Myth and Truth
As I see it, math is viewed by majority of students in America as an
unneeded subject. Indeed, they realize that in their practical life they
will unlikely have to solve a system of linear equations or calculate
an area of a trapezoid. In 99% of the cases they are right, so they are
negatively predisposed to math and consider it to be just a bunch of
facts they have to memorize to pass an exam and then they can happily
forget it. Nothing is left from this knowledge after a year or two.
Looks like the system of education focused on giving the students
prepackaged recipes to as many math topics as possible fails because
students do not go any deeper in understanding these topics, just learn a
rule "if you have to find ... do ...", which, as I mentioned before, is
easily forgotten as unneeded. As one person characterised it, this
system is "a mile wide and an inch deep". Not withstanding some memory
training, this system bears no useful purpose.
So, the need to drop all the mathematical facts onto student's head
counting that something will stick and might be useful in their future
life is a myth.
What I think the focus of math education should be is the development of
the mind. Does not matter what math topic is studied. What matters is,
how does it help to develop abilities of a student's mind. What
differentiates math from other subjects is its completely artificial
character. It's a product of bright minds. No other subject has this
degree of brain participation because all other subjects are dealing
with real objects. Studying chemistry or literature, foreign language or
biology - it's all about what exists in the world, already created.
Studying math allows to get a student involved in the creative process.
Prove this theorem, solve this problem, construct this geometrical
figure - these are creative assignments not available anywhere else.
Artificial character of math allows to train the student's mind to be
logical, analytical, creative, intelligent. Math is an ultimate form of
art because it is completely a product of artful minds.
If I were a teacher, I would start my first lesson from asking students
about why should they study math. I am sure there will be very few, if
any, who would answer that it's needed to develop their mind or simply
to get smarter. Most likely, I would hear no answer at all. And then I
would compare studying math to develop the mind based on problem solving
approach to exercises in gym to develop the muscles, stamina and other
physical abilities. What gym is for physical development, math is for
intellectual development. I realize that not many students will be
interested in their intellectual development, after all people in
general are not satisfied with their physical appearance but quite
satisfied with how their brain functions, but whoever will be
interested, will benefit tremendously.
So, mathematics is a tool to develop a student's mind. This is the
truth. And to accomplish this, a student must be challanged with proofs
of theorems, problem solving, logical approach to theory and creativity.
This approach is not in the focus of most schools and UNIZOR.COM
attempts to compensate this.
Our goal with Unizor is not as much to change an entire system of
education, we realize impossibility of this, but to present a choice for
those real knowledge seekers, who are still there in the world,
deprived from proper intellectual development by existing educational
system.
Wednesday, March 27, 2013
Friday, September 28, 2012
Introduction to Unizor Education
Unizor features a curriculum for advanced high school course of mathematics. The word "advanced" we mean in terms of depth of learning, rather than in terms of number of topics covered. We started with such topics as "Math Concepts", "Algebra", "Geometry", "Trigonometry" and "Vectors". In the future we plan to add more material to existing topics as well as include some new ones, like "Analysis".
Not everything is fully completed yet, but substantial part of the Web site is ready for use and new material is added daily.
Most important characteristic that distinguishes this course from many other on-line math courses is its emphasis on problem solving, rigorousness of presentation and proof of theorems as the main methodology, rather than memorization of formulas and properties, unfortunately typical in many cases. A short statement that encompasses this idea is "Knowledge vs. Skill" or "Why? vs. How?". Unizor emphasizes "Knowledge" and "Why?" rather than "Skill" and "How?". For instance, we don't just give a formula to solve some type of equations, we derive it. We don't just say that medians in a triangle intersect in one point, we prove it. And we offer hundreds of problems related to presented theoretical material because solving them is the main tool to develop intellectual power. Students forget formulas soon after completing their education, but ability to solve problems will be useful in all aspects of their future life, personal as well as professional.
