Can you find the chord AB length? | (Circle) |
Вставка
- Опубліковано 18 вер 2024
- Learn how to find the chord AB length. Important Geometry and Algebra skills are also explained: Pythagorean theorem; Point-line distance formula; perpendicular bisector 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
• Can you find the chord...
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 chord AB length? | (Circle) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindAB #Chord #PointLineDistanceFormula #Circle #PerpendicularBisectorTheorem #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.
An other possible solution could be applying the tangent secant theorem first from point C
CB*CA = 3*3
And then from point P
PA*PB = 1*1
Setting AB=x , CA=y and being CP=2√ 10 as shown in the video
we have 2 equations with 2 unknowns
(X+y)*y=3*3
(2√ 10 - y)*(2√ 10 - X - y)=1*1
1) xy + y² = 9
2) 40 - 2√ 10x -4√ 10y + xy + y² = 1
substituting 1) in the 2)
24 - √ 10x - 2√ 10y = 0
x = (24 - 2√ 10y)/√ 10 and substituting in the 1) we have the quadratic equation
√ 10y² - 24y + 9√ 10 = 0
for y = (12 - 3√ 6)/√ 10
x = 6/5√ 10
y≈1.471, x≈4.648 ok
y≈6.119, x≈−4.648 (rejected)
At 0:55 - one cannot say that the point C is the origin of the coordinate system. All that is given is that the center O of the circle has coordinates (3, 3) in some coordinate system. It may require some more justification.
Yes, that should be specified! Knowing O alone with no notion of scale is of no help.
Angle of chord AB :
tan α = 2/6 --> α = 18,43495°
Apothem of chord AB:
a = (3-½2).cosα = 6/√10 =1,897 cm
Pytagorean theorem:
(½c)²+a² = R²
(½c)² = 3²- 3,6 = 5,4 = 3*9/5
c = 6√3/√5 = 4,6476 cm ( Solved √ )
Let's find the length of AB:
.
..
...
....
.....
From the given coordinates of the points O and C we can conclude that the radius of the circle and the side length of the square turn out to be r=3 and s=6, respectively. We also can conclude that the line CP is represented by the following function:
y = (yP − yC)*(x − xC)/(xP − xC) + yC = (2 − 0)*(x − 0)/(6 − 0) + 0 = x/3
The circle on the other hand is represented by the following function:
(x − xO)² + (y − yO)² = r²
(x − 3)² + (y − 3)² = 3² = 9
Therefore we can calculate the coordinates of the points of intersection like this:
(x − 3)² + (y − 3)² = 9
(x − 3)² + (x/3 − 3)² = 9
x² − 6*x + 9 + x²/9 − 2*x + 9 = 9
(10/9)*x² − 8*x + 9 = 0
10*x² − 72*x + 81 = 0
x = [72 ± √(72² − 4*10*81)]/(2*10) = [72 ± √(5184 − 3240)]/20 = (72 ± √1944)/20 = (72 ± 18√6)/20 = (36 ± 9√6)/10
y = x/3 = (12 ± 3√6)/10
According to the sketch we can conclude:
xA = (36 − 9√6)/10 yA = (12 − 3√6)/10
xB = (36 + 9√6)/10 yB = (12 + 3√6)/10
Now we are to calculate the length of AB:
AB² = (xB − xA)² + (yB − yA)² = (18√6/10)² + (6√6/10)² = (6√6/10)²*(3² + 1) = (3√6/5)²*10
⇒ AB = (3/5)*√6*√10 = (3/5)*2*√3*√5 = (6/5)√15 ≈ 4.648
Best regards from Germany
That's the one, congratulations ! The example finally gives the chance to use about Analytic Geometry !
....to talk about A.G.
