Calculate the Radius of the circle in the square | (Important Math skills explained) |
HTML-код
- Опубликовано: 16 сен 2024
- Learn how to find the Radius of the circle inscribed in a square. Important Geometry skills are also explained: circle theorem; Pythagorean Theorem; similar triangles; Two-tangents theorem. 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
• Calculate the Radius o...
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
Calculate the Radius of the circle in the square | (Important Math skills explained) | #math #maths
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindRadius #Circle #Radius #Square #GeometryMath #PythagoreanTheorem
#MathOlympiad #SimilarTriangles #TwoTangentsTheorem
#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 #ExteriorAngleTheorem #PythagoreanTheorem #IsoscelesTriangles #AreaOfTriangleFormula #AuxiliaryLines
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #AreaOfTriangles #CompetitiveExams #CompetitiveExam
#MathematicalOlympiad #OlympiadMathematics #LinearFunction #TrigonometricRatios
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
Olympiad Question
Find Area of the Shaded Triangle
Geometry
Geometry math
Geometry skills
Right triangles
Exterior Angle Theorem
pythagorean theorem
Isosceles Triangles
Area of the Triangle Formula
Competitive exams
Competitive exam
Find Area of the Triangle without Trigonometry
Find Area of the Triangle
Auxiliary lines method
Pythagorean Theorem
Square
Diagonal
Congruent Triangles
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.
nice and simple method sir.
Thanks for liking ❤️
Here's a very quick solution. Extend AE and BC to meet at K. Triangles EDA and ABK are similar 3-4-5 triangles, with a ratio of ED:AB = 3:4. The radius of a circle inscribed in EDA is the triangle's area ((3*4)/2 = 6) divided by half the perimeter ((3+4+5)/2 = 6), or 1, so the radius of the circle inscribed in ABK is 4/3.
(Note: To derive the above formula for the inradius of a triangle, connect the incenter to the triangle's vertices and add the areas of the three component triangles.)
smart guy
Thanks ❤️
I was curious that your solution generalized would, given x=3, y=4, and z=5, be r = (y/x)*(xy/(x+y+z)), while I got r = (y/x)*((x+y-z)/2), so I did a bit of algebra.
To prove that xy/(x+y+z) = (x+y-z)/2, given the triangle with sides x, y, and z is a right triangle, thus x^2 + y^2 = z^2:
Let (x+y) = a
Then xy/(x+y+z)
= xy/(a+z)
= xy(a-z)/((a+z)(a-z))
= xy(a-z)/(a^2 - z^2)
= xy(x+y-z)/((x+y)^2 - (x^2 + y^2))
= xy(x+y-z)/(x^2 + 2xy + y^2 - x^2 - y^2)
= xy(x+y-z)/2xy
= (x+y-z)/2
Since your equation xy/(x+y+z) was for the inradius of a right triangle, that means my (x+y-z)/2 also works to find the inradius of a right triangle. Certainly a known formula, but I think it's cool that I arrived at it by accident!
Even simpler. The radius of the incircle of a right angled triangle is r = (a + b - c)/2 (c = hyp). The three sides of your extended triangle are 4, 16/3 and 20/3. Result follows.
After having extended BC and AE to point P, we can notice that ∠DAP and ∠APB are alternate angles, same for ∠DEA and ∠EAB.
Therefore, △DAE ~ △BPA and their corresponding legs AB and ED are in a 4:3 ratio. So BP = (4) 4/3 = 16/3 and AP = (5) 4/3 = 20/3.
The inradius _r_ of the incircle in a triangle with sides of length _a, b, c_ is given by _r = √((s − a) (s − b) (s − c) / s)_ where _s = ½ (a + b + c)._
In our case, we then have s = ½ (4 + 16/3 + 20/3) = 8 and thus r = √((8 − 4) (8 − 16/3) (8 − 20/3) / 8)) which simplifies to 4/3 units.
