Cylinder, Cone and Sphere Problems: Problems with Solutions

Solution:
Let [tex]O_1[/tex] and [tex]O_2[/tex] be the centers of the spheres and let [tex]P[/tex] be any common point of the two spheres. Since [tex]PO_1=PO_2=5[/tex], the point [tex]P[/tex] is equidistant from [tex]O_1[/tex] and [tex]O_2[/tex], so it lies in the plane through the midpoint [tex]M[/tex] of [tex]O_1O_2[/tex] perpendicular to [tex]O_1O_2[/tex]. Hence the common circle lies in this plane.
The line [tex]O_1M[/tex] is perpendicular to this plane, so [tex]M[/tex] is the foot of the perpendicular from the center [tex]O_1[/tex] to the plane of the circle. Therefore [tex]M[/tex] is the center of the common circle and [tex]MP[/tex] is its radius.
[tex]O_1M=\frac{O_1O_2}{2}=\frac{8}{2}=4[/tex]
In the right triangle [tex]O_1MP[/tex] ([tex]\angle O_1MP=90^{\circ}[/tex]), by the Pythagorean theorem
[tex]MP=\sqrt{O_1P^2-O_1M^2}=\sqrt{5^2-4^2}=\sqrt{9}=3[/tex]
The radius of the circle is [tex]3[/tex] in.

Solution:
The radius of the base of the can and the radius of each ball are [tex]r=2[/tex] in. The five balls placed one on top of another have total height [tex]5\times 4=20[/tex] in, equal to the height of the can, so they fit exactly in it.
Volume of the can:
[tex]V=\pi r^2h=\pi\times 2^2\times 20=80\pi[/tex]
Volume of the five balls:
[tex]V_1=5\times \frac{4}{3}\pi r^3=5\times \frac{32\pi}{3}=\frac{160\pi}{3}[/tex]
The empty part has volume [tex]V-V_1[/tex], so its share of the volume of the can is
[tex]\frac{V-V_1}{V}=\frac{80\pi-\frac{160\pi}{3}}{80\pi}=\frac{\frac{80\pi}{3}}{80\pi}=\frac{1}{3}[/tex]
Hence [tex]\frac{1}{3}[/tex] of the volume of the can remains empty.

Solution:
Let [tex]K[/tex] and [tex]L[/tex] be the midpoints of [tex]AB[/tex] and [tex]CD[/tex]. In a rectangle the perpendicular bisector of [tex]AB[/tex] is also the perpendicular bisector of [tex]CD[/tex], so it is the line [tex]KL[/tex]. As the perpendicular bisector of the chord [tex]AB[/tex], it passes through the center [tex]O[/tex] of [tex]k[/tex].
The inscribed angle [tex]ABC[/tex] is right, so [tex]AC[/tex] is a diameter of [tex]k[/tex]:
[tex]AC=\sqrt{AB^2+BC^2}=\sqrt{8^2+6^2}=10\Rightarrow R=\frac{AC}{2}=5[/tex]
Rotating the circle about its diameter lying on [tex]KL[/tex] gives a ball of radius [tex]R=5[/tex]:
[tex]V_1=\frac{4}{3}\pi R^3=\frac{4}{3}\pi\times 5^3=\frac{500\pi}{3}[/tex]
Rotating the rectangle about [tex]KL[/tex] gives a cylinder with radius [tex]KB=\frac{AB}{2}=4[/tex] and height [tex]KL=BC=6[/tex]:
[tex]V_2=\pi\times KB^2\times KL=\pi\times 4^2\times 6=96\pi[/tex]
Therefore
[tex]\frac{V_1}{V_2}=\frac{500\pi}{3\times 96\pi}=\frac{125}{72}[/tex]

Solution:
Let [tex]O[/tex] and [tex]O_1[/tex] be the centers of the larger and the smaller base, and let [tex]AA_1[/tex] be a generatrix ([tex]A[/tex] on the larger base, [tex]A_1[/tex] on the smaller one). In the axial section the quadrilateral [tex]OAA_1O_1[/tex] is a right trapezoid with [tex]OA=R[/tex] and [tex]O_1A_1=r[/tex].
Drop the perpendicular [tex]A_1H[/tex] to [tex]OA[/tex]. Then [tex]A_1H=12[/tex] is the height and [tex]AH=R-r[/tex]; from the right triangle [tex]AHA_1[/tex] by the theorem of Pythagoras:
[tex]R-r=\sqrt{13^2-12^2}=5[/tex]
The lateral surface area of a truncated cone is [tex]S=\pi(R+r)l[/tex], so
[tex]\pi(R+r)\times 13=91\pi\Rightarrow R+r=7[/tex]
Solving the system:
[tex]\begin{cases}R+r=7\\R-r=5\end{cases}\Rightarrow R=6, \ r=1[/tex]
The radii are [tex]6[/tex] in and [tex]1[/tex] in.