We would like our students to think about math for their intellectual development as they think about gym for physical development. Mathematical objects (equations, triangles, vectors etc.) are the language for problem solving inasmuch as weights are tools for lifting in a gym. The multitude of mathematical objects is used for balanced and harmonious intellectual development, same as many different tools in a gym are used for balanced and harmonious physical development of all groups of muscles. The time students usually spend on their physical development (gym, basketball, soccer etc.), in our opinion, should be comparable to time they spend on their intellectual development using the art of Mathematics. And what better develops the young mind than solving problems? Think about spending at least an hour a day learning theoretical material and solving problems presented on our site.
Unizor is designed for individualized learning, facilitated in the environment of homeschooling or "flipped" classroom, which allows different types of users: students as objects of education, their parents as a party interested in the results of education in homeschooling environment or teachers, delegated a supervisory role by parents, in a "flipped" classroom or group study environment.
In Unizor environment students will use lectures, problems, proofs and exams provided on the side. Homeschooling parents or teachers in their supervisory capacity will be more involved in a supervision of the educational process. They will have access to the results of all exams taken by students and will be able to enroll their students in different courses as well as to mark certain courses as completed, if the scores on exams taken by their students are satisfactory. In a "flipped" classroom or group study environment facilitated by Unizor teachers will be able to provide additional support for students requiring it.
Educational material on Unizor is provided in two forms: video presentation and notes for each one. Theoretical material is more or less identical in both forms. Most theorems are proven also in both forms, in lectures and in notes for them. Numerous problems are usually stated in notes without solutions, but solutions are presented in the corresponding lectures in an oral form. All material is controlled by a hierarchical menu and is intended to be studied in the order presented on the Web site, starting from "Math Concepts".
Unizor is free and has no advertizing. It is created just to spread the Knowledge and turn the mathematical education towards development of creativity and analytical thinking - qualities needed throughout an entire life, rather than memorization of formulas and properties, which are usually forgotten right after graduation.
Not everything is fully completed yet, but substantial part of the Web site is ready for use and new material is added daily.
Most important characteristic that distinguishes this course from many other on-line math courses is its emphasis on problem solving, rigorousness of presentation and proof of theorems as the main methodology, rather than memorization of formulas and properties, unfortunately typical in many cases. A short statement that encompasses this idea is "Knowledge vs. Skill" or "Why? vs. How?". Unizor emphasizes "Knowledge" and "Why?" rather than "Skill" and "How?". For instance, we don't just give a formula to solve some type of equations, we derive it. We don't just say that medians in a triangle intersect in one point, we prove it. And we offer hundreds of problems related to presented theoretical material because solving them is the main tool to develop intellectual power. Students forget formulas soon after completing their education, but ability to solve problems will be useful in all aspects of their future life, personal as well as professional.
We would like our students to think about math for their intellectual development as they think about gym for physical development. Mathematical objects (equations, triangles, vectors etc.) are the language for problem solving inasmuch as weights are tools for lifting in a gym. The multitude of mathematical objects is used for balanced and harmonious intellectual development, same as many different tools in a gym are used for balanced and harmonious physical development of all groups of muscles. The time students usually spend on their physical development (gym, basketball, soccer etc.), in our opinion, should be comparable to time they spend on their intellectual development using the art of Mathematics. And what better develops the young mind than solving problems? Think about spending at least an hour a day learning theoretical material and solving problems presented on our site.
Unizor is designed for individualized learning, facilitated in the environment of homeschooling or "flipped" classroom, which allows different types of users: students as objects of education, their parents as a party interested in the results of education in homeschooling environment or teachers, delegated a supervisory role by parents, in a "flipped" classroom or group study environment.
In Unizor environment students will use lectures, problems, proofs and exams provided on the side. Homeschooling parents or teachers in their supervisory capacity will be more involved in a supervision of the educational process. They will have access to the results of all exams taken by students and will be able to enroll their students in different courses as well as to mark certain courses as completed, if the scores on exams taken by their students are satisfactory. In a "flipped" classroom or group study environment facilitated by Unizor teachers will be able to provide additional support for students requiring it.
