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Name: Nodar Nadibaidze

Solution: The equation of a circle in the standard form is:
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2

(1)

r – Radius of a circle
(a;b) – Center coordinates
We should represent the given equation in the text as (1):
𝑥 2 + 6𝑥 + 𝑦 2 − 4𝑦 = −4
𝑥 2 + 6𝑥 + 9 − 9 + 𝑦 2 − 4𝑦 + 4 = 0
So if we want our equation to be the same as (1), we should write it as complete square, like
𝑎2 + 2𝑎𝑏 + 𝑏 2 = (𝑎 + 𝑏)2
...
Now we have 𝑥 2 + 6𝑥 + 9, that is the same as
(𝑥 + 3)2 ) and transfer -4 to the left side of the equation and write it after 𝑦 2 − 4𝑦 (Now, 𝑦 2 −
4𝑦 + 4 = (𝑦 − 2)2 )
...

The center = (−3; 2) , the radius 𝑟 = 3
...
14)
Put the value of r in (2):
𝐶 = 2𝜋3 = 6𝜋
Area of a circle:
𝐴 = 𝜋𝑟 2

(3)

Put the value of r in (3):
𝐴 = 𝜋32 = 9𝜋
Answer: The center = (−3; 2) , 𝑟 = 3 , 𝐶 = 6𝜋 , 𝐴 = 9𝜋
...
So
𝐴𝐷 = 𝐷𝐵 =

𝐴𝐵
2

The tangent segments to a circle from an external point are
equal
...
In our case 𝑂𝐸 ⊥ 𝐶𝐸
...


We can use the Pythagorean theorem for any right triangle
...

𝐸𝑂2 + 𝐶𝐸 2 = 𝐶𝑂2
𝐸𝑂 = 𝑂𝐷 = 𝑟, 𝐶𝐷 = ℎ, 𝐶𝑂 = 𝐶𝐷 − 𝑂𝐷 = ℎ − 𝑟
Put those values into the above equation:
𝑟 2 + 𝐶𝐸 2 = (ℎ − 𝑟)2
𝐶𝐸 2 = (ℎ − 𝑟)2 − 𝑟 2
𝐶𝐸 2 = ℎ2 − 2ℎ𝑟 + 𝑟 2 − 𝑟 2
𝐶𝐸 2 = ℎ2 − 2ℎ𝑟
𝐶𝐸 = √ℎ2 − 2ℎ𝑟
We should use the Pythagorean theorem one more time in ∆𝐴𝐶𝐷:
𝐴𝐷2 + 𝐶𝐷2 = 𝐴𝐶 2
𝐴𝐷 =

𝐴𝐵
2

,

𝐶𝐷 = ℎ,

𝐴𝐶 = 𝐴𝐸 + 𝐸𝐶 =

𝐴𝐵
2

Put those values into the above equation:
𝐴𝐵2
4

𝐴𝐵

+ ℎ2 = ( 2 + √ℎ2 − 2ℎ𝑟)

2

𝐴𝐵 2
𝐴𝐵 2
+ ℎ2 =
+ 𝐴𝐵√ℎ2 − 2ℎ𝑟 + ℎ2 − 2ℎ𝑟
4
4
𝐴𝐵√ℎ2 − 2ℎ𝑟 = 2ℎ𝑟
2ℎ𝑟

𝐴𝐵 = √ℎ2

−2ℎ𝑟

We expressed the length of the base using height h and radius r
...
) it is, it doesn’t effect on
the result
...
We want to find the ratio V1: V2
...
Let AEFKG be my and AABCD be ny
...

Volume of a frustum pyramid is
1

𝑉 = 3 𝐻(𝑄 + √𝑄𝑞 + 𝑞)

(1)

𝐻 – Height
𝑄 𝑎𝑛𝑑 𝑞 – Area of bases
The given plane parallel to the base will divide our
pyramid into two frustum pyramids
...


In upper pyramid height is PR= h/2, AEFKG= my and let ALMNO be A
...
Now we can write:
1

ℎ

1

ℎ

=

1 ℎ
∗ (𝑚𝑦+√𝑚𝑦∗𝐴+𝐴)
3 2
1 ℎ
∗ (𝐴+√𝐴∗𝑛𝑦+𝑛𝑦)
3 2

𝑉1 = 3 ∗ 2 (𝑚𝑦 + √𝑚𝑦 ∗ 𝐴 + 𝐴)
𝑉2 = 3 ∗ 2 (𝐴 + √𝐴 ∗ 𝑛𝑦 + 𝑛𝑦)
𝑉1
𝑉2

(2)

In order to find above ratio, we should express A in terms of y, because A is unknown
...

