Geometry Basics: Star problems

Every angle of equilateral triangle $ABP$ is $60^\circ$.

The square has a right angle at $A$, and $P$ is inside it, so angle $DAP$ is

All square sides have length $AB$, while $AP=AB$.

Thus $AD=AP$, making triangle $ADP$ isosceles.

Its two base angles, $ADP$ and $APD$, are each

At $D$, segment $DP$ splits the square's right angle.

Therefore angle $CDP$ is

Similarly, angle $PBC$ is $30^\circ$, and $BP=BC$.

Triangle $BPC$ therefore has angle $BCP=75^\circ$, leaving angle

The angles of triangle $DPC$ sum to $180^\circ$, so angle $DPC$ is

Each deduction follows from equal sides and angle sums, so the argument does not depend on the appearance of the drawing.

The axes split the quadrilateral into four triangles, each with perpendicular base and height 4 and 3.

Each triangle has area $\frac{1}{2}\times4\times3=6$, so the total area is $4\times6=24$ square units.

For an interior integer-coordinate point, its vertical coordinate can only be $-2,-1,0,1,$ or $2$.

The rows at heights $3$ and $-3$ contain only boundary vertices, and all other integer-height rows are outside.

At height 0, the horizontal coordinate is strictly between $-4$ and $4$, giving 7 points.

Moving from the top vertex down toward the horizontal axis, the width grows in direct proportion to the vertical distance traveled.

At height 1, the right and left boundaries are $8/3$ and $-8/3$, so the possible horizontal coordinates are $-2,-1,0,1,2$: 5 points.

At height 2, the boundaries are $4/3$ and $-4/3$, giving 3 points.

Reflection across the horizontal axis gives another 5 and 3 points at heights $-1$ and $-2$.

Thus the count is

The strict inequalities exclude all boundary points.

Removing two corner lengths from each side leaves a square base of side $12-2x$ cm.

Folding raises the walls to height $x$ cm, so the volume is $x(12-2x)^2$ cubic centimeters.

The cut must satisfy $x>0$ and $12-2x>0$.

Because $x$ is a whole number, the only possibilities are 1, 2, 3, 4, and 5.

A cut of 6 would leave no base, so it is not an allowed box.

For $x=1$, the volume is

For $x=2$, it is

For $x=3$, it is

The remaining choices give $4\times4\times4=64$ and $5\times2\times2=20$ cubic centimeters.

The largest value is 128, achieved only by $x=2$.

This is a proof of the maximum because the list includes every permitted cut size.

Increasing the wall height helps volume, but shrinking the base works in the opposite direction; the two effects must be considered together.

For part (b), folding does not change the area of the remaining cardboard.

The base and four walls together have area $144-4x^2$ square centimeters.

As the positive cut size increases, more cardboard is removed, so this area is largest at $x=1$, giving $144-4=140$ square centimeters.

Volume depends on both base size and height, while surface area here measures remaining cardboard; these are different objectives and have different optimal cuts.

At one corner, the cut removes two original boundary segments of length 1 cm.

It exposes two new segments, also of length 1 cm.

The total boundary length does not change.

This works at all four corners, so the remaining perimeter is $2a+2b$ cm.

The four removed squares have total area 4 square centimeters.

The stated numerical equality is therefore $ab-4=2a+2b$, or

Adding 4 to both sides gives

This rearrangement turns the geometric condition into a factor problem.

Both factors are positive whole numbers because $a$ and $b$ are at least 3.

Ignoring order, the only positive factor pairs of 8 are 1 and 8, or 2 and 4.

Adding 2 to each factor gives side pairs $(3,10)$ and $(4,6)$.

Both work: their remaining areas are $30-4=26$ and $24-4=20$, while their boundary lengths are $2(3+10)=26$ and $2(4+6)=20$.

The exhaustive factor list proves there are no other rectangles.

The initial water volume is $10\times10\times4=400$ cubic centimeters.

Up to the cube's top, the available horizontal area for water is the tank's base area minus the cube's footprint: $100-36=64$ square centimeters.

Water reaching exactly to the cube's 6 cm top would therefore occupy $64\times6=384$ cubic centimeters.

Since 400 is greater than 384, the cube must be fully covered.