Educational material on Unizor is provided in two forms: video presentation and notes for each one. Theoretical material is more or less identical in both forms. Most theorems are proven also in both forms, in lectures and in notes for them. Numerous problems are usually stated in notes without solutions, but solutions are presented in the corresponding lectures in an oral form. All material is controlled by a hierarchical menu and is intended to be studied in the order presented on the Web site, starting from "Math Concepts".
Unizor is free and has no advertizing. It is created just to spread the Knowledge and turn the mathematical education towards development of creativity and analytical thinking - qualities needed throughout an entire life, rather than memorization of formulas and properties, which are usually forgotten right after graduation.
Sunday, November 27, 2011
Road to Infinity
I have recorded and posted to http://www.unizor.com some material about infinity. More precisely, about different infinities. I've placed it into a new topic "Quantity" in the Math Concepts menu item. There are a few mini-theorems and exam problems as well.
Welcome to Unizor, the road to infinity!
Welcome to Unizor, the road to infinity!
Wednesday, November 16, 2011
Problems in Geometry Series 3
I have uploaded a new lecture to http://www.unizor.com/ with the following geometrical problems.
1. Prove that in an equilateral triangle sum of distances of any internal point to all three sides is constant and is equal to an altitude of a triangle.
2. Prove that a parallelogram with congruent diagonals is a rectangle.
3. Prove that a parallelogram with perpendicular diagonals is a rhombus.
4. Prove that a parallelogram with a diagonal being an angle bisector is a rhombus.
5. Given a rhombus. From a point of intersection of its diagonals we dropped perpendiculars to all four sides. Prove that points of intersection of these perpendiculars with sides form a rectangle.
6. Prove that angle bisectors of a rectangle form a square.
7. Given a square ABCD. Let points A', B', C' and D' be midpoints of sides CD, DA, AB and BC correspondingly. Prove that segments AA', BB', CC' and DD' form a square with each side having a length of 2/5 of the length of each of these segments.
8. Given a square ABCD. Let points A', B', C' and D' be positioned on sides AB, BC, CD and DA correspondingly, such that segments AA', BB', CC' and DD' are congruent. Prove that A'B'C'D' is a square.
9. What condition should a quadrangle satisfy, if a new quadrangle with vertices at midpoints of each its side form
(a) parallelogram,
(b) rhombus,
(c) rectangle,
(d) square.
1. Prove that in an equilateral triangle sum of distances of any internal point to all three sides is constant and is equal to an altitude of a triangle.
2. Prove that a parallelogram with congruent diagonals is a rectangle.
3. Prove that a parallelogram with perpendicular diagonals is a rhombus.
4. Prove that a parallelogram with a diagonal being an angle bisector is a rhombus.
5. Given a rhombus. From a point of intersection of its diagonals we dropped perpendiculars to all four sides. Prove that points of intersection of these perpendiculars with sides form a rectangle.
6. Prove that angle bisectors of a rectangle form a square.
7. Given a square ABCD. Let points A', B', C' and D' be midpoints of sides CD, DA, AB and BC correspondingly. Prove that segments AA', BB', CC' and DD' form a square with each side having a length of 2/5 of the length of each of these segments.
8. Given a square ABCD. Let points A', B', C' and D' be positioned on sides AB, BC, CD and DA correspondingly, such that segments AA', BB', CC' and DD' are congruent. Prove that A'B'C'D' is a square.
9. What condition should a quadrangle satisfy, if a new quadrangle with vertices at midpoints of each its side form
(a) parallelogram,
(b) rhombus,
(c) rectangle,
(d) square.
Problems in Geometry Series 2
I have uploaded a new lecture to http://www.unizor.com/ with the following problems.
1. Prove that a median to a hypotenuse in the right triangle is equal to half of that hypotenuse.
Hint: extend the median through hypotenuse by its length.
2. Prove a converse theorem, that if a median in a triangle is equal to half a side it falls on, then an angle, where it starts, is the right angle.
3. Prove that in a right triangle a median and an altitude to its hypotenuse form an angle, equal to difference between this triangle's acute angles.
