Math Competitions Problem Solutions

July, 2019
Math in Music 2 video: 
https://youtu.be/9HRARNpyhxo

Edited "Math in Music" article:
http://david.balmin.com/mathinmusic.pdf

June, 2019
SD Honors Math 2018, Part I, 24
https://youtu.be/mM3FjGsmJh4

Problem 6 of 2007 IMO:

https://youtu.be/Y1EE_xRG6HA

SD Honors Math  2018, Part I, 19

https://www.youtube.com/watch?v=kx0q0Vvclro

IMO 2017 Problem 3 (revised)

https://youtu.be/R2xa82wqzTE

July - August, 2018
I  have posted the solutions of problems 3 and 6 of 2018 IMO in YouTube:

https://youtu.be/vyGh3yx0Hs4 

https://youtu.be/3X90CrNeiNE 

https://youtu.be/V36-OD_CoMU 

June 2, 2018
I  have posted the solution of USAMO 2018, Problem 3. It's an interesting fact and the solution is pretty! 
https://youtu.be/45SidE3dVTU 

December 12, 2017
I  have published an article "Solving Angles in a Parallelogram"

https://advancinginmath-349b7.firebaseapp.com/Papers/AnglesInAParallelogram.pdf

September 26, 2017
I  have posted the solutions of all 6 problems from 2017 International Math Olympiad (IMO) in YouTube.

https://advancinginmath-349b7.firebaseapp.com/IMOvideos/Links.htm

August 1, 2017

I posted Math in Music video in YouTube: 

https://youtu.be/DXy50lT14aM

June 23, 2017
Сompleted two weeks of All-Star Academy Math Competition Camp in San Diego. The vast majority of students have graduated with flying colors! Outstanding job guys! Keep up the good work. 

February 1, 2017
I gave one lecture to the combined Gauss & Cauchy group at San Diego Math Circle. The topic was: "Preparation for AMC-10 and AMC-12 tests".

December 7, 2016
I gave two lectures at San Diego Math Circle in October and November 2016. The topic was: "Preparation for the AMC-8 test".

December 7, 2016
I was coaching two Math Club groups of students: at Oak Valley Middle School and at the San Diego Russian School in September and October 2016. 

May 16, 2016
Completed this school year at the Russian School on the campus of UCSD university in San Diego.

September 26, 2015
I am working as a coach of the Math Club team at Oak Valley Middle School in San Diego. This group of motivated students have ambitious goals to succeed at the various math competition tests.

December 17, 2015
I edited my paper "The Maximum Area of a Cyclic Polygon Theorem": 
http://advancinginmath.appspot.com/Papers/CyclicPolygonsTheorem1.pdf

and the solution of Problem 25 of 2011 AMC 12A (at the end of this blog).

August 3, 2015
MathLinks Math Summer camp was a success! I really enjoyed working with two groups of extremely talented students. Good luck to them in their pursuing brilliant future in math, sciences, engineering and whatever other areas of their interests might be!

May 17, 2015
I had the final math lesson of this school year at the Russian School on the campus of UCSD university in San Diego. All the students enjoyed the challenging and interesting math problems and learned a thing or two about math theory too.

January 19, 2015

I teach advanced math on Sundays at the Russian-language school on the campus of UCSD university in San Diego. I have already given three lessons there to the group of enthusiastic and talented students. These meetings are going in the creative and informal atmosphere. So far, we have discussed  mostly counting / combinatorial topics and problems. At our next meeting, we will start tackling some geometry.

November 14, 2014
I met with the students of Saint Augustine High School Math Circle who plan to participate in advanced Math Competitions later this year.

November 1, 2014
I delivered two lectures to students of San Diego Math Circle:
Euler group (7th and 8th grades) and Fermat group (5th and 6th grades).
Subject: AMC 8 test preparation. Students of both groups participated very actively in the process of math problem solving and in the review of math theory used in AMC 8 tests. I had a lot of positive feedback from students and their parents after the lectures.

August 3, 2014
I was teaching two groups of students at the MathLinks Summer Camp starting on July 15 and ending on August 1 (during the first and third weeks - middle-school level, during the second week - high-school level) . Most students in both groups were quite advanced in school mathematics and highly motivated. All students were eager to learn more advanced math and to solve challenging math problems. They learned a lot about math and math problem-solving methods during those three weeks of camp (and had good time too!). 

June 7, 2014
I published video lesson with Introduction to Riemann's Hypothesis for school students.
http://www.youtube.com/watch?v=6qqWWshQQmA

June 7, 2014
I developed Android phone application: Riemann's Zeta-Function Calculator.
https://www.youtube.com/watch?v=dTwlUZ67PSo

October 22, 2013
Simplified the solution of Problem #4 in my Math E-Books Volume 1. 
These books are available for downloading at: http://advancinginmath.appspot.com/

October 20, 2013
My meetings with Saint Augustine High School Math Circle group in San Diego go very well!

July 28, 2013
I was teaching the group of 14 students (grades 7 - 9) at the MathLinks Summer Camp between July 8 and July 26, 2013. It was a very strong group of young people who were eager to learn advanced math and solve challenging math problems. They learned a lot about math and math problem-solving methods during those three weeks of camp (and had good time too!). 

