Can you find the Radius of the Yellow circle? | (Squares) |
HTML-код
- Опубликовано: 26 июн 2024
- Learn how to find the Radius of the Yellow circle. Important Geometry and Algebra skills are also explained: Pythagorean theorem; circle theorem; area of the square 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
• Can you find the Radiu...
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
Can you find the Radius of the Yellow circle? | (Squares) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindRadius #Square #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
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.
I am so happy for making parts of this channel 3 months ago. Thanks to your lessons , I was able to solve this on my own like it was nothing. 👍💯 you're the best.
Fantastic!🌹
Glad to hear that!
You are very welcome!
Thanks for the feedback ❤️
The point P is the centre of the outer square. So another circle of size equal to the yellow one can be drawn in the diagonally opposite corner.
So a measure taken from point T through O to centre O’ and to T’ projected on to a side of the outer square measure r + r/sqrt2 +rsqrt2 + r and so value is r= 4.10
Should be r+ r/sqrt2 + r/sqrt2 + r = 14 r= 14/(2+sqrt2) = 4.100
R + R cos45° = (14 - √49)
R (1+cos45°) = 7
R = 7 / (1 + 1/√2)
R = 4,1 cm ( Solved √ )
Alternative method using incircle formula A = RS:
1. Join BD to form triangle ABD.
BD = diagonal of square ABCD = 14sqrt2
Area of triangle ABD = (1/2)(14)(14) = 98 = A
Semiperimeter of triangle ABD = (1/2)(14 + 14 + 14sqrt2) = 7(2 + sqrt2) = S
2. Yellow circle is incircle of triangle ABD.
Hence radius R = A/S = 98/[7(2 + sqrt2)] = 7(2 - sqrt2)
Thanks for the feedback ❤️
14-7√2=7(2-√2)≈4,095≈4,1
Excellent!
Thanks for sharing ❤️
Formula for inradius of an inscribed circle in a triangle: Inradius = area/semiperimeter. Simply setup the triangle , get the semi-perimeter and divide that into the area of the triangle. Triangle area is 98. Hyptenuse of triangle is 19.7989. Tangent to circle. Semi-perimiter is 23.8994. I get an inradius of 4.10.
Excellent!
Thanks for sharing ❤️
considering the square as a grid which begins at (0,0) and ends at (14,14), we can initialize the circle's equation at (x-r)^2+(y+r-14)^2=r^2.
it gives us 3 points along the circle: (0,14-r), (r,14) and (7,7)
first 2 points plugged in the equation give us r^2=r^2. the third will give us eventually r^2-28r+98=0, which gives us two solutions, of which we only care about r
Excellent!
Thanks for sharing ❤️
My reasonning is based on 2 circles inscribed in a square: 2 circles of radius R, tangent to each other and each tangent to the sides of the square. Consecutively, there is a FORMULA to get R from the square side length (n): R = n - n√2/2. Then R = 14 - 14√2/2 = 4.100505 cm
Excellent!
Thanks for the feedback ❤️
DQ = 14-r , DB = SQRT(14^2 + 14^2) = 14*SQRT(2) , so DP = 7*SQRT(2), DP=DQ=14-r, so 7*SQRT(2) = 14-r ==> r= 14-7*SQRT(2) = 7*(2-SQRT(2)) = 4.1
Excellent!
Thanks for sharing ❤️
wait. . . DP = DQ, really?
I'm really sure DQ > DP 🍵🍵🍵
@@andryvokubadra2644 the circle is tangent to both lines DA and DB
@@franzgorg I see but the problem when DP > DQ, an elementary school's student must agree DP > DQ 😁😁😁
@@andryvokubadra2644 "tangent" means that the triangle DOQ is the identical to the triangle DOP, I don't understand why you are saying that DP > DQ. Can you please explain?
Very well explained
Glad to hear that! 🌹
Thanks for the feedback ❤️
First a rough estimate to insure against a maths gremlin: PA is 7*sqrt(2) and r is a bit less than half of that, so around 3*sqrt(2)ish or 4 ish..
