Euclidean Geometry
Logic and Problem-Solving
Euclidean Logical Thinking for Digital Content Creators 3. Finding Ideas with Euclidean Thinking Part 1 : Thinking in Ratios
Hello, this is Jonghwa Park
In this session, we’ll explore a method to generate ideas
by setting a standard, comparing it with other elements
and considering appropriate ratios
One effective way to understand the world rationally and logically
is to think by comparing elements against a standard
For example, why do we use numbers?
The reason we use numbers is to compare
A person who is 180 cm tall is about 10 cm taller than someone who is 170 cm tall
We use numbers
to compare, understand, and grasp such differences
The ancient Greeks seemed to think the same way
Instead of focusing on absolute values, they often relied on relative ratios
to make wise judgments
Today, let’s explore their thinking techniques together
Applying Ratios and Comparisons to Problem Solving
In English, there is something called a rational number
The term “rational” from rational number originates from
the Latin word “ratio,” which means proportion or ratio
Given two integers a and b, a rational number is written as b over a
A number expressed as a ratio of integers is a rational number
Around 500 BCE, Pythagoras believed
that the world was composed of numbers
However, the only numbers he recognized were rational numbers
He thought the most logical and rational way to understand the world
was by perceiving it through numbers
This belief likely contributed to the dual meaning of “rational,”
as both a mathematical term and a descriptor of logical reasoning
Thinking in terms of ratios and proportions is actually a highly effective technique for problem-solving
Let’s delve deeper by examining some problems
Problem 1: In the diagram below, what is the length of x
This problem involves two right triangles
By analyzing the diagram
you can observe that triangle ABC is similar to triangle ADE
If we flip and align triangle ADE, we can confirm their similarity
Since the two triangles are similar
Length of AC : Length of AB = Length of AE: Length of AD
Using this ratio, we can calculate the following
We can conclude that x = 5
Problem 2: In the diagram below, line segments AF, BE, and CD are parallel
If AF = 5, AB = 6, BC = 2, and CD = 9
What is the length of BE?
To solve this, draw a line segment FG
parallel to AC
From drawing this parallel line FG
we can determine that AF = BH = CG = 5
Triangles FHE and FGD are similar
So, FH:HE=FG:GD
6:HE=8:4
So, HE=3
Finally, BE=BH+HE
which is 5+3
The answer is 8
This problem demonstrates how creating similar triangles
helps in applying ratios for problem-solving
Using similarity and applying ratios through comparison
is a representative technique for generating ideas in problem-solving
In music, there is the concept of perfect pitch
It’s the ability to identify a note, such as Sol or Fa
accurately just by hearing it
However, not many people possess perfect pitch
Most people identify notes relatively, using a reference note
because it is easier and more effective
Similarly, problem-solving often involves setting a standard, making comparisons
and applying ratios
You can experience this process by solving problems
Problem 3: In the following scenario, a right triangle ABCD is folded at point B
so that AB aligns perfectly with BC
What is the area of the shaded region?
The problem can be approached as follows
By folding triangle ABE onto triangle BDE
the two triangles become congruent, meaning they have the same area
What we are trying to find is the area of triangle CDE
Since triangles BDE and CDE share the same height
their areas have a ratio of 3:1
Thus, the areas of triangles ADE, BDE, and CDE
are in the ratio 3:3:1
It's 3/7, 3/7, and 1/7 of the total area
For the large triangle ABCD, its total area is calculated as 3×4×1/2
which is 6
Therefore, the areas of triangles ABE, BDE, and CDE
are each 18/7, 18/7, and 6/7
The area of the shaded region is 6/7
Problem 4: Inside a regular octagon with an area of 40, point P is given
For triangles ABP and EFP
The problem asks for the combined area of the two areas
To determine the area of triangles ABP and EFP
consider the auxiliary lines shown on the screen
The combined area of triangles ABP and EFP is half of the area of rectangle ABEF
1/2 of the area
The only information provided is the total area of the octagon
If we can determine what percentage
of the total octagon’s area the rectangle ABEF occupies
then half of that area will give us the desired answer
Now it's time for us to examine the octagon
The basic information we have of an octagon is
as can be seen on the screen, we can divide it into
8 equal triangular sections from the center
If we draw the rectangle ABEF onto this diagram, this is what we can see
two of the triangles that divided the rectangle into 8 sections
also is divided rectangle ABEF into two sections
Rectangle ABEF spans two of these triangular sections
which means it occupies a quarter of the total octagon’s area
Since the area of the two triangles
is 1/2 of rectangle ABEF
this is 1/4 of the octagon's total area
A ratio we can find in problem is like this
For example, in the given triangle
let's think about the two triangles' area ratio
The ratio of the areas of trangle OAC and OBC
As the two triangles have the same height
The lines AC and CB ratio would be the area ratio
Let us look at another one
When the following triangle's area is 10
What is the area of b?
The area of a triangle is the length x width x 1/2
So, when the length is the same, the height becomes the ratio of the area
Let us find the ratio of the area
If we say that the area of the given triangle is S
The area of A is 3/7 x S
The area of B + C is 4/7 x S
And as B and C have the same length and width, the area is the same
So, the area of B = C = 4/7 x 1/2
S x 2/7
As the area of the entire triangle is 10
The area of B is 20/7
Problem 5: In the following figure, AD:DB = 1:2
and AE:EC = 1:1
The three sections divided from triangle ABC
What is the ratio of the areas of triangles ADE, BDE, and BCE?