Hint: use a previous theorem.
4. Given triangle ΔABC. Segment AD bisects angle ∠BAC (point D lies on side BC). Straight line through point D is parallel to side AC, intersecting side AB at point E. Straight line through point E is parallel to side BC, intersecting side AC at point F. Prove that segments AE and CF are congruent.
5. Given an angle ∠MXN. Inside it an angle ∠PYQ is positioned in such a way that MX is parallel to PY, NX is parallel to QY and distance between MX and PY is equal to distance between NX and QY. Prove that a bisector of angle ∠MXY is a bisector of angle ∠PYQ.
6. Prove that any segment, that connects two bases of a trapezoid, is divided by a median in two congruent parts.
7. Given a triangle ΔABC. Let vertex B be at the top and side AC be a base. Let point X be an intersection of two bisectors of angles at the base. Straight line through point X parallel to a base intersects its left side AB and right side BC at points M and N correspondingly. Prove that segment MN is equal to a sum of segments AM and CN.
8. Straight lines are drawn through all three vertices of a triangle, forming another (bigger) triangle. Prove that this bigger triangle is divided by sides of a smaller one into 4 triangles, each congruent to a small triangle and each side of a bigger triangle is twice as big as parallel to it side of a smaller triangle.
9. Prove that in an isosceles triangle sum of two distances from any point on a base to two legs is constant and is equal to an altitude from any of two end points of a base to a leg.
10. Change a condition of a previous theorem to use a point on continuation of a base (outside of a triangle). Formulate a theorem in this case and prove it.
1. Prove that a median to a hypotenuse in the right triangle is equal to half of that hypotenuse.
Hint: extend the median through hypotenuse by its length.
2. Prove a converse theorem, that if a median in a triangle is equal to half a side it falls on, then an angle, where it starts, is the right angle.
3. Prove that in a right triangle a median and an altitude to its hypotenuse form an angle, equal to difference between this triangle's acute angles.
Hint: use a previous theorem.
4. Given triangle ΔABC. Segment AD bisects angle ∠BAC (point D lies on side BC). Straight line through point D is parallel to side AC, intersecting side AB at point E. Straight line through point E is parallel to side BC, intersecting side AC at point F. Prove that segments AE and CF are congruent.
5. Given an angle ∠MXN. Inside it an angle ∠PYQ is positioned in such a way that MX is parallel to PY, NX is parallel to QY and distance between MX and PY is equal to distance between NX and QY. Prove that a bisector of angle ∠MXY is a bisector of angle ∠PYQ.
6. Prove that any segment, that connects two bases of a trapezoid, is divided by a median in two congruent parts.
7. Given a triangle ΔABC. Let vertex B be at the top and side AC be a base. Let point X be an intersection of two bisectors of angles at the base. Straight line through point X parallel to a base intersects its left side AB and right side BC at points M and N correspondingly. Prove that segment MN is equal to a sum of segments AM and CN.
8. Straight lines are drawn through all three vertices of a triangle, forming another (bigger) triangle. Prove that this bigger triangle is divided by sides of a smaller one into 4 triangles, each congruent to a small triangle and each side of a bigger triangle is twice as big as parallel to it side of a smaller triangle.
9. Prove that in an isosceles triangle sum of two distances from any point on a base to two legs is constant and is equal to an altitude from any of two end points of a base to a leg.
10. Change a condition of a previous theorem to use a point on continuation of a base (outside of a triangle). Formulate a theorem in this case and prove it.
Problems in Geometry Series 1
I have uploaded a new lecture to http://www.unizor.com/ with the following geometrical problems.
1. Construct a triangle by two sides and a median to the third side.
2. Construct a quadrangle by all four sides and a median segment connecting midpoints of opposite sides.
Hint: use parallel shift.
3. Construct a trapezoid by one of its interior angles, two diagonals and a median (segment between midpoints of two non-parallel sides).
Hint: use parallel shift.
