Japanese Sangaku Geometry | Find area of the Green Triangle |
HTML-код
- Опубликовано: 11 сен 2024
- Learn how to find the area of the Green Triangle. Area of the circle is known. Important Geometry and Algebra skills are also explained: Pythagorean theorem; area of the circle formula; area of the triangle formula. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Japanese Sangaku Geome...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Japanese Sangaku Geometry | Find area of the Green Triangle | #math #maths #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#Sangaku #Japanese #FindGreenTriangleArea #CircleArea #EquilateralTriangle #Circle #GeometryMath #PythagoreanTheorem
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
Andy Math
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the length AB
Pythagorean Theorem
Right triangles
Intersecting Chords Theorem
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.
Real nice! 🙂
Glad you like it!
Thanks ❤️
Let x be the side of the green triangle. Since the area of the circle is π, then its radius is 1 and the side of the large triangle is 3 / √3 = √3.
Extend one side of the green triangle to the circumference. Due to rotational symmetry the extended segment consists of x + √3 / 2 + x.
So, by the chord theorem, x (x + √3 / 2) = (√3 / 2)² ⇒ 4x² + 2√3 x − 3 = 0 ⇒ x = √3 / (1 + √5) ⇒ green area = 3⁄32 √3 (3 − √5) ≈ 0.124 sq. u.
Thank you PreMath! 🙏
Excellent!
You are very welcome!
Thanks for sharing ❤️
Circle O:
A = πr²
π = πr²
r² = 1
r = 1
Draw radii OA and OE (using existing points) and draw radius OG, where C is on OG, and OG bisects AB at H.
As ∆DEF is equilateral, and O is the center of the circle it's inscribed in, OE is an angle bisector of ∠DEF. Thus ∠CEO is 30°, and ∆OCE is a 30-60-90 special right triangle. As OE = r = 1, OC = OE/2 = 1/2.
Let s = the side length of the green equilateral triangle ∆ABC. As OG is an angle bisector of ∠BCA and thus bisects ∆ABC, ∆AHC and ∆CHB are congruent 30-60-90 special right triangles, and so AH = HB = s/2 and CH = s√3/2.
Triangle ∆AHO
AH² + HO² = OA²
(s/2)² + (1/2+s√3/2)² = 1²
s²/4 + (1+√3s)²/4 = 1
s² + (1 + 2√3s + 3s²) = 4
4s² + 2√3s - 3 = 0
s = [-(2√3) ± √(2√3)²-4(4)(-3)]/2(4)
s = -√3/4 ± [√12+48]/8
s = -√3/4 ± √60/8
s = -√3/4 ± √15/4
s = √15/4 - √3/4 ==> s > 0
A = s²√3/4
A = (√15/4-√3/4)²(√3/4)
A = (15/16+3/16-2√45/16)(√3/4)
A = (18/16-6√5/16)(√3/4)
A = 18√3/64 - 6√15/64
A = (9√3-3√15)/32 ≈ 0.124
Excellent!
Thanks for sharing ❤️
The original challenge of this sangaku is to find the ratio of the sides of the two equilateral triangles, which is 1:2φ 😉
Excellent!
Thanks for sharing ❤️
Let phi be the golden ratio implies A(triangle(ABC)) = (3 * sqrt(3)/16) * (phi - 1)^2. This problem is one of many Japanese Temple Geometry Problems.
Excellent!