Thanks ❤️
AE=5 so FE=r+1. Then OE²=2r²+2r+1. Now, from point E, draw a line parallel to CB; it will intersect OH at point I. OE²-CH²=OI²; 2r²+2r+1-(4-r)²=OI²=(r-1)², therefore, r=4/3
Nice . I too solved in this method.❤
Thanks ❤️
Lots of solutions!
Appreciate the enthusiasm on comments.
Thanks ❤️
Nice! CP = k → tan(δ) = 3/4 = 4/(4 + k) → k = 4/3 → CP = 20/3 = (4 - r) + (16/3 - r) → r = 4/3
Thanks ❤️
Thanks for sharing with us
Thanks ❤️
Solution by tangent double angle formula, tan(2Θ) = (2tan(Θ))/(1 - tan²(Θ)): Note that
Thanks❤️
GB = r so AG = 4 - r => AF = 4 - r => FE = 1 + r. EO^2 = (r - 1)^2 + (4 - r)^2. Also, EO^2 = r^2 + (1 - r)^2. Equate the two and simplify.
easy solutionb:
scalability factor needs to be consdiered.
for a right angle tringle with sides 3,4,5 -> incraidus is 1 (sum of sides -hypotenous)/2 = (3+4-5)/2=2/2=1.
for triangle 3,4 with right angle, hypotenous is 5 . with extra extended side 1, , we have 4/3 as extended base.
so by 3 is a factor. so scale up units by 3 times.
now, 3,4,5 becomes 9,12,15 then bse extend = 4, remaing base = 4*3=12 m, totla base 16.
height 4 * = 12, 12:16 whih is 4 scale for 3:5
So scal eecome 1/3 * 4 = 4/3.
keeping scale aside, now biger triangle sides are 3,4.
so biggher hypotenous becomes 5 and incircle become 1.
when scale is kept back, 1 * 4/3 = 4/3 is the required inradius
ability to imagine scaling of shapes
and remmebring 3,4,5 general properties with inradius being 1. is the bsic logic.
we can compute all without putting pen on paper
Completing the triangle was brilliant!
Thanks ❤️
I have much more simple ... less than 1 minute to solve
(0,0) lower right corner
(r,r) center of circle (I mirror x axis)
slope of line = -3/4 slope of radius to line +4/3
x=r+r.3/5; y=r+r.4/5 = (8r/5,9r/5) intersection with line
y=4-3x/4 equation of line
9r/5 = 4 - 24r/20
3r = 4 r = 4/3
Great! I solved it by the equation of the circle intersecting with the line. Given the number of points available I obtained a series of simultaneous equations, which I used to find the unknowns: the center of the circle (8/3, -8/3), the point of tangency (28/15, -8/5) and the radius (4/3).
very nice sharing Sare❤❤❤
Thanks ❤️
angle DEA = EAB, since DC and AB are parallel. Segment OA bisects angle EAB. tan(angle FAO) = r/(4-r) = tan(1/2 atan (4/3)
I went a different way.
Square's area is 16.
The 3,4,5 triangle part has area of 6, so the quadrilateral ECBA is 10.
This may be split into four triangles: AOE, BOC, AOB, and EOC - all of which have r as their height or part of their height.
Their areas are 2.5r, 2r, 2r and 2-0.5r respectively.
6r + 2 for an area of 10.
6r = 8
r = 8/6 = 4/3
I suppose you could call r 1.333..., but 4/3 is more accurate.