May 23, 2013
I had the first introductory meeting with Saint Augustine High School Math Circle group in San Diego. 10 students attended this meeting. It was a great experience. All students actively participated in solving two math problems and discussions of math history and theory.

January 19, 2013
I delivered two lectures to the 10th-graders (the Gauss group) and the 12th-graders (the Cauchy group) of San Diego Math Circle at the UCSD. Subjects: "AMC 10 and 12 Test Preparation".


December 15, 2012
I gave a lecture to the 12th-graders (the Cauchy group) at the San Diego Math Circle. The topic was "Some Proofs of Math Olympiad Problems" with the emphasis on using Calculus for the strict proofs of the solutions in several Math Olympiad and AMC problems. It was a very advanced group of students. Some of them had participated in Math Olympiads and they actively participated in the discussions during my lecture.

November 8, 2012
I delivered two lectures to students of San Diego Math Circle:
Euler group (7th and 8th grades) and Fermat group (5th and 6th grades).
Subject: AMC 8 test preparation.
Both lectures had subject "AMC 8 test preparation". 

September 12 2012
Posted Youtube videos with solutions of AIME II Problems #15 and #7 :
http://www.youtube.com/watch?v=uWADhXq8n7U
http://youtu.be/WRP--6Db1qw

July 26, 2012
Yesterday was the closing day of MathLinks Summer Camp (it lasted 3 weeks). I had 11 students in my group, grades 7 - 9. It was both enjoyable and productive. The students in my group have learned the great deal about math and math problem solving during those three weeks.

Watch my videos (search for "David Balmin Math" in YouTube).

Yes, you can!

I provide step-by-step instruction for solving math problems from the annual AMC 8/10/12 (American Mathematics Competitions, grades 8, 10, 12) and AIME (American Invitational Mathematics Examination) tests.
Studying for and taking AMC and AIME tests significantly improve students' math skills. Many top universities request AMC scores as part of the enrollment application process. In order to be successful in these tests, students need solid experience in solving math problems that have level of difficulty expected in these tests.
Generally, we will focus on problem solving and will cover any topic of school math that is necessary to solve a problem. As a result, the student's knowledge and understanding of school math and problem solving skills will be improved!

Download my e-books from the "Training for Math Competition Tests" series here: 
http://advancinginmath.appspot.com/

Each book contains detailed step-by-step solutions of several math competitions problems that substitutes for 1-3 private lessons, depending on the student’s level and pace of studying.

In these books, I use the same methodology based on the principles of math problem solving that I use in private training sessions. These principles are outlined and explained in each volume of this series.

Read other posts below in this blog that illustrate the process of solving actual problems from the selected AMC tests.

Keep in mind that these problems were among the most complex problems in AMC 10/12 tests and may look intimidating at first glance, but, after a detailed step-by-step review, these and other AMC/AIME problems' solutions will become clear and straightforward to you.

Note: In the actual training sessions and in the YouTube video lessons, I use normal mathematical notation (not like "z^8" or "sqrt(3)")!

Click the link below to make donations via PayPal  to support this work on helping school students to perfect their skills in math problem solving:
http://advancinginmath.appspot.com/donations/index.htm

Email to dbalmin@gmail.com

Sincerely,
David Balmin

Solution of the problem #22 in AMC 12 A 2012

AMC (American Math Competitions) Test Problem #22
This is the simplified formulation of this problem.

A straight line segment connects every two midpoints of the edges within each face of a cube. A collection of distinct planes is such that the intersection of the union of all these planes with the surface of the cube consists of all such segments. What is the difference between the maximum and the minimum numbers of planes that such collection can have?

(A)   8   (B) 12   (C) 20   (D) 23   (E) 24

Solution

This diagram shows 3 of the total 6 faces of the cube. Altogether, there are 6 x 6 = 36 segments described in this problem. Let’s call the segments that are parallel to the edges “edge-parallel” and the remaining segments - “diagonal”. So, there are 2 x 6 = 12 “edge-parallel” segments and 4 x 6 = 24 “diagonal” segments.

Our goal is to identify the collection of the minimum number of distinct planes that satisfy the conditions of this problem (we will call it “the minimal collection of planes”) and the collection of all possible distinct planes, such that intersection of each plane with the surface of the cube consists of several such line segments (we will call it “the maximal collection of planes”).

The diagram shows midpoint A of one of the edges of the cube. Let's identify all the planes containing all possible pairs of segments that emanate from point A and lie in two perpendicular faces of the cube. In this diagram, numbers 1,2,3 are assigned to 3 such segments in each of these two faces of the cube.

We can identify the following types of planes by placing 3 x 3 = 9 pairs of segments in four groups, so that all the symmetrical pairs are in one group. (Note: considering the symmetry of all edges of the cube, it is clear that the midpoint of any other selected edge belongs to the same types of planes as point A.)

Case I:      Segments 2-2
Case II:     Segments 1-1 and 3-3
Case III:    Segments 1-2, 2-1, 3-2, 2-3
Case IV:    Segments 1-3 and 3-1

Now, let's calculate the numbers of planes in each group.

