Can you find the chord AB length? | (Radius) |

Learn how to find the chord AB length. Important Geometry and Algebra skills are also explained: Pythagorean theorem; circle theorem; 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? | (Radius) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindAB #Chord #Semicircles #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.

[@sorourhashemi3249]

Thanks. I love it. Challenging❤

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️

[@phungpham1725]

1/ Focus on the triangle AOP of which the perimeter = 5+5+4= 14 so, by Heron theorem: the area of triangle APO = sqrt( 7x2x2x3)=2sqrt21 -> 1/2 AC.OP= AC.5/2= 2sqrt21-> AC= 4sqrt21/5 -- > AB = 8 sqrt21/5😅😅😅

[@imetroangola4943]

It's great that you saw this solution! 👏🏻👏🏻👏🏻

[@phungpham1725]

@@imetroangola4943 Thank you so much!

[@egillandersson1780]

I did it the same way !

[@marioalb9726]

Pytagorean theorem: R² = 3²+r² = 3²+4² --> R= 5cm Cosine law for isosceles triangle AOP: r² = 2R²(1-cosα) cosα = 1- r²/ 2R² = 1-4²/(2*5²) α = 47,15636° Chord AB: c = 2 R sinα = 7,332 cm (Solved √)

[@marioalb9726]

Pytagorean theorem: R² = 3²+r² = 3²+4² --> R= 5cm Cosine law for isosceles triangle AOP: r² = 2R²(1-cosα) cosα = 1- r²/ 2R² = 1-4²/(2*5²) α = 47,15636° β = ½(180°-α) = 66,42182° Chord AB: c = 2 R sinα = 7,332 cm c = 2 r sinβ = 7,332 cm

[@montynorth3009]

OP = 5 as calculated from 3,4,5 triangle. Triangle APO with height y. y^2 = 4^2 - x^2. Also, y^2 = 5^2 - (5 - x)^2. Therefore 16 - x^2 = 25 - 25 +10x - x^2. 16 = 10x. x = 1.6 Then y^2 = 4^2 - 1.6^2 in triangle ACP. y^2 = 13.44. y = 3.666. AB = 2y = 7.332.

[@marcgriselhubert3915]

We use an orthonormal center O and first axis (OQ). The radius of the yellow semi circle is 5 (see triangle OQP), so its equation is x^2 + y^2 = 25. The equaton of the red circle is (x -3)^2 + (y -4)^2 = 16 or x^2 +y^2 - 6.x -8.y +9 = 0. We surch the intersection. By difference we have 6.x + 8.y -34 = 0 or y = (-3.x +17)/4, then we replace y by this value in x^2 + y^2 = 25 and obtain 25.x^2 -102.x -111 = 0. Deltaprime = 5376 = (2^8).21 So x = (51 -16.sqrt(21))/25 which is the abscissa of A, or x = (51 + 16.sqrt(21))/25 which is the abscissa of B We obtain the ordinates of these two points with y = (-3.x +17)/25. Finally: A((51 -16.sqrt(21))/25; (68 +12.sqrt(21))/25) and B((51 + 16.sqrt(21))/25; (68 - 12.sqrt(21))/25) Then VectorAB((32.sqrt(21))/25; (-24.sqrt(21))/25) = (8.sqrt(21))/25.Vector U with VectorU(4; -3) and norm(VectorU) = 5 Then AB = [(8.sqrt(21))/25].5 = (8/5).sqrt(21).

[@prossvay8744]

In circle Point O Let R is the Radius of cirmicircle PQ=4 In ∆ OPQ OQ^2+PQ^2=OP^2 3^2+4^2=QP^2 So PQ=5 (4+x)((4-x)=(OA)OB) (1) x(10-x)=(OA)(OB) (2) (1)&(2) 16-x^2=10x-x^2 So x=16/10=8/5 M middle AB MP^2+OA^2=AP^2 (8/5)^2+OA^2=4^2 So OA=4√21/5 So AB=2(4√21/5)=8√21/5 units=7,33units.❤❤❤

[@michaeldoerr5810]

I am glad that I have learned of a geometry problem that is tricky if you do not know how to use clever geometry with chord length theorems!!! Also for yesterday's I have noticed that PreMath has hearted a ton of comments offering advanced explanations than the straightforward one given in yesterday's video. I think that PreMath could make a playlist of comments that showcase solutions that have been hearted. Maybe a compilation even!!!

[@jamestalbott4499]

Thank you!