There are $400-384=16$ cubic centimeters of water above its top.

Above the cube, that excess spreads across the full 100-square-centimeter base area.

Its depth is $16/100=4/25$ cm.

The final depth is therefore $6+4/25=154/25$ cm.

Notice that using the 64-square-centimeter water area at every height would give the wrong answer once the water rises above the cube.

For part (b), the needed water volume is exactly 384 cubic centimeters.

Before the cube is inserted, that water covers the entire tank base, giving initial depth $384/100=96/25$ cm.

Let $f$ be the fraction of a full turn that remains after the cut.

The restrictions on the angle guarantee $0<f<1$.

The remaining area is $f\pi(R^2-r^2)$ square centimeters, because it is the same fraction $f$ of the outer disk's area minus the inner disk's area.

The combined curved length is $f(2\pi R+2\pi r)=2f\pi(R+r)$ cm.

Equating the numerical values and writing the difference of squares as a product gives

The common factor $f\pi(R+r)$ is positive, so division by it is valid.

We obtain $R-r=2$.

Conversely, if the width is 2 cm, substituting $R-r=2$ makes the two expressions identical for every $f$ between 0 and 1.

Thus the width condition is both necessary and sufficient.

The conclusion concerns numerical values measured in the specified units; it does not identify area and length as the same kind of measurement.

On each large face, the cubes with exactly one painted face form an interior square of side $n-2$.

There are therefore $6(n-2)^2$ such cubes.

This count is always divisible by 3, including the value 0 when $n=2$.

A cube with exactly two painted faces lies on a large edge but not at a corner.

The large cube has 12 edges, each contributing $n-2$ such cubes, for a total of $12(n-2)$.

Exactly three painted faces occur on the 8 corner cubes.

These categories are disjoint and include every painted cube.

Consequently, the number with two or three painted faces is $12(n-2)+8$.

When divided by 3, this number leaves remainder 2: the first term is divisible by 3, while 8 leaves remainder 2.

If exactly half of the painted cubes had one painted face, their number would equal the number in the other two categories.

A multiple of 3 cannot equal a number leaving remainder 2 on division by 3.

This contradiction rules out every allowed $n$; no trial of large cube sizes is needed.

The right-hand rectangles share their width, so the ratio of bottom height to top height is $45/30=3/2$.

This same ratio holds in the left column.

Its bottom area is therefore $18\times3/2=27$ square centimeters.

The total area is 120 square centimeters.

Let the right column's width be $w$ cm.

Because the row heights are whole numbers, $w$ must be a positive divisor of both 30 and 45.

Its only possibilities are 1, 3, 5, and 15.

The top height is $30/w$.

Dividing the top-left area by that height gives left width

This is a whole number only for $w=5$ or $w=15$ among the listed choices.

If $w=5$, the left width is 3 and the row heights are 6 and 9.

The whole rectangle is 8 by 15, with perimeter 46 cm.

If $w=15$, the left width is 9 and the heights are 2 and 3.

The whole rectangle is 24 by 5, with perimeter 58 cm.

Both constructions realize every given area and integer requirement.

Since all common divisors were checked, the list is complete, and 46 cm is the minimum.

Group the unit squares into a 7 by 9 array of 2 cm by 2 cm blocks.

Color the blocks in an alternating dark-and-light pattern, like a checkerboard, starting with a dark corner.

Each block contains four unit squares of its color.

There are 63 blocks.

Because both 7 and 9 are odd, the alternating pattern has 32 dark blocks and 31 light blocks.

One way to see this is that successive rows have 5 and 4 dark blocks: four rows contribute 5 each and three rows contribute 4 each.

The board therefore has 128 dark unit squares and 124 light unit squares.

Along any fixed row, colors repeat in runs of two: dark, dark, light, light, and so on, possibly starting at another position in the pattern.

Any four consecutive squares contain two of each color.

The same is true along each column.

Every allowed tile therefore covers exactly two dark and two light squares, even if its ends do not align with the 2 cm blocks.

Any collection of these tiles covers equal numbers of dark and light squares.

The board has unequal numbers, so no exact covering exists.

Although the area would require 63 tiles, the coloring reveals an additional obstruction.