Solution of the problem #24 in AMC 10 A 2011

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Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedral?
(A) 1/12    (B) sqrt(2)/12    (C) sqrt(3)/12    (D) 1/6    (E) sqrt(2)/6

Solution:

Let’s assign the following numbers to the vertices of the unit cube:
numbers 1, 2, 3, 4 to the four vertices of the bottom face of the cube (clockwise) and numbers 5, 6, 7, 8 to the vertices of the top face of the cube.

Let's position the two tetrahedra that satisfy the conditions of this problem so that the first tetrahedron has vertices 4, 5, 7, 2 and the second  tetrahedron has the remaining four vertices 1, 3, 6, 8. The pyramid with vertices 4, 5, 7, 8 shares one face with the first tetrahedron (the triangle with vertices 4,5,7).

Obviously, the other three faces of the first tetrahedron have their adjacent pyramids, congruent to the pyramid with vertices 4, 5, 7, 8. If we remove all four adjacent pyramids from the cube, the only part left will be the first tetrahedron.

It makes calculation of the volume of each of the two tetrahedra simple: it is the volume of the cube minus four times volume of the pyramid with vertices 4, 5, 7, 8.
The volume of this pyramid is the area of its face with vertices 4, 5, 8 (S = 1*1* 1/2) multiplied by the length of its altitude (the edge between vertices 7 and 8 whose length is 1) and divided by 3. It is: S*1*1/3 = 1*1*1/2*1/3 = 1/6. The volume of the cube equals 1*1*1 = 1. So, the volume of the tetrahedron is: 1 – 4*1/6 = 1/3.

Let’s observe that the face of the first tetrahedron with vertices 4, 5, 7 cuts the three edges of the other tetrahedron in the middle of each edge between the vertices: (1,8), (6,8), (3,8). It is easy to see if we realize that the edges of the second tetrahedron are the other diagonals of the square faces of the cube! For example, two diagonals (1,8) and (4,5) of the square (1,5,8,4) are the edges of two different tetrahedra!

We can see from this observation that each tetrahedron cuts off four top pyramids from the other tetrahedron, and each top pyramid is similar to the entire tetrahedron with the ratio of their edges = 1/2. Therefore, the ratio of the volumes of each top pyramid to the entire tetrahedron is (1/2)^3 = 1/8.
So, each top pyramid of a tetrahedron cut by the other tetrahedron has volume equal to 1/8 of the volume of tetrahedron. Now, we need to subtract the volumes of four top pyramids from the volume of any one tetrahedron to calculate the volume of intersection of two tetrahedra. It is: 1/3 - 4*(1/3)*(1/8) = 1/6   (D)

Solution of the problem #24 in AMC 12 B 2011

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Let P(z) = z^8 + (4*sqrt(3) + 6)*z^4 - (4*sqrt(3) + 7). What is the minimum perimeter among all the 8-sided polygons in the complex plane whose vertices are precisely the zeros of P(z)?

(A) 4*sqrt(3) +  4   (B) 8*sqrt(2)   (C) 3*sqrt(2) + 3*sqrt(6)  

(D) 4*sqrt(2) + 4*sqrt(3)   (E) 4*sqrt(3) + 6

Solution:

This is a nice problem because it requires that you work with its data creatively.

If we connect the dots A,B,C,D,E,F,G,H,A in this order, we will get the concave 4-fold symmetrical 8-gon. Due to symmetry, all sides of this 8-gon are congruent. Let's prove that this 8-gon has the smallest perimeter among all other possible 8-gons that connect these eight dots. Each side of this 8-gon has length = sqrt(2). Other possible sides can have lengths of DF = sqrt(2) or EG = sqrt(3) + 1 > sqrt(2).
Q.E.D.