Alternative solution:
Label angle PCD= alpha and OCP = beta
We have: beta= angle OCD-alpha= 45 degrees-alpha
--> tan (beta)= tan(45- alpha) = ((1-1/3)/(1+1/3))=1/2
-> CM= 2OM=2d
-> sq OM +sqCM=sqCO
sqd+sq(2d)=sq(3sqrt2)
5sqd=18-> sqd=18/5
-> sqAM=9-(18/5)=27/5
AM=sqrt(27/5)
AB=2.(sqrt(27/5)=4.65 units😅😅😅
Second alternative:
Focus on the triangle CEP:
We have: CE=6sqrt2, CP=2sqrt10, and PE=4
So area of the triangle CPE
= 1/2 CP.CEsin beta=12
-> sin beta=1/ sqrt5
-> OM/CO=d/3sqrt2=1/sqrt5
-> d= 3.sqrt (2/5)-> sqd=18/5
-> sqAM= 9-18/5=27/5
AB=2. sqrt(27/5)=4.65
We use an orthonormal center C and first axis (CD)
The equation of the circle is(x -3)^2 + (y -3)^2 = 3^2 or x^2 + y^2 -6.x - 6.y +9 = 0
The equation of (CD) is y = x/3. At the intersection we have x^2 + (x^2)/9 -6.x - 2.x +9 = 0 or (10/9).(x^2) -8.x + 9 = 0. Deltaprime = 16 - 10 = 6
So x = (4 -sqrt(6))/(10/9) = (9/10).(4 - sqrt(6)) which is the abscissa of A, or x = (4 + sqrt(6))/(10/9) = (9/10).(4 + sqrt(6)) wich is the abscissa of B
A and B are on (CD), then we have A((9/10).(4 - sqrt(6)); (3/10).(4 - sqrt(6))) and B((9/10).(4 + sqrt(6)); (3/10).(4 + sqrt(6)))
Then VectorAB((9/10).(2.sqrt(6)); (3/10).(2.sqrt(6))) = (3/10).(2.sqrt(6)).VectorU, with VectorU(3; 1)
Then AB = (3/10).(2.sqrt(6)).norm(VectorU) = (3/10).(2.sqrt(6).sqrt(10) = (6/5).sqrt(15).
(The formula giving the distance point - line is not necessary here, although I use it frequently.It's only a problem of intersection circle -line)
You didn’t set up the origin is at C. You assumed that when *solving* the equation.
From the set up it is possible C is not the origin. It could be for instance (1, 1) or (-20, -20).
Therefore everything you do is invalid.
Unless you make C is the origin in the set up.
solve for the coord of pts A and B using coord geometry
(x--3)^2 + (y`--3)^2 = 3^2
and y = 1/3x
then use the distance formula
Yes, I used this method as well. The mathematics are a little trickier this way, but it gives the right answer alright!
CP = sq.rt. 2^2 + 6^2 (pythagoras)
CP = 6.3246.
Drop perpendicular from O to cross CP at point R and bisect CD at point S.
RS will equal half of PD = 1.
Therefore OR = 3 - 1 = 2.
Triangles CDP & ORM are similar.
So CP / 6 = OR / OM.
6.3246 / 6 = 2 / OM.
OM = 1.8974.
Bisecting chord theory.
(OM + r) (r - OM ) = AM x MB.
(1.8974 + 3) (3 - 1.8974) = AM^2. (AM = MB)
4.8974 x 1.1026 = AM^2.
AM^2 = 5.4.
AM = 2.324.
AB = 2 x 2.324.
AB = 4.65.
Why CDP and ORM are similar?
@@soli9mana-soli4953
Let angle CPD = alpha.
Then angle CRS = alpha ( PD // RS.)
Then angle ORM = alpha. (directly opposite.)
The other acute angles in both triangles are 90 minus alpha.
@@montynorth3009thank you, I saw
Draw a line from O to C. Distance works out to 4. 2626 at 45 degrees. Find angle PCD which is 18.43. Subtract from 45 degrees. Leaves angle OCM of 26.57 degrees. Find the sine of OC which turns out to be d = 1.9066. Then just use the general chord formula . I use: chord length = 2 × √(r2 − d2) = 4.6324 Pretty close to suggested solution here. Keep everything simple as you can, I always say. Yeah, the suggested solution procedure is elegant, but it's time-consuming.
