The fresco "The School of Athens" by Renaissance artist Raphael depicts philosophers in a grand architectural setting. It includes over 50 figures in a variety of poses and locations within the fresco. The archways are painted in a semicircular shape. The fresco is one of Raphael's most famous works and an excellent example of Renaissance artistic techniques and ideals.

1. “The School of Athens” is the fresco made by the Italian Renaissance artist Raphael (1483–1520).
The archs are in a semi-circle form.
1) In a coordinate system draw a circle, when the circle‘s radius is equal to 4, and the center is in
O( -2; 3) point. Calculate the circumference of the given circle and its area.
Write down the circumference of the circle.
2) In a coordinate system draw the given semi-circles:
𝑥2
+ 𝑦2
= 9, when 𝑦 ≥ 0 and 𝑥2
+ (𝑦 − 3)2
= 9, when 𝑦 ≤ 3.
Calculate the coordinates of their collision points.
The answers:
1) 𝐶 = 8𝜋, 𝑆 = 16𝜋. (𝑥 + 2)2
+ (𝑦 − 3)2
= 16.
2)
(
3√3
2
;
3
2
) ir (−
3√3
2
;
3
2
)

2. Albrecht Dürer (1471-1528 m.) – the most popular German Renaissance artist,
copper carver and art theoretic.
“Melancholia” is one of the most popular masterpieces created by the artist.
„Melancholia“
In A. Dürer’s “Melancholia” you can see a
magical square. In every line, column, and
diagonal you can sum up every number and
get the same answer.
1) Fill in the magical square.
2) When was “Melancholia” created?
The answer is hidden in the mystical
square. Two numbers one next to the
other in the same line.
16 13
10 11
6 12
4 14
Answers:
1) 2) 1514 year

3. Jacek Yerka (Polish surrealist)
“The Tower of Subconsciousness“ and “The Garden“
Tasks:
Once upon a time a master asked his servants to scoop some water from a magical well and fill a
reservoir with it. Three servants offered their service to complete this mission. Each of them took a
bucket and started scooping water from the magical well. The surface of the part of the tower which
had to be filled was of a shape of a perfect hexagon, which sides were each 20 m, and the height of
the reservoir was 15 m.
1. Which spatial geometrical shape meets the part of the tower which has to be filled?
2. How many litres are needed to fill the reservoir fully? Give the answer using an integral
number.
3. How many buckets would fill the reservoir fully if it is known that the diameter of a bucket‘s
surface is 26 cm, the diameter of the bucket‘s top is 36 cm, and the height of the bucket is 40
cm. Consider the value of Pi as ≈3,14. Give the answer using an integral number.
4. The first servant alone would fill the reservoir in 70 h, the second in 72 h, and all three
together in 25 h. How much time would it take for the third servant to fill the reservoir if he
was doing it all by himself? Give the answer in units worth.
Answers:
1. Vertical hexagonal prism.
2. 𝑉 = 15588000 𝑙
3. 519600 buckets
4. 85 h.

4. Osman Hamdi Bey (Turkish artist)
„Girl reading the Quran“
Tasks:
1. The length of the a side of the rhombus highlighted in the painting is 0,5 m, and the shorter
diagonal is 60 cm. long. Find the area of the rhombus.
2. What would the area of the rhombus be, if the area of four hexagonal tiles was deducted
from it? Here MN is the median of ABC.
3. What percentage of the area of the rhombus meets the area of triangle BNM? Give the answer
in exactness of 0,01%.
4. In what percentage is the area of all hexagonal tiles bigger than the area of all equilateral
triangles which are in the rhombus?
Answers:
1. 𝑆𝑟𝑜𝑚𝑏ℎ𝑢𝑠 = 2400 𝑐𝑚2
= 0,24 𝑚2
.
2. 𝑆 = 2400 − 1350√3 𝑐𝑚2
3. 4,06 %
4. 200%

5. Math in Arts
Greek Amphora
A classic Greek amphora is composed of two parts – a sphere A, from which
the bottom segment is cut off, and a neck formed by a sphere B rotating
around the vertical axes. The sphere B has the same diameter as the sphere A,
and its center is positioned on the opposite side of the square formed on the
diagonal of the square inscribed into the circle of the bottom’s sphere plane
projection (look at the picture below).
Problem
With the diameter of the bottom sphere of the
amphora be equal to 40 cm, Find the area of the
amphora’s projection on the vertical plane.
Solution
After noticing that segments x and y have the same
area, we can conclude that the amphora’s
projection area is twice as large as the inscribed
square.
Let d be the diameter of the sphere.
r = d/2
a = r / √2
S = 2 (2a)2
= 2 (r √2)2
= 4 r2
= 4 (d/2)2
= d2
=
= 402
cm2
= 1600 cm2

6. Perspective in Painting
Artists use perspective to
distinguish close objects
from remote objects. This
is done by reducing an
object’s visible size
proportionally to the
distance from the object.
Let’s consider a
masterpiece from Xanthos
Hadjisoteriou, an
acclaimed Greek Cypriot
painter.
Problem
Assuming all houses on this
picture have windows of
the same size, find the real
height of the belfry in the
far-left corner if the height
of the woman on the front
is 160 cm.
Solution
Let’s assume that the view
point in this picture is at
the bottom center of this
picture. This is not exactly
true, but will work for our
approximations.
Let’s draw vectors from the
view point to all objects we
want to take into account.
The table below shows
horizontal and vertical sizes
of each vector taken from
PowerPoint, where the
vectors were drawn, and
the calculated length of
each vector.

7. Vector # Horizontal size (“) Vertical size (“) Length (‘)
1 2.2 2 2.973
2 2.25 3.06 3.798
3 0.88 2.94 3.069
4 0.95 3.46 3.588
5 0.12 5.64 5.641
6 0.18 5.97 5.973
7 2.75 5.76 6.383
8 2.78 7.06 7.588
1. Knowing that sizes of windows residing on vectors 3-4 and 5-6 are the same, define the proportion
between the visible size and distance from the view point:
a. Window 3-4 visible height = .52”
b. Window 5-6 visible height = .33”
c. Proportion between distance and visible height is:
<length of vector-5> / <length of vector-3>
:
<window 3-4 visible height> / <window 5-6 visible height>
or
5.641 / 3.069 : .52 / .33, which means that
“Objects that are 1.838 times further seem 1.576 times smaller”
2. The belfry is 6.383/2.973= 2.147 times further than the woman. Hence, it seems 1.576/1.838*2.147=
1.841 times smaller.
3. Woman’s visible height: 3.06” – 2” = 1.06”
4. Belfry’s visible height: 7.06” – 5.76” = 1.30”
5. Belfry’s real height: 1.30/1.06*1.841=2.258 is larger than woman’s or is:
160*2.258=3m 61cm
Conclusion
Since in the real life a belfry is much higher, either the assumption about the view point is incorrect, or the
artist diverges from true proportions to produce special impressions.