Let’s add top part to become the whole pyramid
...
Here is the property of similarity:

√AEFKG
√AABCD

=

𝑇𝑃
𝑇𝑆

Put the values:

√𝑚𝑦
√ny

=

𝑇𝑃
𝑇𝑆

𝑇𝑃 √m
=
𝑇𝑆 √n
Let TP be √m𝑥 and TS be √n𝑥


...
So:
𝑃𝑅 =

𝑃𝑆
2

𝑃𝑆 = 𝑇𝑆 − 𝑇𝑃 = √n𝑥 - √m𝑥

,

𝑃𝑅 =

√n𝑥 − √m𝑥
2

We will need TR:
𝑇𝑅 = 𝑇𝑃 + 𝑃𝑅 = √m𝑥

+

√n𝑥 − √m𝑥
2

=

√n𝑥+ √m𝑥
2

Property of similarity:
(𝑇𝑅)2
(𝑇𝑆)2

=

𝐴
AABCD

2
√n𝑥+ √m𝑥
)
2
2

(

(√n𝑥)

=

𝐴
ny

𝑛𝑥 2 +2√𝑛𝑚𝑥2 +𝑚𝑥2
4𝑛𝑥 2

=

𝐴
ny

After reducing fraction to a simpler form (x^2) and multiplying both sides of equation by n,
we have:
𝑛+2√𝑛𝑚+𝑚
4

=

𝐴
y

From here to isolate A, we should multiply both sides of equation by y:

𝐴=

𝑛+2√𝑛𝑚+𝑚

Now we can put the value of A in (2):

4

y

𝑉1
𝑉2

𝑉1
𝑉2

𝑉2

Answer:

=

=

𝑉1

1 ℎ
𝑛+2√𝑛𝑚+𝑚
𝑛+2√𝑛𝑚+𝑚
∗ (𝑚𝑦+√𝑚𝑦∗
y+
y)
3 2
4
4
1 ℎ 𝑛+2√𝑛𝑚+𝑚
𝑛+2√𝑛𝑚+𝑚
∗ (
y+√
y∗𝑛𝑦+𝑛𝑦)
3 2
4
4

𝑚𝑦+√𝑚𝑦∗

𝑛+2√𝑛𝑚+𝑚
𝑛+2√𝑛𝑚+𝑚
y+
y
4
4

𝑛+2√𝑛𝑚+𝑚
√𝑛+2√𝑛𝑚+𝑚y∗𝑛𝑦+𝑛𝑦
y+
4
4

=

𝑛+2√𝑛𝑚+𝑚 𝑛+2√𝑛𝑚+𝑚
+
4
4

𝑚+√𝑚∗

𝑛+2√𝑛𝑚+𝑚
√𝑛+2√𝑛𝑚+𝑚∗𝑛+𝑛
+
4
4

𝑉1
𝑉2

=

𝑛+2√𝑛𝑚+𝑚 𝑛+2√𝑛𝑚+𝑚
+
4
4

𝑚+√𝑚∗

𝑛+2√𝑛𝑚+𝑚
𝑛+2√𝑛𝑚+𝑚
+√
∗𝑛+𝑛
4
4

Solution:
a) Let’s draw a line on the vertex B and the midpoint of the CD slope N
...
∆NBC=∆NED based on the
second sign of the equality of triangles, because
CN=DN (N is midpoint of CD), ∠BNC=∠END as
vertical angles and ∠NCB=∠NDE as alternate
interior angles (BC and AD parallel lines are
crossed by CD transversal)
...


Therefore, MN midsegment of trapezoid is the midsegment of ∆ABE triangle
...


b) MN is parallel to the base edges and 𝐴𝑀𝐵𝐶𝑁 = 𝐴𝐴𝑀𝑁𝐷
...

Area of trapezoid is:
𝐴=

𝑎+𝑏
∗ℎ
2

h – Height
a, b – bases
In our case 𝐴𝑀𝐵𝐶𝑁 =

𝑥+𝑏
2

∗ ℎ1 , 𝐴𝐴𝑀𝑁𝐷 =

𝑥+𝑎
2

𝐴𝑀𝐵𝐶𝑁 = 𝐴𝐴𝑀𝑁𝐷
𝑥+𝑏
𝑥+𝑎
∗ ℎ1 =
∗ ℎ2
2
2
We should multiply both sides of equation by 2
...
To do that, we need to express ℎ1 in terms of
ℎ2
...
Let the height of ∆BEC be h
...
It means, that
these triangles are similar to each other and we can use the properties of similarity:
𝑏
𝑥
𝑏
𝑎

ℎ

= ℎ+ℎ

1

ℎ

= ℎ+ℎ

1 +ℎ2

(2)

(∆BEC~∆MEN, “~” sign of similarity)

(3)

(∆BEC~∆AED)

From (2):
ℎ𝑥 = 𝑏ℎ + 𝑏ℎ1
ℎ(𝑥 − 𝑏) = 𝑏ℎ1
𝑏ℎ

ℎ = 𝑥−𝑏1
We expressed h in terms of h1
...

ON is parallel to BC and AD
...