To determine the ratio of the areas of these three triangles
we need to establish a base and height as a reference
Using AC as the base, triangle ABC
can be divided into triangles ABE and BCE
The areas of triangles ABE and BCE are equal
because they share the same base and height
In the second figure, the ratio of the areas of triangles ADE and BDE is 1:2
To sum up, if we say that the area of triangle ADE is S
The area of triangle BDE is 2S
And the area of triangle BCE equals the sum of the areas of triangles ADE and BDE, which is 3S
Thus, the ratio of the areas of triangles ADE, BDE
and BCE is 1:2:3
Let us continue on with the next question
Problem 6: In the following diagram, AD:DB = 1:2
and BE:EC = 3:5
If the area of quadrilateral ADEC is 39
what is the total area of triangle ABC?
To solve, divide triangle ABC into segments based on the given ratios
BE:EC = 3:5, so is we call the area of triangle ABE, S
The area of triangle AEC is Sx5/8
the area of triangle ABE is Sx3/8
AD:DB = 1:2, so triangles ADE and BDE
divide the area of eighths of S by a ratio of 1 to 2
So, the area of triangle ADE is 3S/8 x 1/3
1S/8
The area of triangle BDE is 3S/b x 2/3
1S/4
In the problem, the area of quadrilateral ADEC is 39
So we can find S through the following equation
From this, S=52
Identifying Information You Want to Compare
Let me ask you a question
When given rectangles ABCD and ACEF as follows
What do you think the area of rectangle ACEF would be?
Comparing the two rectangles, let’s focus on triangle ACD
First, the area of rectangle ABCD is twice the area of triangle ACD
Similarly, the area of rectangle ACEF
if we divide it into two parts as shown below
is also twice the area of triangle ACD
Therefore, the area of rectangle ACEF is equal to the area of rectangle ABCD
and the area is 4 multiplied by 8
which equals 32
This way, comparing based on a specific reference and identifying ratios
is an idea frequently used in problem-solving
What we need to find is setting a reference point for comparison
Sometimes finding a reference not explicitly stated in the problem
can also be the key point
Problem 7: In the following figure, the area of rectangle ABCD is 36
the area of triangle ABP is 6
and the area of triangle ADQ is 9. Find the area of triangle APQ
To solve this problem, some people use variables
and quickly establish relationships
However, if you blindly create equations and perform complex calculations
you end up doing difficult computations
Such difficult calculations often fail to work out
and frequent mistakes happen during complex calculations
We need to identify more information
Additionally, sometimes it’s necessary to find the key points not explicitly shown in the problem
First, let’s draw a diagonal as follows
Since the area of rectangle ABCD is 36
the area of triangle ABC, which is half of the rectangle, is 18
Since the area of triangle ABP is 6, the area of triangle APC is 12
and the ratio of BP to PC is 6 to 12
which simplifies to 1:2
Let’s express this as a:2a
Since the area of triangle ADQ is 9
the area of triangle ACQ is also 9
and the ratio of CQ to DQ is 1:1
Let’s write this as b:b
Looking at this setup, since the area of rectangle ABCD is 36
3a×2b=36
and a×b=6
The area of triangle CPQ is 2a x b x 1/2
which is a x b
and a x b = 6
The area of triangle APQ, which we need to find
is the total area
minus the areas of ABP, ADQ, and CPQ
Therefore, from 36, subtracting 6, 9, and 6
the result is 15, which is the answer
Given rectangles ABCE and CEFG as follows
when the shaded area is 11, find the area of rectangle ABCD
In the given figure, if we draw a line as shown below
we see that the areas of triangles ACH and DEH are equal
Our strategy is to prove that the areas of triangles DGP and EFP are equal
If we can show this, the shaded area will equal the area of triangle ACD
and if this area is 11, the area of rectangle ABCD is confirmed as 22
We can deduce that triangles DGP and EFP are congruent
because EF and DG are both 2 and angles DPG and EPF are vertical angles and thus equal
Therefore, the area of rectangle EFGH equals the area of triangle ACH
and in the end, the shaded area equals the area of triangle ACD
As observed earlier, the area of rectangle ABCD is 22
A psychologist once said this:
If Koreans stopped comparing and stopped worrying about money, they’d be happy
Those who talk about happiness often point out comparison
as the root of unhappiness and the starting point of misery
But remember, comparison is also the root of problem-solving and the starting point for answers
the root of problem-solving and the starting point for answers
The key to problem-solving is finding what to compare and thinking in terms of ratios
For those of you creating digital content
use comparison not as an emotional concept but as a thinking skill
and apply it effectively to solve problems
Applying Comparison and Ratios in Problem-Solving Rational number: A number that can be expressed as a ratio of two integers, representing proportion and ratio Similar shapes are often created to compare and use ratios
When solving problems, it's important to set a reference and use comparison and ratios When comparing two shapes with the same height, the ratio of their bases becomes the ratio of their areas
Identifying Information You Want to Compare Comparison is the root of problem-solving and the starting point for answers
Focusing on specific aspects to compare and identify ratios What we need to find is setting a reference point for comparison
Sometimes finding a reference not explicitly stated in the problem can also be the key point