4. Construct a quadrangle by three sides and two interior angles adjacent to one of known sides.
Hint: use parallel shift.
5. Construct a trapezoid by its two bases and two legs.
Hint: use parallel shift.
6. Construct a quadrangle by its four sides, provided one diagonal bisects one angle it is connected to.
Hint: use symmetry relative to an axis.
7. A billiard table has a rectangular shape. There are two balls on it in some position. Determine direction of the movement of one ball in such a way, that, reflecting from all four sides, it will hit the second ball.
Hint: use symmetry relative to table borders on each side.
1. Construct a triangle by two sides and a median to the third side.
2. Construct a quadrangle by all four sides and a median segment connecting midpoints of opposite sides.
Hint: use parallel shift.
3. Construct a trapezoid by one of its interior angles, two diagonals and a median (segment between midpoints of two non-parallel sides).
Hint: use parallel shift.
4. Construct a quadrangle by three sides and two interior angles adjacent to one of known sides.
Hint: use parallel shift.
5. Construct a trapezoid by its two bases and two legs.
Hint: use parallel shift.
6. Construct a quadrangle by its four sides, provided one diagonal bisects one angle it is connected to.
Hint: use symmetry relative to an axis.
7. A billiard table has a rectangular shape. There are two balls on it in some position. Determine direction of the movement of one ball in such a way, that, reflecting from all four sides, it will hit the second ball.
Hint: use symmetry relative to table borders on each side.
Monday, October 10, 2011
Construction Problems in Geometry
Construction problems in geometry are extremely important. They help to develop student's creativity, logical thinking and ability to combine analysis of a problem at hand with synthesis of existing methods into concrete solution. Solving problems is the most important purpose of UNIZOR.COM. Please examine these problems and try to solve them. The accompanying lecture contains all the solutions, but watch it only after you spent sufficient amount of time trying to solve these problems yourself.
1. Construct a sum of two angles.
In details, this construction problem has the following meaning, applicable to other problems as well (so, we will not repeat it).
Given two angles and a ray somewhere on a plane.
Using a given ray as one leg, construct another ray from the same starting point such that these two rays (one given and one constructed) form an angle, which has a measure equal to a sum of measures of two given angles.
2. Construct a difference between a bigger and a smaller angle.
3. Given a sum and a difference between two angles, construct these angles.
4. Divide an angle into 4, 8 and 16 congruent parts.
5. Construct a triangle by its two sides and an angle opposite to a bigger side.
Is it always a unique triangle? Analyse the number of solutions of this problem.
6. Construct a triangle by its two sides and an angle opposite to a smaller side.
Is it always a unique triangle? Analyse the number of solutions of this problem.
7. Construct an isosceles triangle by a base (a side, generally speaking, non-congruent to two others) and one of two other sides.
8. Construct an isosceles triangle by a base (a side, generally speaking, non-congruent to two others) and one of two congruent angles (usually called angles at the base or base angles).
9. Construct an isosceles triangle by one of two congruent sides and an angle between congruent sides (usually called an angle at the vertex).
10. Construct an isosceles triangle by one of two congruent sides and a base angle it forms with a base (a side, generally speaking, non-congruent to two others).
11. Construct a right triangle by its two legs (i.e. sides, perpendicular to each other, sometimes called also catheti).
12. Construct a right triangle by one of its legs (also known as cathetus) and a hypotenuse.
13. Construct a right triangle by a cathetus and an acute angle it forms with a hypotenuse.
14. Construct an isosceles triangle by an altitude to a base and one of two congruent sides.
15. Construct an isosceles triangle by an altitude to a base and an angle opposite to a base (i.e. vertex angle).
16. Construct an isosceles triangle by a base and an altitude towards one of two congruent sides.
17. Construct a right triangle by a hypotenuse and an acute angle it forms with one of catheti.
18. Given an angle and a point inside it. Construct a straight line crossing this point and cutting congruent segments from sides of a given angle.
19. Construct two segments by their given sum and difference.
20. Divide a given segment into 4, 8 and 16 congruent parts.
21. Given a straight line and two points anywhere on a plane. Construct a point on a given line equidistant to two given points.