Case I. This type of plane divides the cube into two congruent rectangular parallelepipeds (shown in the diagram below) that intersects with the surface of the cube at 4 “edge-parallel” segments.

It is clear that there are 3 distinct, pairwise perpendicular, planes of this type: two vertical planes and one horizontal plane. These 3 planes cover 3 different sets of 4 “edge-parallel” segments. So, altogether, they cover all 12 “edge-parallel” segments.

Case II. This type of plane cuts off a tetrahedron whose one vertex is also a vertex of the cube (as shown in the diagram below).

It satisfies the conditions of the problem because its intersection with the surface of the cube consists of 3 “diagonal” segments adjacent to one of the vertices of the cube. Since there are 8 vertices of the cube, there are 8 such distinct planes that cover 8 different sets of 3 “diagonal” segments. So, altogether, they cover all 24 “diagonal” segments.

Case III. This type of plane cuts off a triangular prism whose one edge is also one full edge of the cube (as shown in the diagram below).

It satisfies the conditions of the problem because its intersection with the surface of the cube consists of 2 “diagonal” segments and 2 "edge-parallel" segments. Since there are 12 edges of the cube, there are 12 such distinct planes that, altogether, cover all 24 “diagonal” segments and all 12 "edge-parallel" segments (each "edge-parallel" segment is shared by two such planes).

Case IV.  This type of plane also satisfies the conditions of this problem, but it's more difficult to see than for the first three types. This is probably why this is problem #22 of AMC 12.

This type of plane cuts off a solid shape shown in the diagram below.

This solid shape is a heptahedron (meaning that it has 7 faces) that has a regular hexagon as its base and 3 alternating pairs of triangles and pentagons as its side faces. For the aficionados and connoisseurs of geometry, we can notice that it has 15 edges and 10 vertices (even though this information is not relevant to the solution of this problem).

The edges of the hexagon at the base of this heptahedron are 6 “diagonal” segments. Each of these 6 “diagonal” segments belongs to exactly one of 6 faces of the cube. Consequently, considering the symmetry of "diagonal" segments, faces, edges, and everything else in the cube, there are 4 such planes that, altogether, cover 4 different sets of 6 “diagonal” segments. So, they cover all 24 “diagonal” segments.

Strictly speaking, we must prove that the entire set of such 6 segments (in which one “diagonal” segment is properly selected from each face of the cube)  belongs to one plane. It can be done, for example, by calculating the coordinates x,y,z of the ends of each of these segments and by using the formula that verifies that all these points belong to one plane. The coordinates of midpoints of the properly selected 6 edges of the cube are shown in the diagram below.

A plane is a graph of the linear function z = ax + by + c.
The plane that contains the first 3 midpoints, (0,1,0), (1,0,0), (2,0,1), is the graph of the following linear function:
z = x + y -1
(the values of a, b, c can be easily calculated by plugging the coordinates of first 3 midpoints in the places of of x, y, z in the general equation z = ax + by + c).
Now, we can simply check that the coordinates of the remaining 3 midpoints, (2,1,2), (1,2,2), (0,2,1), satisfy this equation. They do.

Here is another proof of this fact. If we draw a straight line from each end of each of the six diagonal segments to vertex A of the cube (as shown on the picture below), two of these line segments belong to the left face of the cube, two other segments - to the front face of the cube, and the last two segments - to the bottom face of the cube. Each of these line segments is the hypotenuse of a right triangle congruent to the triangle marked with green color on the diagram. Thus, the length of each hypotenuse is equal to sqrt(1 + 1/4) = sqrt(5)/2. Hence, they form six congruent isosceles triangles that form a regular hexagonal pyramid. It proves that six diagonal segments belong to one plane since they are the edges of the base of this pyramid.

Note: if we connect the same six points to the opposite vertex of the cube, A1, it will form the other regular hexagonal pyramid, congruent to the first one. Its lateral edges belong to the right, back, and top faces of the cube. This is another proof that all the midpoints 1 - 6 belong to one plane: these six points are equidistant from points A and A1, and we can use the theorem that the locus of all points equidistant from two fixed points in 3-dimensional space is the plane perpendicular to the line segment connecting these two points.

Let's summarize our calculations. One type of planes covers all 12 "edge-parallel" segments:
      I.             3 planes that cut off the rectangular parallelepipeds

and three other types of planes can each cover all 24 “diagonal” segments:

II.                    8 planes that cut off the tetrahedra

III.                  12 planes that cut off  the triangular prisms (these planes also cover all 12 "edge-parallel" segments)

IV.                 4 planes that cut off the heptahedra

Types I and IV are the clear winners to be included in the minimal collection of planes, since the total number of planes in this collection is 3 + 4 = 7. Of course, they must be counted in the maximal collection of planes as well. So, they can be eliminated from the calculations of the difference between the maximum and the minimum numbers of planes.

As a result, the only two types of planes that belong to the maximal collection and do not belong to the minimal collection are types II and III. The total number of distinct planes in these two types is 8 + 12 = 20.

The answer: (C).

Solution of the problem #24 in AMC 10 A 2011

-

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

-  
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

-

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)