Make a point in the circle called H such that PHO is a right triangle with sides r (the hypotenuse), 7-r. and 7-r.
(7-r)^2 + (7-r)^2 = r^2
49 - 14r + r^2 + 49 - 14r + r^2 = r^2
2r^2 - 28r + 98 = r^2
r^2 - 28r + 98 = 0
(28+or-sqrt(784 - 4*98))/2 = r
(28+or--sqrt(392)/2 = r
(28+or-2*sqrt(98))/2 = r
14+or-sqrt(98) = r
14+or-7*sqrt(2) = r
Unusually, the positive answer must be discarded, as it would be larger than the side length of the original square.
14 - 7*sqrt(2)
14-9.9=4.1
r=4.1 (rounded) which fits with the original rough estimate.
Excellent!
Thanks for the feedback ❤️
Here's a super simple solution you can do in your head: Draw BP, which obviously equals 7 sqrt 2 since it's the diagonal of a 7x7 square. BT must equal that because it's a second tangent from the same point to the circle. Now simply subtract that value from 14 and you have your r value. Tada :-)
Thanks for the feedback ❤️
Perpendicular distance from O to Line PE = 7 - r
Perpendicular distance from O to line PF = 7 - r
Then: 2(7 - r) ² = r² solve to find r = 7(2-√2) = 4.1
Excellent!
Thanks for sharing ❤️
AC = 14√2; PC = 7√2 → AP = 14√2 - 7√2 = 7√2 = r(√2 + 1) → r = 7(2 - √2)
Excellent!
Thanks for sharing ❤️
Possibly the simplest solution as it involves only basic trigonometry.
With the additional knowledge of (√2 and) 1/√2, then with a calculator, it is faster to first divide by √2:
7 = r(1 + 1/√2) -> r = 7 / 1.707 = 4.10
Without a calculator, the above is easier because you multiply by (2-1.414 = 0.586) in stead of dividing by 1.707.
DP = CP
DP = 7sqrt2
DP = DQ (tangents
r = 14 - 7sqrt
Thank you!
You are very welcome!🌹
Thanks ❤️
I did it in a different way. First I realised that the edge of the small square is (1+0.707) times the radius of the circle = 7
Then radius = r = 7/1.707 = 4.1 units.
Thank you
You are very welcome!
Thanks ❤️
Thanks Sir
Wonderful method for solve
I’m very enjoy
With glades
❤❤❤❤❤❤
Most welcome, dear 🌹
Thanks for the feedback ❤️
Square PFCE:
Aₛ = s²
49 = s²
s = √49 = 7
Draw AC. As ∠PFC = ∠ADC =90° and ∠FCP = ∠DCA, ∆PFC and ∆ADC are similar triangles. As the diagram is symmetrical about AC, ∆CEP is congruent to ∆PFC and ∆CBA is congruent to ∆ADC.
Triangle ∆PFC:
PF² + FC² = PC²
7² + 7² = PC²
PC² = 49 + 49 = 98
PC = √98 = 7√2
As FC = DC/2, by similarity, PC = AC/2. As AP+PC = AC, AP = PC = 7√2.
As BA and AD are tangent to circle O at T and Q respectively, ∠OTA = ∠AQO = 90°. As OT = OQ = r and ∠TAQ = 90°, then ∠QOT = 90° as well and OTAQ is a square with side lengths r.
Triangle ∆AQO:
AQ² + OQ² = OA²
r² + r² = OA²
OA² = 2r²
OA = √(2r²) = √2r
AP = AO + OP
7√2 = √2r + r
r(√2+1) = 7√2
r = 7√2/(√2+1)
r = 7√2(√2-1)/(√2+1)(√2-1)
r = (14-7√2)/(2-1)
r = 14 - 7√2 = 7(2-√2) ≈ 4.10 units
Excellent!
