Cube in a cone

Take a cross section of the cone so that the intersecting plane is perpendicular to the base of the cone and contains a space diagonal of the cube. The resulting shape will be a triangle with a rectangle inscribed. The rectangle's dimensions are $height = e$ and $length = e\sqrt{2}$ , where e is the cube's edge length. (The length is $e\sqrt{2}$ because it is the face diagonal of the cube.) Dividing the triangle symmetrically in half, there is a right triangle with a rectangle inscribed. This rectangle has same height but length $\frac{e\sqrt{2}}{2}$ since it was cut in half. The large triangle has vertical leg of length $240\sqrt{2}$ (height of cone) and horizontal leg $120$ (cone radius). Now triangle similarity can be used to solve for $e$ , using the largest triangle and the topmost smaller one: $\frac{240\sqrt{2}}{120} = \frac{240\sqrt{2}-e}{\frac{e\sqrt{2}}{2}}$ . Simplifying, it becomes $2\sqrt{2} = \frac{480\sqrt{2}-2e}{e\sqrt{2}}$ and then $4e = 480\sqrt{2}-2e$ . Some more quick calculations shows $e = 80\sqrt{2}$ , so the answer is $80+2 =$ 82 .

Let the side length of the cube be $x$ .

Consider a cross-section of the cone with a plane going through $O$ , any other base vertex of the cube and the vertex of the cone. This is essentially a triangle with a parallelogram inscribed inside. This parallelogram's longer side is the cube face diagonal with length $x\sqrt{2}$ ; the shorter side is the cube edge with length $x$ . The triangle height is equal to $240\sqrt{2}$ and base length to $120\times 2 = 240$ .

Consider half of this triangle bounded by its height ( figure ). We can find the hypotenuse to be $360$ . As can be seen from the figure, finding $x$ is reduced to solving a system of equations with three variables: $\begin{cases}(240\sqrt{2} - x)^{2} + (x\sqrt{2}/2)^{2} = a^{2} \\ (120 - x\sqrt{2}/2)^2 + x^2 = b^2 \\ a + b = 360 \end{cases}$

Solving this system, we get $x = 80\sqrt{2}$ , so the answer is $82$ .

First of all, let us define some terms. I shall call $a$ the side length of the cube, whereas $d$ shall be the diagonal of any of the squares forming the cube.

Let us imagine a cone with a circular base of radius 120 centimeters, a base $B$ with a height of $240\sqrt{2}$ . Let that cone have a second base $B_{1}$ at a height $a$ , with a radius $\frac{d}{2}$ .

We also know that $d^{2}=2a^{2} \Rightarrow d=a\sqrt{2}$ , according to the Pythagorean Theorem. That means that the radius of $B_{1}$ equals $\frac{a\sqrt{2}}{2}$ .

In any cone, the ratio of the main base and any base at a certain height $H_{1}$ equals the ratio of the original height of the cone $H$ and the difference between between $H$ and $H_{1}$ ( $\frac{B}{B_{1}}=\frac{H^{2}}{(H-H_{1})^{2}}$ ).

In our case, $B=120^{2}\pi, B1=\frac{a^{2}}{2}\pi, H=240\sqrt{2}$ and $H_{1}=a$ . Therefore, we have

$\frac{120^{2}\pi}{\frac{a^{2}}{2}\pi}=\frac{(240\sqrt{2})^{2}}{(240\sqrt{2}-a)^{2}} \Rightarrow \frac{2\cdot 120^{2}}{a^{2}}=\frac{(240\sqrt{2})^{2}}{(240\sqrt{2}-a)^{2}}$ .

By rooting both sides of the equation, we get

$\frac{120\sqrt{2}}{a}=\frac{240\sqrt{2}}{240\sqrt{2}-a} \Rightarrow \frac{1}{a}=\frac{2}{240\sqrt{2}-a} \Rightarrow 240\sqrt{2}-a=2a \Rightarrow 240\sqrt{2}=3a \Rightarrow a=80\sqrt{2}$ .

Therefore, the desired sum is $82$ .

Consider the area above the cube. Using similar triangles, we may write $\dfrac{240 \sqrt{2} - s }{\dfrac{s \sqrt{2}}{2}} = \dfrac{ 240 \sqrt{2} } {120}$ .

Thus, we get $s = 80 \sqrt{2}$ and the answer is $80+2=82$ .

denote x as the sile length of the cube. the radius of the circular base above the cube is $\frac{x}{\sqrt{2}}$ use the properties of congruent triangles we have : $\frac{\frac{x}{\sqrt{2}}}{120}$ = $\frac{240\sqrt{2} - x}{240\sqrt{2}}$

solve this equation we get x = $80\sqrt{2}$

We look at the cone from the top (apex) and name the vertices of the square seen as $EFGH$ . To simplify this problem, we will take a cross section of the cone along FH (a diagonal of the visible square). Naming, $EF=x$ we have this diagram .

Lines EH and OD are parallel, so $\angle VHE$ and $\angle VDO$ are equal (corresponding angles). $\angle OVD$ and $\angle EVH$ are equal as well, leading to the conclusion that $\bigtriangleup VEH \sim \bigtriangleup VOD$ . As $FH=x\sqrt{2}$ and $EH=\frac{FH}{2}$ , $EH=\frac{x\sqrt{2}}{2}$ . We can now make the following proportion and solve for $x$ , which is the side length of the cube: $\frac{VE}{EH}=\frac{VO}{OD}$ $\frac{240\sqrt{2}-x}{\frac{x\sqrt{2}}{2}}=\frac{240\sqrt{2}}{120}$ $\frac{240\sqrt{2}-x}{\frac{x}{\sqrt{2}}}=2\sqrt{2}$ $({240\sqrt{2}-x})({\frac{\sqrt{2}}{x}})=2\sqrt{2}$ $\frac{480}{x}=3\sqrt{2}$ $\frac{160}{x}=\sqrt{2}$ $x=80\sqrt{2}$ The desired sum is then $\boxed{82}$ .

Let the length of the side of the cube be x.

Then the height of the cube is x, so the distance from the top of the cube to the top of the cone is (240*sqrt(2)-x).

The four vertices of the cube that touch the cone lie on a circle that is in a plane parallel to the plane containing the base of the cone; the diameter of this circle is the diagonal of the face of the cube, or x*sqrt(2).

If we now look at the two-dimensional figure we get when viewing the cone directly from the side, the entire cone and the portion of the cone above the cube appear as similar triangles. From the heights and bases of these two similar triangles, we get

(240 sqrt(2)-x)/(240 sqrt(2)) = (x*sqrt(2))/120

Solve this proportion to find the value of x.

So, x= 48*sqrt(2) and the ans is 48+2 = 50.