[@MrPaulc222]

Because OQ+ is a tangent tine to the small circle, PQO is a right triangle. PQ = 4 because it's r. OQ = 3 (given). OP = 5 because it is the hypotenuse of a 3,4,5. It it also R. Call AB 2x. Call its midpoint, M. It is a chord in both circles. In the red circle, intersecting chords give y(8 - y) = x^2 where y is the distance from M to the red circumference along a line MO. PM is 4 - y. OM is 5 - (4 - y) = y + 1. Imagine a full yellow circle. 2R = 10. Intersecting chords give (5 + y + 1)*(4 - y) = x^2 Simplify a bit for (6 + y)(4 - y) = x^2. In the red circle the chords are y(8 - y) = x^2. Therefore, (6 + y)(4 - y) = y(8 - y). Expand and tidy up: 24 - 2y - y^2 = 8y - y^2 24 - 2y = 8y. 24 = 10y. y = 2.4 or 12/5 if preferred. Red circle is now (12/5)*(8 - (12/5)) = x^2. Tidy up to (12/5)(28/5) = x^2. (336/25) = x^2 sqrt(336)/(sqrt(25)) = x, so sqrt(336)/5. As AB = 2x, AB = 2*sqrt(336)/5. In decimal, that approximates to 7.33. I now looked. Yes, our labelling differed, as you might expect, and I threw in a couple of extraneous calculations, but the essentials were all there.

[@quigonkenny]

As PQ is a radius of circle P, PQ = 4. As PQ = 4 and OQ = 3, ∆OQP is a 3:4:5 Pythagorean triple right triangle and OP = 5. As OP is a radius of semicircle O, OA = OB = OP = 5. Let ∠POB = θ. As AB is a chord for both semicircle O and circle P, and as a line that bisects a chord perpendicularly must be collinear with the center of the circle, OP bisects AB and thus ∠AOP = ∠POB = θ. By the law of cosines: cos(θ) = (OB²+OP²-PB²)/2(OB(OP) cos(θ) = (5²+5²-4²)/(2(5²)) cos(θ) = (25+25-16)/50 = 34/50 = 17/25 cos(2θ) = 2cos²(θ) - 1 cos(2θ) = 2(17/25)² - 1 cos(2θ) = 2(289/625) - 1 = 578/625 - 1 cos(2θ) = -47/625 AB² = OA² + OB² - 2OA(OB)cos(2θ) AB² = 5² + 5² - 2(5²)(-47/625) AB² = 25 + 25 + 50(47/625) AB² = 50 + 94/25 = (1250+94)/25 AB² = 1344/25 AB = √(1344/25) = (8√21)/5 ≈ 7.33 units

[@manaspratimdas5758]

I'm pleased to see many different approaches towards this problem. Let me put mine: Well, AB is the common chord to both circles of radii 5 and 4. We know the formulla, squared length of a chord, AB**2=4(r**2-d**2) We know r1 and r2 =(5,4) . We also know d1+d2=5, hence we can determine d1/and/or/d2. By putting it back in AB**2 formulla we gan get the AB.

[@SinergiasHolisticas]

Love it!!!!!!!!!!

[@uwelinzbauer3973]

First I also found that pythagorean 3,4,5 triple. Then I divided the isosceles 5,5,4 triangle OPB into two equal right triangles with Hypothenuse 5 and one side 2. Then angle alpha = asn(2/5). Then using y/5 = sin (2 * alpha) leads to length of chord AB ≈ 7.33 Thanks for sharing this nice geometry question. Wish you a happy Sunday 😊

[@DB-lg5sq]

Merci beaucoup pour votre effort

[@santiagoarosam430]

OQ=3 ; QP=4---> OP=5 ---> Cuerda AB=2c; M es su punto medio y "s" la flecha ---> Potencia de M respecto a la circunferencia roja =(4+s)(4-s)=16-s² =c²= Potencia de M respecto a la circunferencia amarilla =s(2*5 -s)=10s-s² = 16-s²---> s=16/10=8/5---> c²=16-s²= 16-(8/5)² ---> c=4√21/5---> 2c=8√21/5 =AB. Gracias y saludos.