Thanks for sharing ❤️
Great 🎉🎉🎉
@ 7:59 we form the conjugate that the Vulcan Spock from Star Trek used too mind meld a telepathic link with any organism. ...I mean same process. 🙂
Excellent!😀
Thanks for the feedback ❤️
Very simple. Point P is the center of the big square (easy)
Then we use an orthonormal center A and first axis(AD)
The equation of the circle is (x -R)^2 + (y - R)^2 = R^2
or x^2 + y^2 -2.R.x - 2.R.y =R^2 = 0
P(7; 7) is on the circle, so: 49 + 49 -14.R -14.R +R^2 = 0
or R^2 -28.R +98 = 0. Deltaprime = 14^2 - 98 = 98 =49.2
So R = 14 -7.sqrt(2) or R = 14 +7.sqrt(2) which is rejected as beeing superior to te side length of the big square.
Finally: R = 14 - 7.sqrt(2)
Excellent!
Thanks for sharing ❤️
One could use the following path :
Area of Triangle ABD = 196 / 2 = 98
01) (14R + 14R + 14Rsqrt(2)) = 196
02) 14R * (1 + 1 + sqrt(2)) = 196
03) R * (2 + sqrt(2)) = 14
04) R = 14 / (2 + sqrt(2))
05) R ~ 4,100505
Thanks for the feedback ❤️
Acircle is inscribed in right triangle ABD, R= (AB+AD-BD)/2=(28-14\/2)/2=7(2-\/2)=4,1.
Excellent!
Thanks for sharing ❤️
Yes, no need to go thru other methods if we can assume the solver should know this basic formula.
What do formula you use?
@@andryvokubadra2644
Finding the radius of a circle inscribed in a right triangle.
The Formula is the following :
Let BD = a
Let AD = b
Let BD = c
R = (a + b - c) / 2
In the present case
a = 14
b = 14
c = 14*sqrt(2)
a + b = 28
R = [28 - 14*sqrt(2)] / 2
R = 14*(2 - sqrt(2) / 2
R = 14*(0,586) / 2
R = 7*(0,586)
R = 4,1005,,,
@@andryvokubadra2644 Формула выводится для прямоугольного треугольгика. Для других треугольнтков формула другая, радиус равен площадь разделить на периметр. Здесь сумма плошадей трёх треугольников, образованных сторонами треугольника и биссектрисами углов, сходящихся в центре окружности. высотами, в этих треугольниках. являются радиусы вписанной окружности. проведенные из её центра, перпендикулярно сторонам.
7sqrt(2)=(1+sqrt(2))r, r=7
sqrt(2)/(sqrt(2)+1)=7(2-sqrt(2)).😊
Excellent!
Thanks for sharing ❤️
14√2-√49√2-r√2=r → r=14-7√2 =4,10050.... Gracias y un saludo cordial.
Excellent!👍
Thanks for sharing ❤️
Blue square area=49
Side of the square=7
Connect P to Q (Q on AD)
So AQ=DQ=7
PQ=DF=7
Connect A to P
In ∆ APQ
AP^2=7^2+7^2
AP=AO+OP
AO=r√2 ; OP=r
2(7^2)=(r√2+r)^2
So r=14-7√2=4.1 units.❤❤❤
Excellent!
Thanks for sharing ❤️
You confused me, with your construction of PQ as a perpendicular on AD, since Q already exists and is not in a spot where PQ is perpendicular...