Now we can write properties of similarity:
𝑀𝑂
𝑎

=

𝐵𝑂

𝑀𝑂

;

𝐵𝐷

=

𝑏

𝐴𝑂
𝐴𝐶

𝑂𝑁

;

𝑎

𝐶𝑂

=

𝐴𝐶

𝑂𝑁

;

𝑏

=

𝑂𝐷
𝐵𝐷

Let’s add first and fourth and second and third equations:
𝑀𝑂
𝑎

+

𝑂𝑁
𝑏

𝐵𝑂

=

𝑀𝑂
𝑎

+

𝐵𝐷

+

𝑂𝑁
𝑏

𝑂𝐷

𝑀𝑂

𝐵𝐷

𝑏

+

𝑀𝑂

=1

𝑏

𝑂𝑁

+

𝑎

=

𝑂𝑁
𝑎

𝐴𝑂
𝐴𝐶

+

𝐶𝑂
𝐴𝐶

=1

Let’s add these two equation:
𝑀𝑂

+

𝑎
𝑀𝑁

+

𝑎

1

𝑂𝑁
𝑏

+

𝑀𝑁

𝑀𝑂
𝑏

+

𝑂𝑁
𝑎

= 1+1

=2

𝑏

1

𝑀𝑁 ( + ) = 2
𝑎
𝑏
𝑀𝑁 =

2
1 1
(𝑎+𝑏)

The names of given lengths:
2𝑎𝑏

a) Arithmetic mean; b) Quadratic mean; c) Geometric mean; d) Harmonic mean (the same as 𝑎+𝑏)

Solution:

Here is the sketch on the left
...


First we should draw AC diagonal and remove BF
and ED (look at the photo on the right)
...

Let’s write area of all triangles:

𝐴𝐵𝐶𝐹 =

𝐶𝐹 ∗ ℎ2
𝐹𝐷 ∗ ℎ2
; 𝐴𝐵𝐹𝐷 =
2
2

𝐴𝐸𝐵𝐷 =

𝐸𝐵 ∗ ℎ3
𝐴𝐸 ∗ ℎ3
; 𝐴𝐴𝐸𝐷 =
2
2

EB=AE and the height is same for both triangle – It means that AEBD= AAED
CF=FD and the height is same for both triangle – means that ABCF= ABFD

𝐴𝐸𝐵𝐹𝐷 = 𝐴𝐸𝐵𝐷 + 𝐴𝐵𝐹𝐷
𝐴𝐸𝐵𝐷 = 𝐴𝐴𝐸𝐷 ;

𝐴𝐵𝐹𝐷 = 𝐴𝐵𝐶𝐹

So:

𝐴𝐸𝐵𝐹𝐷 = 𝐴𝐴𝐸𝐷 + 𝐴𝐵𝐶𝐹
Now let’s add (1) and (2) equations:

(2)

𝐴𝐴𝐸𝐶𝐹 + 𝐴𝐸𝐵𝐹𝐷 = 𝐴𝐸𝐵𝐶 + 𝐴𝐴𝐹𝐷 + 𝐴𝐴𝐸𝐷 + 𝐴𝐵𝐶𝐹

(3)

We need to look at the first sketch and write the following:

𝐴𝐴𝐸𝐶𝐹 = 𝐴𝐴𝐸𝐺 + 𝐴𝐸𝐻𝐹𝐺 + 𝐴𝐻𝐶𝐹
𝐴𝐸𝐵𝐹𝐷 = 𝐴𝐸𝐵𝐻 + 𝐴𝐸𝐻𝐹𝐺 + 𝐴𝐺𝐹𝐷
𝐴𝐸𝐵𝐶 = 𝐴𝐸𝐵𝐻 + 𝐴𝐵𝐻𝐶 ; 𝐴𝐴𝐹𝐷 = 𝐴𝐴𝐺𝐷 + 𝐴𝐺𝐹𝐷
𝐴𝐴𝐸𝐷 = 𝐴𝐴𝐺𝐷 + 𝐴𝐴𝐸𝐺 ; 𝐴𝐵𝐶𝐹 = 𝐴𝐵𝐻𝐶 + 𝐴𝐻𝐶𝐹
Put these values into (3):
𝐴𝐴𝐸𝐺 + 𝐴𝐸𝐻𝐹𝐺 + 𝐴𝐻𝐶𝐹 + 𝐴𝐸𝐵𝐻 + 𝐴𝐸𝐻𝐹𝐺 + 𝐴𝐺𝐹𝐷 = 𝐴𝐸𝐵𝐻 + 𝐴𝐵𝐻𝐶 + 𝐴𝐴𝐺𝐷 + 𝐴𝐺𝐹𝐷 + 𝐴𝐴𝐺𝐷 +
𝐴𝐴𝐸𝐺 + 𝐴𝐵𝐻𝐶 + 𝐴𝐻𝐶𝐹
We have same values on both sides, so after simplifying:

2𝐴𝐸𝐻𝐹𝐺 = 2𝐴𝐵𝐻𝐶 + 2𝐴𝐴𝐺𝐷
𝐴𝐸𝐻𝐹𝐺 = 𝐴𝐵𝐻𝐶 + 𝐴𝐴𝐺𝐷



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