22. Construct a point equidistant to three vertices of a given triangle.
23. Given an angle and a straight line that crosses both its sides. Construct a point on this line equidistant to two sides of a given angle.
24. Construct a point equidistant to three sides of a given triangle.
25. Given a straight line PQ and two points M and N outside it, but on the same side from it. Construct a point X on the line PQ such that angles ∠PXM and ∠QXN are congruent.
26. Construct a right triangle by one of its catheti and a sum of hypotenuse and the other cathetus.
27. Construct a triangle by a side, an angle it makes with another side and a sum of two other sides.
28. Given an angle and two points, one on each side of an angle. Construct a point equidistant from both sides of an angle and, at the same time, equidistant from two given points on angle's sides.
Notice, that distances to angle's sides should be equal to each other, but not necessarily equal to distances to given points.
1. Construct a sum of two angles.
In details, this construction problem has the following meaning, applicable to other problems as well (so, we will not repeat it).
Given two angles and a ray somewhere on a plane.
Using a given ray as one leg, construct another ray from the same starting point such that these two rays (one given and one constructed) form an angle, which has a measure equal to a sum of measures of two given angles.
2. Construct a difference between a bigger and a smaller angle.
3. Given a sum and a difference between two angles, construct these angles.
4. Divide an angle into 4, 8 and 16 congruent parts.
5. Construct a triangle by its two sides and an angle opposite to a bigger side.
Is it always a unique triangle? Analyse the number of solutions of this problem.
6. Construct a triangle by its two sides and an angle opposite to a smaller side.
Is it always a unique triangle? Analyse the number of solutions of this problem.
7. Construct an isosceles triangle by a base (a side, generally speaking, non-congruent to two others) and one of two other sides.
8. Construct an isosceles triangle by a base (a side, generally speaking, non-congruent to two others) and one of two congruent angles (usually called angles at the base or base angles).
9. Construct an isosceles triangle by one of two congruent sides and an angle between congruent sides (usually called an angle at the vertex).
10. Construct an isosceles triangle by one of two congruent sides and a base angle it forms with a base (a side, generally speaking, non-congruent to two others).
11. Construct a right triangle by its two legs (i.e. sides, perpendicular to each other, sometimes called also catheti).
12. Construct a right triangle by one of its legs (also known as cathetus) and a hypotenuse.
13. Construct a right triangle by a cathetus and an acute angle it forms with a hypotenuse.
14. Construct an isosceles triangle by an altitude to a base and one of two congruent sides.
15. Construct an isosceles triangle by an altitude to a base and an angle opposite to a base (i.e. vertex angle).
16. Construct an isosceles triangle by a base and an altitude towards one of two congruent sides.
17. Construct a right triangle by a hypotenuse and an acute angle it forms with one of catheti.
18. Given an angle and a point inside it. Construct a straight line crossing this point and cutting congruent segments from sides of a given angle.
19. Construct two segments by their given sum and difference.
20. Divide a given segment into 4, 8 and 16 congruent parts.
21. Given a straight line and two points anywhere on a plane. Construct a point on a given line equidistant to two given points.
22. Construct a point equidistant to three vertices of a given triangle.
23. Given an angle and a straight line that crosses both its sides. Construct a point on this line equidistant to two sides of a given angle.
24. Construct a point equidistant to three sides of a given triangle.
25. Given a straight line PQ and two points M and N outside it, but on the same side from it. Construct a point X on the line PQ such that angles ∠PXM and ∠QXN are congruent.
26. Construct a right triangle by one of its catheti and a sum of hypotenuse and the other cathetus.
27. Construct a triangle by a side, an angle it makes with another side and a sum of two other sides.
28. Given an angle and two points, one on each side of an angle. Construct a point equidistant from both sides of an angle and, at the same time, equidistant from two given points on angle's sides.
Notice, that distances to angle's sides should be equal to each other, but not necessarily equal to distances to given points.
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