Thanks for sharing ❤️
The radius of the circle is 1, then OF = 1 and OC = 1/2
Let's use an orthonormal, center O, first axis parallel to (DE). We have C(0, -1/2)
The equation of (CB) is y +1/2 = -tan(60°).x or y = -sqrt(3).x -1/2
The equation of the circle beeing x^2 + y^2 = 1, at point B (intersection of the circle and (CB)) we have: x^2 + (-sqrt(3).x -1/2)^2 = 1
We expand: 4.(x^2) +sqrt(3).x -3/4 = 0. Delta = 3 -4.4.(-3/4) = 15, so x = ( -sqrt(3) + sqrt(15))/8 is the abscissa of B (the other solution is rejected as negative). Then the ordinate of B is y = -sqrt(3). [(-sqrt(3 + sqrt(15))/8] -1/2 = (3 - 3.sqrt(5))/8 -1/2 = (-1 -3.sqrt(5))/8
Finally we have B((-sqrt(3) +sqrt(15))/8; (-1 -3.sqrt(5))/8) and VectorCB((-sqrt(3) + sqrt(15))/8; (3 - 3.sqrt(15))/8)
BC^2 = (3 + 15 -6.sqrt(5) + 9 + 45 -18.sqrt(5))/ 64 = (72 -24.sqrt(5))/ 64 = (9 -3.sqrt(5))/ 8
BC is the side length of the equilateral triangle ABC, the area of this triangle is (sqrt(3)/4).(BC^2)
So the area is (sqrt(3)/4).((9 -3.sqrt(5))/8 )) = (9.sqrt(3) -3.sqrt(15)) / 32 = ((3.sqrt(3)/32).(3 - sqrt(5)).
Thanks for sharing ❤️
Let's enjoy this easter egg:
.
..
...
....
.....
First of all we calculate the radius R of the circle:
A = πR²
π = πR²
1 = R²
⇒ R = 1
May S be the side length of the triangle DEF. Since the triangle CEF is a right triangle, we can apply the Pythagorean theorem:
CF² = EF² − CE² = S² − (S/2)² = S² − S²/4 = 3*S²/4 ⇒ CF = (√3/2)*S
According to the intersecting chords theorem we obtain:
CD*CE = CF*(2*R − CF)
(S/2)*(S/2) = (√3/2)*S*[2*R − (√3/2)*S]
S²/4 = √3*S*R − 3*S²/4
S² = √3*S*R
⇒ S = √3*R = √3
May s be the side length of the triangle ABC and may M be the midpoint of AB. We can obtain CM in the same way we calculated CF:
CM = (√3/2)*s
The triangle AOM is also a right triangle, so we apply the Pythagorean theorem again:
AO² = MO² + AM²
AO² = (CO + CM)² + AM²
AO² = (CF − FO + CM)² + AM²
R² = [(√3/2)*S − R + (√3/2)*s]² + (s/2)²
1² = [(√3/2)*√3 − 1 + (√3/2)*s]² + (s/2)²
1² = [3/2 − 1 + (√3/2)*s]² + (s/2)²
1² = [1/2 + (√3/2)*s]² + (s/2)²
1 = 1/4 + (√3/2)*s + 3*s²/4 + s²/4
0 = −3/4 + (√3/2)*s + s²
s
= −√3/4 ± √(3/16 + 3/4)
= −√3/4 ± √(3/16 + 12/16)
= −√3/4 ± √(15/16)
= −√3/4 ± √3√5/4
Since s>0 the only useful solution is:
s = √3(√5 − 1)/4
Finally we can calculate the area of the green triangle:
A(ABC)
= (1/2)*AB*CM
= (1/2)*s*(√3/2)*s
= (√3/4)*s²
= (√3/4)*3*(√5 − 1)²/16
= (3√3/64)*(5 − 2√5 + 1)
= (3√3/64)*(6 − 2√5)
= (3√3/32)*(3 − √5)
≈ 0.124
Best regards from Germany
Excellent!
Thanks for sharing ❤️
Nice, tho' I think it was easier to work with a half-side of the smaller triangle.
s = (√15 - √3)/8
... is 0.26762
And then the area is merely
AREA = 1/2 2s * √3s = √3s²
... is 0.12404
In any case, its all Easter Eggs
As always
BEAUTIFUL ❤
Excellent!
Glad to hear that!
Thanks ❤️
Thank you!
You are very welcome!
Thanks ❤️
I used the trigonometric difference identity:
Sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
£=angle between verticle radius to O then to B.
Sin(30°-£)=sin30cos£-cos30sin£
Sin£=x
Besides not using a calculator (until the final numeric value .1240471...), is there a benefit to crunching more numbers?
It's good to practice more algebraic geometry.
Thank you.
Good one!
Glad to hear that!
Thanks ❤️
1) Pi*R^2 = 1*Pi ; R^2 = 1 ; R = 1 ; The Point O is the Center of the Circle and the Center of the Equilateral Triangle.