Solution:
Let [tex]r[/tex] be the radius of the base and [tex]l[/tex] the generatrix. The development of the lateral surface is a circular sector with radius [tex]l[/tex] and central angle [tex]72^{\circ}[/tex], and the length of its arc equals the circumference of the base:
[tex]2\pi r=\frac{72^{\circ}}{360^{\circ}}\times 2\pi l=\frac{2\pi l}{5}\Rightarrow l=5r[/tex]
The total surface area is the sum of the base area and the lateral surface area:
[tex]S=\pi r^2+\pi rl=\pi r^2+5\pi r^2=6\pi r^2[/tex]
From [tex]6\pi r^2=6\pi[/tex] we get [tex]r^2=1[/tex], i.e. [tex]r=1[/tex] in.

Solution:
Let [tex]R[/tex] and [tex]r[/tex] be the radii of the larger and the smaller base of the truncated cone and [tex]h[/tex] its height. One base of the cylinder is the smaller base of the truncated cone and the other lies in the plane of the larger base, so the cylinder has radius [tex]r[/tex] and height [tex]h[/tex] (the distance between the two base planes).
The volumes of the cylinder and of the truncated cone are
[tex]V_1=\pi r^2h,\qquad V_2=\frac{1}{3}\pi h\left(R^2+Rr+r^2\right)[/tex]
From [tex]V_2=7V_1[/tex]:
[tex]\frac{1}{3}\pi h\left(R^2+Rr+r^2\right)=7\pi r^2h\Rightarrow R^2+Rr+r^2=21r^2\Rightarrow R^2+Rr-20r^2=0[/tex]
Dividing by [tex]r^2[/tex] and setting [tex]x=\frac{R}{r}[/tex], we get
[tex]x^2+x-20=0\Rightarrow x_1=4, \ x_2=-5[/tex]
Since [tex]x\gt 0[/tex], [tex]x=4[/tex]: the radius of the larger base is [tex]4[/tex] times the radius of the smaller base.

Solution:
Let [tex]O[/tex] be the center of the sphere and [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] the dimensions of the box. Two opposite faces lie in parallel planes tangent to the sphere, and the distance from [tex]O[/tex] to a tangent plane equals the radius [tex]2[/tex].
The sphere is inside the box, so [tex]O[/tex] lies between the two planes and the distance between them is [tex]2+2=4[/tex]. This holds for each of the three pairs of opposite faces:
[tex]a=b=c=4[/tex]
So the box is a cube whose edge equals the diameter of the sphere, and its volume is [tex]V=4^3=64[/tex] cubic inches.

Solution:
The axial section is an isosceles trapezoid [tex]ABCD[/tex] with bases [tex]AB=2R=60[/tex], [tex]CD=2r=28[/tex] and legs [tex]AD=BC=l[/tex] (the generatrices), where [tex]\angle ADC=120^{\circ}[/tex]. Since [tex]AB\parallel CD[/tex], we have [tex]\angle DAB=180^{\circ}-120^{\circ}=60^{\circ}[/tex].
Let [tex]DH\perp AB[/tex], [tex]H\in AB[/tex]. In the isosceles trapezoid [tex]AH=\frac{AB-CD}{2}=R-r=16[/tex], and from the right triangle [tex]AHD[/tex]
[tex]l=AD=\frac{AH}{\cos 60^{\circ}}=\frac{16}{\frac{1}{2}}=32[/tex]
The lateral surface area and the areas of the two bases are
[tex]\pi(R+r)l=\pi\times 44\times 32=1408\pi,\quad \pi R^2=900\pi,\quad \pi r^2=196\pi[/tex]
Hence the total surface area is
[tex]S=1408\pi+900\pi+196\pi=2504\pi[/tex]