AQ = DQ
PQ = DF, Really? 🤔🤔🤔
An elementary school's student must agree if your assume is totally wrong 🍵🍵🍵
(14✓2-7✓2)÷ 2=7√2÷2=7/√2
My way of solution ▶
Ablue= 49 cm²
a= 7 cm
the length of the diagonal would be:
d= 7√2 cm
the diagonal of the big square is equal to:
D= 14√2 cm
D= d+2r+x
14√2= 7√2+2r+x
7√2= 2r+x..........(1)
the radius of the circle is r
if we draw a square in this cirlcle with a= r
d= √2 r
⇒
√2 r= r+x............(2)
√2 r= r+x
√2 r-r = x
r(√2-1)= x
if we put this in equation (1) we get:
7√2= 2r+x
7√2= 2r+r(√2-1)
7√2= r(2+√2-1)
r= 7√2/(1+√2)
r= 7√2*(1-√2)/(1-2)
r= 7√2*(√2-1) cm
r= 14- 7√2
r= 7(2-√2) cm
r ≈ 4,1 cm
Excellent!
Thanks for sharing ❤️
14√2/2=r+√2r...r=7√2/(1+√2)=14-7√2
Excellent!👍
Thanks for sharing ❤️
4.1 units
Excellent!
Thanks for sharing ❤️
14-r=7+(r/sqrt(2))
Thanks for sharing ❤️
r= (14√2-7√2) ÷2= 7✓2÷ 2= 7/√2
r = 14 - 7√2
Excellent
Thanks for sharing ❤️
7/✓2
Я решил вообще не используя сторону большого квадрата 🤷🏻♂️
Ответ: 14-7sqrt(2)
большое спасибо🌹
這個 圓 並不存在於 正方形 的範圍內 這可能是一個錯誤的圖形
Let's find the radius:
.
..
...
....
.....
Since AD and AB are tangents to the circle, we know that ∠AQO=90° and ∠ATO=90°. Additionally we know that ∠QAT=90° and OQ=OT=R (with R being the radius of the yellow circle). Therefore AQOT is a square with a side length being identical to the radius of the circle and we can conclude that O is located on the diagonal AC. Since we have a ratio of √2 between the length of the diagonal of a square and its side length, we obtain:
AC = √2*AD
CP + OP + AO = √2*AD
√2*CE + R + √2*AQ = √2*AD
√2*√A(CEPF) + R + √2*R = √2*AD
√2*√49 + R + √2*R = 14√2
7√2 + R + √2*R = 14√2
R + √2*R = 7√2
R*(√2 + 1) = 7√2
R*(√2 + 1)*(√2 − 1) = 7√2*(√2 − 1)
R*(2 − 1) = 14 − 7√2
⇒ R = 14 − 7√2 ≈ 4.101
Best regards from Germany
Super!👍
Thanks for sharing ❤️
STEP-BY-STEP RESOLUTION PROPOSAL :
01) OT = OQ = R
02) OA^2 = OT^2 + OQ^2
03) OA^2 = R^2 + R^2
04) OA^2 = 2R^2
05) OA = R*sqrt(2)
06) AP = 7*sqrt(2)
07) AP = OA + OP
08) 7*sqrt(2) = R + R*sqrt(2)
09) 7*sqrt(2) = R*(1 + sqrt(2))
10) R = 7*sqrt(2) / (1 + sqrt(2))
11) R ~ 4,1 ln un
12) ANSWER : Radius equal to approx. 4,1 Linear Units.
.
Best wishes from The International Islamic Institute for Studying Ancient Mathematical Thinking, Knowledge and Wisdom in Cordoba Caliphate - Al Andalus
Bravo!👍
Thanks for sharing ❤️🙏
Waaaaaaaaaay too complicated approach.
Extend the line segment FP to intersect AB (call that intersect M). Extend the line segment QO to intersect the FM segment, call that intersect N.
Now, QO is a radius length, and OP is also a radius length (r). The triangle OPN is a right triangle (intersection of perpendicular line segments), specifically a 45, 45, 90 triangle, so the hypotenuse OP = r = sqrt(2) ON.
Since we are told P is the center of the square we can say AB (or side length a) halved is the same as QO + ON
r + r/sqrt(2) = a/2
[2+sqrt(2)] r = a
r = a / [2+sqrt(2)]
r = 14/3.4142
= 4.1 u
Thanks for the feedback ❤️