Can you find the Radius of the Yellow circle? | (Squares) | #math #maths | #geometry

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!

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

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
#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.

[MateusMuila]

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.

[hongningsuen]

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)

[timmcguire]

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 :-)

[yalchingedikgedik]

Thanks Sir
Wonderful method for solve
I’m very enjoy
With glades
❤❤❤❤❤❤

[thewolfdoctor]

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

[ABhaim]

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<14, thus r=14-7sqrt(2)

[LuisdeBritoCamacho]

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

[GillesF]

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

[lasalleman]

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.

[MrPaulc]

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.

[wackojacko]

@ 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. 🙂

[andreasproteus]

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

[marioalb]

R + R cos45° = (14 - √49)
R (1+cos45°) = 7
R = 7 / (1 + 1/√2)
R = 4, 1 cm ( Solved √ )

[quigonkenny]

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

[sergeyvinns]

Acircle is inscribed in right triangle ABD, R= (AB+AD-BD)/2=(28-14\/2)/2=7(2-\/2)=4, 1.

[murdock]

AC = 14√2; PC = 7√2 → AP = 14√2 - 7√2 = 7√2 = r(√2 + 1) → r = 7(2 - √2)

[marcgriselhubert]

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)

[LucasBritoBJJ]

DP = CP
DP = 7sqrt2

DP = DQ (tangents

r = 14 - 7sqrt

[gauravroy]

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.

[santiagoarosam]

14√2-√49√2-r√2=r → r=14-7√2 =4, 10050.... Gracias y un saludo cordial.