Solution:
Let the right angle be at [tex]C[/tex], the shorter leg [tex]CA=2k[/tex], the longer leg [tex]CB=3k[/tex], and let the axis [tex]s[/tex] pass through [tex]B[/tex] parallel to [tex]CA[/tex]. Then [tex]CB\perp s[/tex], so [tex]C[/tex] and [tex]A[/tex] are at distance [tex]3k[/tex] from [tex]s[/tex]. Let [tex]B_1[/tex] be the projection of [tex]A[/tex] on [tex]s[/tex]; then [tex]BB_1=2k[/tex] and [tex]B_1A=3k[/tex].
In the rotation the leg [tex]CB[/tex] sweeps a disc with center [tex]B[/tex] and radius [tex]3k[/tex], the leg [tex]CA[/tex] sweeps the lateral surface of a cylinder with radius [tex]3k[/tex] and height [tex]2k[/tex], and the hypotenuse [tex]AB[/tex] sweeps the lateral surface of a cone with apex [tex]B[/tex], base radius [tex]B_1A=3k[/tex] and generatrix [tex]AB=\sqrt{4k^2+9k^2}=k\sqrt{13}[/tex]. So the solid is this cylinder with the cone removed:
[tex]V=\pi(3k)^2\times 2k-\frac{1}{3}\pi(3k)^2\times 2k=18\pi k^3-6\pi k^3=12\pi k^3=12\pi\Rightarrow k=1[/tex]
Hence [tex]CA=2[/tex], [tex]CB=3[/tex], [tex]AB=\sqrt{13}[/tex]. The surface consists of the disc, the lateral surface of the cylinder and the lateral surface of the removed cone (the other base of the cylinder is the base of the removed cone, so it is not part of the surface):
[tex]S=\pi\times 3^2+2\pi\times 3\times 2+\pi\times 3\times \sqrt{13}=9\pi+12\pi+3\sqrt{13}\pi=(21+3\sqrt{13})\pi[/tex]

Solution:
Let [tex]r[/tex] be the radius of the base and [tex]h[/tex] the height of the cylinder. The axial section is a rectangle with sides [tex]2r[/tex] (a diameter of the base) and [tex]h[/tex] (a generatrix); it is a square, so [tex]h=2r[/tex]. Then the lateral surface area is
[tex]S=2\pi rh=2\pi r\times 2r=4\pi r^2=12\Rightarrow \pi r^2=3[/tex]
So the area of one base is [tex]B=\pi r^2=3[/tex] (there is no need to find [tex]r[/tex] itself), and the total surface area is
[tex]S_1=S+2B=12+2\times 3=18[/tex]

Solution:
Cut the cylinder by planes parallel to the bases into [tex]4[/tex] equal cylinders of height [tex]\frac{12}{4}=3[/tex]. The thread is wound evenly and makes [tex]4[/tex] full turns, so on each small cylinder it makes exactly one full turn, and the four pieces of the thread are equal.
Cut the lateral surface of one small cylinder along the generatrix through the starting point of its piece of thread and unroll it into a plane. We get a rectangle with sides [tex]4[/tex] (the circumference of the base) and [tex]3[/tex] (the height). Since the thread rises uniformly as it goes around, its piece becomes a diagonal of this rectangle. By the Pythagorean theorem its length is
[tex]\sqrt{4^2+3^2}=5[/tex]
Therefore the length of the whole thread is
[tex]4\times 5=20[/tex]

Solution:
Let [tex]r[/tex] be the radius of the pit. Its diameter [tex]2r[/tex] equals the thickness of the flesh, so the radius of the cherry is [tex]R=r+2r=3r[/tex]. The ratio of the volumes of the pit and the cherry is
[tex]\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi (3r)^3}=\frac{1}{27}[/tex]
Since the pit and the flesh have the same density, the masses are proportional to the volumes, so the pit is [tex]\frac{1}{27}[/tex] of the mass of every cherry, whatever its size. Hence the pits in [tex]1[/tex] pound of cherries weigh [tex]\frac{1}{27}[/tex] pound and cost [tex]\frac{2.70}{27}=0.10[/tex] dollars, i.e. [tex]10[/tex] cents.

Solution:
Let the legs be [tex]a[/tex] and [tex]b[/tex] and the hypotenuse [tex]c[/tex]. By the theorem of Pythagoras
[tex]c^2=a^2+b^2[/tex]
A sphere with diameter [tex]d[/tex] has radius [tex]\frac{d}{2}[/tex], so its area is
[tex]4\pi\left(\frac{d}{2}\right)^2=\pi d^2[/tex]
Hence the three spheres have areas [tex]\pi a^2[/tex], [tex]\pi b^2[/tex] and [tex]\pi c^2[/tex]. Multiplying the equality of Pythagoras by [tex]\pi[/tex], we get
[tex]\pi c^2=\pi a^2+\pi b^2[/tex]
i.e. the area of the sphere with diameter the hypotenuse equals the sum of the areas of the spheres with diameters the legs.