Seeing that the answer was 4/3, I was curious if it could be generalized as r=y/x (x=3, y=4 in this example). It can't quite, but I found something interesting. The generalized form can be written as r=(y/x)*((x+y-sqrt(x^2+y^2))/2). Note that y/x is part of it. Focusing on Pythagorean triples, i.e. (3,4,5) mapping to (x,y,z), we can replace sqrt(x^2+y^2) with z, giving us r=(y/x)*((x+y-z)/2). For the (3,4,5) ,triple in this example, (x+y-z)/2 calculates to 1 leaving us with r=(y/x)*(1)=4/3. For the next triple I checked, (5,12,13), (x+y-z)/2 calculated to 2, and the one after that, (8,15,17), calculated to 3. I was hoping for a Fibonacci progression, but (7,24,25) is also 3. (9,40,41) is 4. (11,60,61) and (12, 35, 37) are 5. Continuing down the line (ordered now by increasing x), it looks like an interesting pattern is going to emerge, but alas:
(3,4,5) => 1
(5,12,13) => 2
(7,24,25) => 3
(8,15,17) => 3
(9,40,41) => 4
(11,60,61) => 5
(12,35,37) => 5
(13,84,85) => 6
(15,112,113) => 7
(16,63,65) => 7
(17,144,145) => 8
(19,180,181) => 9
(20,21,29) => 6 (soiled it!)
(20,99,101) => 9
(21,220,221) => 10
Still fun that all of the triples I checked produce integers from (x+y-z)/2. Not a single ".5" in sight. Don't know if that will hold for all of the triples, but I'd like to think it would.
Edit: It does! Given:
a. Pythagoras (x^2 + y^2 = z^2)
b. odd +/- odd = even
c. even +/- even = even
d. odd +/- even = odd
e. odd^2 = odd
f. even^2 = even
We can see that there are 4 possibilities:
1. x and y are odd, therefore z is even, and therefore (x+y-z) = odd + odd - even = even - even = even
2. x and y are even, therefore z is even, and therefore (x+y-z) = even + even - even = even - even = even
3. x is odd and y is even, therefore z is odd, and therefore (x+y-z) = odd + even - odd = odd - odd = even
4. x is even and y is odd, therefore z is odd, and therefore (x+y-z) = even + odd - odd = odd - odd = even
Since all four options are even, (x+y-z)/2 will always be an integer!
I brute forced the solution. Area of square =4*4=16 which equals the sum of ADE, AGO, AFO, EOF, OGHB, EOHC i.e sum of areas of 4 triangles, a square and a trapezoid, where GB=OG=r, AG=AF=CH=4-r, AE=5, EF = 5-(4-r)=r+1, EC=1. Solve for r.
Thanks ❤️
From the picture one knows angle FAO: it has tangent 1/2. The tangent is equal to r/(4-r). Solve the linear system => r =4/3 .
this time i did not calculate the intersection of angle bisectors
but the centerpoint whose perpendicular distance must be equal to
r:
10 l1=3:l2=4:dim x(4),y(4):sw=l1/1E2
20 x1=0:y1=l2-l1:x2=l2:y2=l2:r=sw:dx=x2-x1:dy=y2-y1:nk=dx^2+dy^2:goto 90
30 xm=l2-r:ym=r:zx=dx*(xm-x1):zy=dy*(ym-y1):k=(zx+zy)/nk
40 dxl=k*dx:dyl=k*dy:xl=x1+dxl:yl=y1+dyl:
50 dgu1=((xm-xl)^2+(ym-yl)^2)/l1^2:dgu2=(r/l1)^2
70 dg=dgu1-dgu2
80 return
90 gosub 30
100 r1=r:dg1=dg:r=r+sw:if r>10*l1 then stop
110 r2=r:gosub 30:if dg1*dg>0 then 100
120 r=(r1+r2)/2:gosub 30:if dg1*dg>0 then r1=r else r2=r
130 if abs(dg)>1E-10 then 120
140 print r:mass=4E3/nk:goto 160
150 xbu=x*mass:ybu=y*mass:return
160 x(0)=0:y(0)=0:x(1)=l2:y(1)=0:x(2)=x(1):y(2)=l2:x(3)=0:y(3)=l2:x(4)=x(3):y(4)=l2-l1
170 xba=0:yba=0:for a=0 to 4:ia=a:if ia=4 then else 190
180 gcol9:ia=0
190 x=x(ia):y=y(ia):gosub 150:xbn=xbu:ybn=ybu:gosub 200:goto 210
200 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
210 next a:x=x(4):y=y(4):gosub 150:xba=xbu:yba=ybu:x=x(2):y=y(2):gosub 150:xbn=xbu:ybn=ybu
220 gcol 9:gosub 200:x=xm:y=ym:gosub 150:gcol 10:circle xbu,ybu,r*mass
230
1.33333333
>
run in bbc basic sdl and hit ctrl tab to copy from the results window
çok teşekkürler....süper anlatım....zevkle izliyorum...
