candidate values are

The larger is always positive.

The smaller is positive precisely when $r<2$, zero precisely when $r=2$, and negative when $r>2$.

Thus between the two thresholds, both values produce a pair of points: four in total.

At $r=2$, the values are $0$ and $3$.

They give $(0,-2)$ and $(\pm\sqrt3,1)$, so the total is three, not four.

For $r>2$ only the positive value $u_+$ is allowed, giving two points.

Every retained $u$ produces $x=\pm\sqrt u$ and $y=u-2$ satisfying both equations, so no extraneous or missing points remain.

First suppose the center is $(0,v)$.

A disk containing the arc must contain its vertex and endpoints, so its squared radius is at least both $v^2$ and $4+(4-v)^2$.

If $v\ge5/2$, the first is at least $25/4$.

If $v\le5/2$, the second is at least

Thus every center on the axis needs radius at least $5/2$, and equality can occur only at $v=5/2$.

At this candidate center, the squared distance to $(x,x^2)$, with $t=x^2\in[0,4]$, is

Hence the proposed disk really contains the entire arc.

Now allow a center $(u,v)$ and radius $R$.

For each $x$, both reflected arc points must be in the disk.

Averaging their squared distances gives

Therefore the axis-centered disk with squared radius $R^2-u^2$ also contains the whole arc.

By the lower bound already proved,

In particular $R\ge5/2$, and equality forces $u=0$ and then $v=5/2$.

This proves global optimality and uniqueness without assuming the center is symmetric.

On the circle, substitute $x^2=4-y^2$ into the vertical difference:

For $-2\le y\le2$, the factor $y+2$ is nonnegative.

If $c\le0$, the other factor is at least $1$.

If $0<c\le1/4$, then

Thus every $c\le1/4$ works.

At $c=1/4$, the difference is $(y+2)^2/4$, so equality occurs only at the bottom point $(0,-2)$.

If $c>1/4$, the number $y_*=2-1/c$ lies strictly between $-2$ and $2$.

Choose any height $y$ strictly between $-2$ and $y_*$.

Circle points at that height exist, and the first factor is positive while the second is negative.

Thus the proposed bound fails.

This proves sharpness, not merely that $1/4$ works.

For intersections, the bottom point always qualifies.

If $c=0$, the second factor is never zero.

If $c\ne0$, an additional height can only be $y_*$.

For $c<0$ it exceeds $2$; for $0<c<1/4$ it is below $-2$; at $c=1/4$ it repeats the bottom height.

For $c>1/4$ it is an interior height and gives exactly two further points, with

Hence the count changes from one to three immediately above the sharp value.