Draw a line from circle center down, vertical to the bottom of the square. That wine will be two units above the diagonal line and one unit below it. Then draw a line from circle center to the diagonal line. This will give you a right triangle with the hypotenuse being 2 in the longside being X and the short side being X/3. Solving for Pythagoras that yields X equal to 3 *sqrt(2/5). This sets up another Pythagorean triangle with the hypotenuse being 3. Then all you need to do is solve for the unknown member which will yield. 3*sqrt(15)/5. Multiply that times two and you get your solution which in approximated decimal format is 4.65.
Point C as the origin was not given in the problem.
Assuming c as origin
You're absolutely right-this problem would have been easy to solve if the initial drawing had indicated the side length of the square as 6 units or provided the coordinates of point C as (0,0).
you have to determine it yourself 😅 it's not that difficult to use your brain,come on!!
The circle eqn is:
(X-3)²+(Y-3)²=9
Plug X/3 in for Y, and solve for X.
After lots of Algebraic calcs (including quadratic eqn) we find
X=1.4 & 5.8
Plugging back into Y=X/3, we get coordinates of A, B:
A (1.4, 0.467)
B (5.8, 1.93)
Enter Mr Pythagoras:
(AB)²=(5.8-1.4)² + (1.93-0.467)²
AB=4.64
After obtaining the equation of the line CP, I proceeded to equate it to the equation of the circle to obtain the XY coordinates of points A & B to find the displacement between them.
tg(∠PCD) = 1/3. tg(∠OCD) = 1. tg(∠OCP) = 1/2. OM = x. In ▲OCM: x² + (2x)² = (6/√2)². x² = 18/5.
In ▲AOM: AM = √(3² - 18/5) = 3√3/√5. AB = 6√3/√5 = 6√15/5.
El radio es 3; M es el punto medio de AB y ésta corta el diámetro vertical en G ---> CP =√(6²+2²) =2√10 ---> Razón de semejanza entre GMO y PDC =s =OG/CP =[3-(2/2)]/2√10=√10/10---> MO=6s=3√10/5 ---> Potencia de M respecto a la circunferencia =AM²=[3-(3√10/5)]*[3+(3√10/5)]---> AM =3√15/5---> AB=2*AM =6√15/5.
Gracias y saludos
Well ... then there is the line-intersects-circle method, converting the center of the circle to (0, 0)
y = x/3 - 2. is the line of the AB diagonal
y² = 3² - x² ... is the circle itself
square the first formula, and make it equal to the second ...
x²/9 - 4x/3 + 4 = 9 - x². ... then rearrange the bits and normalize to non-fractional
10x² - 12x - 45 = 0
Solve the quadratic the usual way. x = [2.8045, -1.6045). and plug those in
to get the pair of y's. y = [-1.0652, -2.5348]
There we go. Now pythagoras of the 'delta x, delta y' to get the AB length
AB = √( (2.8045 + 1.6045)² + (-1.0652 + 2.5348)² )
AB = 4.6475
Which is the same answer as The Professor's.
1/ Draw the diameter RS//CD and intersects CP at point T.
We have PS= 1-> ST= 3 ( slope= 1/3😊)
2/ Focus on the triangle OMT.
sq OM=sqOT-sqMT
Because of similarity, OM= MT/3
--> sqd+sq(3d)=sq6
10sqd= 36-> sqd=36/10
-> sq AM =9-36/10=54/10
AM = 3.sqrt (3/5)
AB = 6. sqrt(3/5)=4.65😅😅😅
The answer is [6*sqrt(15)]/5. And golly that is a really clever use of coordinate and plane geometry. I shall use that for practice!!!
