*Notes to a video lecture on http://www.unizor.com*

__Geometry+ 07__

*Problem A*

Given any circle with a center at point

*, its diameter*

**O***and any point*

**MN***on this circle not coinciding with the ends*

**P***,*

**M***of a given diameter.*

**N**Let point

*be a projection of point*

**Q***on a diameter*

**P***.*

**MN**This point

*divides diameter*

**Q***into two parts:*

**MN***and*

**MQ = a**

**QN= b**Prove that

(1) Radius of a

*circle is an arithmetic average of*

**OP***and*

**a***.*

**b**(2) Projection segment

*is a geometric average of*

**PQ***and*

**a***.*

**b**(3) Based on these proofs, conclude that geometric average of two non-negative real numbers is less or equal to their arithmetic average.

*Proof*:

(1) Diameter

*of this circle*

**MN**is

*.*

**2r=a+b**Therefore,

*-*

**r = ½(a+b)***arithmetic average*of

*and*

**a***.*

**b**(2) Δ

*∝ Δ*

**MPN***∝ Δ*

**PQN**

**MQP**as right triangles with congruent angles.

**⇒**

*c/b=a/c***⇒**

**c² = a·b**

**⇒**-

*c = √a·b**geometric average*of

*and*

**a**

**b**(3) Cathetus

*is smaller than hypotenuse*

**c***in Δ*

**r***.*

**PQO**Therefore,

**√a·b = c ≤ a = ½(a+b)**End of the proof.

*Problem B*

Prove the following theorem:

If a median and an angle bisector in a triangle coincide, then this triangle is isosceles.

*Important*

Triangle Δ

*has two sides congruent to corresponding two sides of triangle Δ*

**ACM***because they share side*

**BCM***and*

**CM***since*

**AM=BM***is a median.*

**CM**They also have congruent angles ∠

**ACM****=∠**since

*BCM**is an angle bisector.*

**CM****But we cannot use a theorem about triangles Δ**.

*and Δ***ACM***being congruent by two sides and an angle because***BCM**__the angle is not between two congruent sides__See the lecture

*Geometry+ 01*of this course,

*Problem A*as an illustration of a case when two sides of one triangle are equal to two sides of another one and all angles of the first triangle are equal to all angles of the second, yet these triangles are not congruent.

*Hint*

Extend segment

*beyond point*

**CM***to point*

**M***such that*

**N***.*

**CM=MN**Prove that Δ

*= Δ*

**AMN***.*

**BCM**Then prove that Δ

*is isosceles.*

**ACM***Problem C*

Given a

*convex*polygon with

*vertices.*

**N**Draw all the possible diagonals in it.

(1) Assuming that no three diagonals intersect at the same point, how many points of intersection between diagonals will exist?

(2) What is the sum of all angles inside this polygon formed by all its intersecting diagonals and sides?

*Hint C*

For

*the number of intersections is*

**N=6***.*

**15**Consider quadrilaterals formed by any four vertices.

*Answer C*

The number of intersections between the diagonals is

*[*

**n = (N!)/***]*

**(4!)·(N−4)!**The total sum of all angles inside a polygon is equal to

**(2n+N−2)·π**
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