2) OD = OE = OF = 1
3) OD^2 = CD^2 + OC^2 ; R^2 = CD^2 + OC^2 ; 1 = CD^2 + OC^2
4) OC = R/2 = 1/2 = 0,5
5) CD^2 = (S/2)^2
6) So : 1 = (R/2)^2 + (S/2)^2 ; 1 = (1/2)^2 + (S/2)^2 ; 1 = 1/4 + (S/2)^2 ; S^2/4 = 1 - 1/4 ; S^2/4 = 4/4 - 1/4 ; S^2/4 = 3/4 ; S^2 = 12/4 S^2 = 3 ; S = sqrt(3)
7) Side of Triangle is equal to (sqrt(3)/2) * 2 ; S = sqrt3) ~ 1,732
8) Now, with a little help from Algebraic Geometry Calculator I can Find The Height of Small Green Triangle and Side
9) Height = 0,464
10) Side = 0,532
11) Area = (0,464 * 0,532) / 2 ; Area = 0,1234
12) Answer : The Green Triangle Area equal to approx. 0,1234 Square Units.
You know what? Looking at it, I just realized it's x=1/2 in proportion to that right triangle. But since your figuring for a larger triangle, and the side length of that equilateral triangle isn't 1, it was throwing me off, but that perfectly matches the definition of a Rhombus. And seeing the Sin function, I now perfectly understand why it is that way, thanks to this problem. A Sin function is equal to the right triangle's longer leg, should its hypotenuse be 1. That was what I originally saw in this, but it takes me a while to figure out why I was seeing it.
Although, 1/2(bh)=1/2(ab)sinC. You'd have to do the math for me, to see if it's equal.
And I just checked the math on a Rhombus calculator. It's exact.
I just see so much nuance in that, the Sin is the proportion, so when you do base times height, the sin turns into the height when multiplied by the proportion of the other side you multiplied together, as it turns into the height when you multiply it with its appropriate side.
R=1..l(big)=DF=2cos30=√3..h(big)=√3*cos30=3/2...(1/2+hg)^2+(lg/2)^2=1..(1/2+√3/2lg)^2+lg^2/4=1..1/4+√3/2lg+lg^2=1..lg^2+√3/2lg-3/4=0..lg=(-√3/2+√(3/4+3))/2=(-√3/2+√15/2)/2=(√15-√3)/4...Ag=(√15-√3)/8*(√15-√3)√3/8=(√3/64)(18-2*3√5)=(√3/64)(18-6√5)..mah,ho fatto i calcoli sullo smartphone..
Thanks for sharing ❤️
Thanks for watching this video
I was initially confused by the diagram, because the portion of the circle which contains the smaller triangle appears LARGER than the two other, similar areas of the circle. But in fact all three such areas are the same size, and so a smaller triangle of identical size as the one in the diagram would fit equally in each of the three areas...
OC can also be finded by the theorem of equilateral triangle which is 2:1 theorem
Lindo problema e resolução, parabéns!
Rather troublesome to deal with such small equilateral triangle of side 2s, so 1-s^2=(1/2+sqrt(3)s)^2=1/4+sqrt(3)s+3s^2, 4s^2+sqrt(3)s-3/4=0, 16s^2+4sqrt(3)s-12=0, s= -sqrt(3)+-sqrt(17)/8😮😮😮😮😮😮😮😮,
Thanks for sharing ❤️
@PreMath mistake🫣, 3, not 12, so my root of s looks awful.☹️
I have found the side of the equilateral triangle with sines rule, knowing that
2r = side/sin 60
Thanks for sharing ❤️
super max math
10:26 зная х проще перимножить АС и СР
Thanks for sharing ❤️
360°ARl180°=2 (ABCDEFO+2ABCDEFO-2)
Thanks for sharing ❤️
I get A = 0.124047
Excellent!
Thanks for sharing ❤️
Not getting this I agree up to oc being 0.5. Since radius is 1 ac should also be 0.5 but you are saying it is not.
I get it now that arc drawn from centre not point c
No worries!
Thanks for sharing ❤️
The final calculation ia uselessly complicated.
Christ the Lord is risen! Thanks for the math problem!
You are very welcome!
Thanks ❤️