My way of solution ▶
The equation of this circle is:
(y-y₀)²+(x-x₀)²= r²
here is r= 3
y₀= 3
x₀= 3
⇒
(y-3)²+(x-3)²= 9
The equation of the linear graph:
y= ax
f(2)= 6
x= 6, y= 2
⇒
2= 6a
a= 1/3
⇒
y= x/3
ii) Let's find the points where the graph touches the circle
y= x/3
and
(y-3)²+(x-3)²= 9
(x/3 -3)²+(x-3)²= 9
(x-9)²/9 + x²-6x+x²= 9
10x²-72x+81=0
⇒
x₁= 1,3954 ;
y₁= 0,4651
y₂= 5,5348
and,
x₂= 5,8045 ;
y₁= 4,0653
y₂= 1,9347
⇒
x₁= 1,3954 ; y₁= 0,4651 and
x₂= 5,8045 ; y₂= 1,9347
According to the Pythagoras theorem:
[AB]²= (Δx)²+(Δy)²
Δx= 5,8045 - 1,3954
Δx= 4,4091
Δy= 1,9347 - 0,4651
Δy= 1,4696
⇒
[AB]= √4,4091²+1,4696²
[AB]= √21,599...
[AB]= 4,64756 length units
Another mind using problem.
How do you know that point C is the origin?
Professor, eu calculei a equação da circunferência, depois calculei a equação da reta, fiz a intersecção entre a circunferência e a reta, encontrei os pontos e depois calculei a distância entre os dois pontos. Show !!!!
ALTERNATE SOLUTION :
01) Using the Distance from Point O : (3 ; 3) to a Straight Line CP.
02) Area of Triangle [CDP) = 6 * 2 / 2 = 12 / 2 = 6 sq un
03) Line CP Length = sqrt(4 + 36) = sqrt(40) = 2sqrt(10)
04) d(O , CP) = h
05) Area of Triangle [CDP] = (h * 2qrt(10)) / 2 = 6
06) 6 = h * sqrt(10)
07) h = 6 / sqrt(10)
08) h = 6sqrt(10) / 10 ; h = 3sqrt(10) / 5
09) h ~ 1,897
10) 3 + 1,9 = 4,897
11) 3 - 1,9 = 1,1026
12) X * X = 4,897 * 1,1026. X = MA = MB ; being M the Middle Point from A and B.
13) X^2 = 5,39943
14) X = sqrt(5,39943)
15) AB = 2X = 2 * sqrt(5,329)
16) AB ~ 4,647 Linear Units
R=3???..h=√18sin(45-arctg2/6)=6/√10....AB=2√(9-36/10)=2√5,4=4,64758..
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Green Circle Equation : (X - 3)^2 + (Y - 3)^2 = 9
02) CP Straight Line Equation : Y = X / 3. Slope : m = 2 / 6 = 1 / 3
03) Calculating the Coordinates of the two intercepting Points, A and B :
04) (X - 3)^2 + (X/3 - 3)^2 = 9
05) Two Real Solutions :
a) X = 1,39546 and Y = 0,46515
b) X = 5,80454 and Y = 1,93485
06) Distance from A to B : d(A, B) = 4,64758 lin un. Using "Distance Formula" in Wolfram Alpha!!
Therefore,
OUR BEST ANSWER IS :
Line AB Length equal to approx. 4,65 Linear Units.
Funny, I got 4.6467 inches. When converting, that would be 11.8026 cm.
This is actually not Premath. This is higher maths.
It's good though!
I must be a genius because it took centuries for people to come up with the ideas to solve problems like this. I solved in less than ten minutes. I didn't solve it cuz I needed to apply to some tangible things in this world. That would would've bored me half to death. Nope, I solved because it's an escape from the chaos of reality to a world that doesn't really exist. Not easy, but more fun! 🙂
😊
Now that you solved it, go treat yourself to an Ice cream cone.
@@wackojacko3962Now that you have solved it, can you tell me how one can prove that point c must have the co ordinate(0,0)?
Well, I can say proving that it doesn't exist is irrational. 🙂
I don't like working in terms of square roots. They are meaningless when you are trying to find length, distance, or area. Nobody cuts a board to 5 × square root of 3. Or drives 5 × square root of 2 miles. You dont buy a square root of 5 acres. Also, doing these problems it is more likely outside of the school classroom you will use a calculator, and even if you input a number with the square root symbol, your answer will be in decimal form. Therefore I like to work in decimal form using the numbers carried out as many decimal places as the calculator will let me and in my final answer round off to the nearest decimal of the accuracy required for what I am trying to find.
Gracias señor bonito ejercicio