Solution of the problem #25 in AMC 12 A 2011

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Triangle ABC has angle BAC = 60°, angle CBA ≤ 90°, BC = 1, and AC ≥ AB. Let H, I, and O be the orthocenter, incenter, and circumcenter of triangle ABC, respectively. Assume that the area of the pentagon BCOIH is the maximum possible. What is angle CBA?
(A) 60° (B) 72° (C) 75° (D) 80° (E) 90°

Solution:

The range of possible positions of vertex B on the side AB of angle BAC is restricted by the requirements AC ≥ AB and angle CBA ≤ 90°.

In the extreme case when AC = AB, triangle ABC is equilateral and all three center points H, I and O are at the same point. The pentagon BCOIH is reduced to the triangle BCO in this case.

The other extreme position of vertex B is where angle CBA = 90°. In this case, vertex B is the orthocenter of the right triangle ABC and the circumcenter O is the midpoint of its hypotenuse AC. The pentagon BCOIH is reduced to the quadrilateral BCOI in this case.

So, when vertex B moves within its allowed range along the AB side of angle BAC, the following angles vary accordingly:

60° <= angle CBA <= 90°

and

30° ≥ angle OCA ≥ 0°.

Angle BOC = 120° since it's a central angle of the circumcircle of ABC whose inscribed angle BAC = 60°. Let O’ be the point of intersection of perpendicular bisector of BC with the minor arc BC. Since triangle BOO’ is isosceles and angle BOO’ = 120° / 2 = 60°, it’s actually an equilateral triangle. The same is true for triangle COO’. Thus, the radii of the circumcircles of triangles ABC and BCO  are congruent.

So, arc BO always has constant radius R and constant arc measurement 60°, regardless of the position of vertex B and the corresponding position of the circumcircle O.

Let's prove that both orthocenter H and incenter I also always lie on arc BO. For that, it suffices to prove that angles BHC and BIC have measurement 120° each, since angle BOC has measurement 120° and is subtended by the same chord BC from the same side.

Let's prolong line CH until it crosses side BA at point P and line BH until it crosses side AC at point Q. Since H is the orthocenter of triangle ABC, BQ and CP are altitudes of this triangle. Since angle BAC = 60°, angle ACP = 90° - 60° = 30° and angle CHQ = 60°, so that angle BHC = 180° - 60° = 120°.

Further, I is the incenter, so BI and CI are bisectors of angles B and C of triangle ABC. Since angle A = 60°, the sum of angles B and C is 120° and the sum of angles IBC and ICB is 120° / 2 = 60°, so that angle BIC = 180° - 60° = 120°. We have proved that both the orthocenter H and the incenter I always lie on arc BO.

Let BOMN be a quadrilateral where M and N are two arbitrary points in this order on the arc BO.

Based on the Maximum Area of a Cyclic Polygon Theorem (see its proof in this paper: 
https://advancinginmath-349b7.firebaseapp.com/Papers/CyclicPolygonsTheorem1.pdf), the area of the cyclic quadrilateral BOMN is the largest of all such quadrilaterals when points M and N divide the arc BO = 60° into three equal arcs 20° each (and, accordingly, divide angle BCO into three equal angles 10° each). 

Note that the area of the corresponding pentagon BCOMN is also maximum possible in this case, because the area of the isosceles triangle BCO is fixed (all measurements of its sides and angles are constant: BC = 1, angle BOC = 120° and both angles OBC and BCO are 30°).

If we slide vertex B along the side BA within its allowed range to the point where angle OCA = 10° (which is somewhere inside its range: 30° ≥ angle OCA ≥ 0°), then line CM bisects angle BCA. Since incenter I lies on this angle bisector and also on the arc OB, points I and M are located at the same point.

If angle OCA = 10°, then angle BCA = 40° and angle CBA = 180° - 60° - 40° = 80°. Then, angle CPB (where CP passes through point N) = 180° - 80° - 10° = 90°. Since orthocenter H lies on this altitude and also on the arc OB, points H and N are located at the same point.

The answer is: angle CBA = 80° (D)