(2r/(4-r))/(1-r^2/(4-r)^2)=2r/4-r / 16-8r/(4-r)^2=2r/4-r (4-r)^2/8(2-r)=r(4-r)/4(2-r)=cot A=4/3, 3r(4-r)=16(2-r), 3r^2-28r+32=0, (3r-4)(r-8)=0, r=8, rejected, r=4/3.😊
Thanks ❤️
Once you have the sides of the right triangle in which the circle is inscribed, the radius can be easily calculated with the formula r=(a*b)/(a+b+c). So r= (4*5.33333)/(4+5.33333+6.66666)= 4/3.
Thanks
The method used in this video shows that (a − r) + (b − r) = c or r = ½ (a + b − c), the other formula for the radius of a circle inscribed in a right triangle 😉
La retta y=(3/4)x+q è tangente alla circonferenza x^2+y^2=r^2...impongo la condizione di tangenzaΔ=0..r=(4/5)q..inoltre yQ+r=4,con Q=(r,yQ),per cui 4-r=(3/4)r+q...le 2 equazioni danno r=4/3...(Q Is A)
Thanks ❤️🌹
Yay! I solved the problem.
Bravo ❤️
Radius = 4/3
4r + 4r + 5r + (4 - r) = 20
12r = 16
r = 16/12
r = 4/3
Ma solution : angle = arc-tan(¾) ( ( cos(angle) )^2 )*2 = 1.28 1.28 est la solution Bonne journée
You don't need the Pythagoras Theorem for the right triangle, you can use the similar triangles to calculate the hypotenuse just like you did the horizontal side.
Start with APB similar EAD and EA = 5. The rest is left as an exercise for the reader.
Thanks ❤️
Very nice sir 🎉
With due respect
I have 2 results on squaring and cubing of numbers how can I share it with you 🙏
La retta y=(3/4)x+q è tangente alla circonferenza x^2+y^2=r^2...impongo Δ=0,e yQ+r=4 con Q=(r,yQ)...risulta q=(5/4)r...( condizione di tangenza) e yQ=(3/4)r+q...in sintesi r=4/3
Thanks dear❤️
Sorry I am sharing results here Firstly result on squares
Ni(number whose square we have to find )
N(whose square we know)
Such that n
Very clever, I didn't saw the extension to point P 😢
Great!
Thanks ❤️
First like and comment❤
Thanks ❤️
rad.of circle=64/41 unit.
Good night sir l hope you have a happy life
Thanks dear ❤️
You are the best!
Sorry to hear about the terrible things happened at the University of Nevada yesterday
Our thoughts and prayers for the victims at UNLV
🙏☮️
@@PreMath Certainly sir
When I go to ice cream shop I ask for a chocolate chip Two Tangent Theorem scoop with radius of 1 & 1/2 inches too make sure I get my money's worth. 🙂
And I thought that *I* was a smartarse :)
Mathematics is everywhere...
Thanks ❤️
i think i have easier way
tan(DAE)=3/4
tan(FAG)=4/3
FAG=2.OAG
suppose OAG=@
tan 2@=2.tan@/1-tan^2 @
then tan@=1/2
in triangle OAG>>>>>tan@=r/4-r=1/2
then r=4/3
Thanks ❤️