[@SaikatSarkar-c2f]

🎉

[@jaimeyomayuza6140]

Señor profesor mil gracias por su dedicación y tiempo. Buen ejercicio

[@Birol731]

My way of solution ▶ The tangent [OB] is perpendicular to radius of the red circle: 90° By considering the right trinagle Δ(POQ) [OQ]= 3 [QP] is equal to the radius of the red trinagle: [QP]= r [QP]= 4 ⇒ according to the Pythagorean theorem we can write: [OQ]²+[QP]²= [PO]² 3²+4²= [PO]² [PO]= 5 ii) there is a point between the intersection of [PO] and [AB], let's call it "R" [OP] will divide the length [AB] in two equal parts: [AB]/2 = y iii) for the red triangle, by writing the intersecting chords theorem: [AR]*[RB]= [RS]*[RT] [PS]= [PT]= r [PS]= [PT]= 4 [PR]= x [RS]= 4-x [AR]=[RB]= y ⇒ y²= (4+x)*(4-x) y²= 16-x².............Eq-1 iv) By considering the right triangle ΔAOR: [AO]= R [AO]= 5 [OR]= 5-x [RA]= y by applying the Pythagorean theorem: [AO]²= [OR]²+[RA]² 5²= (5-x)² + y² 25= 25-10x+x²+y² x²-10x+y²=0.........Eq-2 we know that: y= 16-x² ⇒ x²-10x+16-x²=0 10x= 16 x= 8/5 y²= 16-x² y²= 16- (8/5)² y= √336/25 y= 4√21/5 [AB]=2y [AB]= 8√21/5 [AB]≈ 7,332 length units

[@marcelowanderleycorreia8876]

Very tricky question.

[@JinnirasFlair]

Is it right?? OQ=3, QP=4,then OP is obviously 5 Then OB =5, OA= 5 also Then AB will be. √50 or 5√2 Sir plz answer..... 🙋🏻‍♀️🙏🏻

[@AllmondISP]

Why? What is the math behind it?

[@JinnirasFlair]

@@AllmondISP is it actually right way or wrong ?? Cause 5√2 or 7.07🙃

[@JinnirasFlair]

@@AllmondISP like pythagorean theory

[@AllmondISP]

​@@JinnirasFlairit's seems wrong. how did you got to 5 sqrt(2) exactly? What are the numbers you used on your pythagoryean theory?

[@quigonkenny]

AB = 5√2 only if ∠AOB = 90°. In this case, ∠AOB is not 90°.

[@unknownidentity2846]

Let's face this challenge: . .. ... .... ..... Since OQ is a tangent to the circle, we know that the triangle OPQ is a right triangle. So we can apply the Pythagorean theorem: OP² = OQ² + PQ² = 3² + 4² = 9 + 16 = 25 ⇒ OP = √25 = 5 Now let's assume that O is the center of the coordinate system and that OQ is located on the x-axis. Then we obtain the following coordinates: O: ( 0 ; 0 ) A: ( xA ; yA ) B: ( xB ; yB ) P: ( 3 ; 4 ) Q: ( 3 ; 0 ) Since A and B are located on the semicircle and the circle, we can conclude: (x − xO)² + (y − yO)² = OP² (x − xP)² + (y − yP)² = PQ² (x − 0)² + (y − 0)² = 5² (x − 3)² + (y − 4)² = 4² x² + y² = 25 x² − 6*x + 9 + y² − 8*y + 16 = 16 x² + y² = 25 x² + y² − 6*x − 8*y = −9 6*x + 8*y = 34 6*x = 34 − 8*y ⇒ x = 34/6 − 8*y/6 = 17/3 − (4/3)*y x² + y² = 25 [17/3 − (4/3)*y]² + y² = 25 289/9 − (136/9)*y + (16/9)*y² + y² = 25 289 − 136*y + 16*y² + 9*y² = 225 25*y² − 136*y + 64 = 0 y = [136 ± √(136² − 4*25*64)]/(2*25) y = [136 ± √(18496 − 6400)]/50 y = (136 ± √12096)/50 Since yA > yB, we can conclude: yA = (136 + √12096)/50 yB = (136 − √12096)/50 Now we are able to calculate the length of AB: AB² = (xB − xA)² + (yB − yA)² = {[17/3 − (4/3)*yB] − [17/3 − (4/3)*yA]}² + (yB − yA)² = [17/3 − (4/3)*yB − 17/3 + (4/3)*yA]² + (yB − yA)² = [(4/3)*yA − (4/3)*yB]² + (yB − yA)² = (16/9)*(yA − yB)² + (yB − yA)² = (16/9)*(yA − yB)² + (yA − yB)² = (16/9)*(yA − yB)² + (9/9)*(yA − yB)² = (25/9)*(yA − yB)² = (25/9)*[(136 + √12096)/50 − (136 − √12096)/50]² = (25/9)*(136/50 + √12096/50 − 136/50 + √12096/50)² = (25/9)*(√12096/25)² = (25/9)*(12096/625) = 1344/25 ⇒ AB = √(1344/25) = 8√21/5 ≈ 7.332 Best regards from Germany

[@LuisdeBritoCamacho]

Wow!!

[@sandytanner9333]

How do we know that the chord is perpendicular to OP?

[@sowmya_nj]

I have the same doubt

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) OP = R 02) OQ = 3 03) PQ = 4 04) R^2 = 3^2 +4^2 ; R^2 = 9 + 16 ; R^2 = 25 ; R = sqrt(25) ; R = 5 05) Now we know that ; PA = PB = 4 lin un and OA = OB = 5 lin un 06) OP = R = 5 lin un 07) Now I can built a Kite : Quadrilateral [OAPB] with External Sides equal to 4 (PA ; PB) and 5 (OB ; OA) lin un. 08) Notice that OP = 5 lin un 09) Using Heron's Formula I can get the Area of an Isosceles Triangle [OPB] (Sides = (5 ; 5 ; 4)): A = 9,165 sq un 10) Now, dividing : (2 * 9,165) / 5, I get the Distance from the Middle Point of AB (M) to B. 11) MB = 18,33 / 5 = 3,666 12) AB = 2 * 3,666 ; AB = 7,332 lin un Therefore, OUR BEST ANSWER IS : In an Euclidian Affine Space, Line AB must be equal to 7,332 Linear Units.

[@sergeyvinns931]

Построим треугольник OPQ, PQ=4, OQ=3, OP=5, соединим Р с В, получим два треугольника ОСВ и РСВ, у них общий катет , равный АВ/2, который мы ищем. обозначим СР как х и составим два уравнения (АВ/2)^2=16-x^2, (AB/2)^2=25-(5-x)^2. Приравняем правые части двух уравнений, возведём в квадрат слагаемое в скобках, и получим, что х=16/10. Теперь найдём АВ, которое найдём, подставив в любое из двух уравнений значение х. АВ=(8\/21)/5=7,33212111192...

[@ashutoshkumardalei3264]

How AB and OP are perpendicular?

[@giuseppemalaguti435]

R=5..AB=2*4*sinarccos(2/5)=8√(1-4/25)=8√21/5

[@JSSTyger]

Well I may not have it right but gave it a go. I think AB = 7.332. I figured big the big radius to be 5 and drew a parallelagram with sides 5, 5, 4, and 4 and presumed that line OP splits AB exactly in 2. Thats where the big question mark is but it seems right. And based on optics, I knew the answer had to be close to √(50).

[@nandisaand5287]

@ 3:10, how do we know AB is perpendicular to OP?

[@DaRealNoobKing]

because AOP and POB are congruent because their side lengths are R, R and r and they share PO

[@waheisel]

Good question, it looks like PreMath drew the perpendicular symbol without proving AB really is perpendicular to OP. As @DaRealNoobKing says, triangle AOP is congruent to POB. That leads to proving APC is congruent to BPC*. That proves that angles ACP and BCP (which add to 180) are equal, so they are right angles. Using this proof you don't even need to show that AB is perpendicular to OP as AC being equal to BC was proven along the way*.

[@alexbidiak2057]

Interesting, why is a radius OP perpendicular to a chord AB?

[@awcampbell2002]

I was wondering the same thing. It seemed like circular reasoning especially since they always disclsim that this may not be drawn to scale

[@rey-dq3nx]

Why not? You can always rotate radius OP anyway you want to make it perpendicular to AB

[@alexbidiak2057]

​@@rey-dq3nx Sure, but P is fixed on a circumference and we cant know an angle between AB and OP

[@alastairjames8723]

I think APO and BPO are congruent as they both have the same three side lengths and they are reflected about PO

[@alexbidiak2057]

@@alastairjames8723 Agree, for both APO and BPO have AO=BO and OC the same for both but this doesn't mean that AC is equal to CB. I asked my question as the author didn't clarify this

[@misterenter-iz7rz]

A complicated puzzle😢

[@misterenter-iz7rz]

This is a rhombus with size 4,4,5,5,5, it's area is sqrt(21)×4=(AB/2)×5; then AB=8/5×sqrt(21).😅

[@imetroangola4943]

The sides of a rhombus are all congruent, that is, the sides are equal! There is no rhombus 5,5,4 and 4.

[@misterenter-iz7rz]

@imetroangola4943 sorry for misuse of terms， kite instead 😉

[@SaikatSarkar-c2f]

😢