The problem implies that the answer does not depend on the position of C on the diameter, so just move C to the center and it immediately follows that x = r .
Extremely easy !!! Position of point C is not defined, we can choose it , meeting the original conditions. ( Also is not defined position of points A and B ). I chose to unify C with O center of semicircle, then AC and BC becomes the radius of semicircle, and knowing its angle is 60°, then triangle ABC becomes an equilateral triangle, with segment AB becoming horizontal. So X = R X = 5 cm ( Solved √ ) Is not necessary any calculations This result is valid for any position of point C between P and Q.
Other method, unifying C with P, then A, P ,and C are the same point, AB and CB become same segnent, and being it angle 60° , we obtain an equilateral triangje OAB, then X = R= 5 cm.
Construct the circle with PQ diameter. O is the center of the circle. Let BM⊥PQ (construction , M point of the circle and N=BM∩PQ ) Triangle OBM is isosceles (OM=OB=R) and ON height => ON is median => MN=BN. Now let CM (construction). You can prove easily that right triangles BCN=MCN => ∠QCM =60°. Notice that ∠ACP=∠QCM=60° so points A,C,M belong to the same line. Hence ∠AMB=30° => arc AB=60° => x=AB=λ₆=R=5 (side of an equilateral triangle inscribed in a circle with radius R)
The problem could also be solved in this way: let point C be origin and length AC =a & BC =b and OC =c; Then Coordinates of Points A,B & O are respectively (-aCos60°,aSin60°),(bCos60°,bSin60°) & (c,0); So,AB^2=a^2+b^2-ab OA^2=a^2+c^2+ac=5^2 OB^2=b^2+c^2-bc=5^2 On equating OA^2=OB^2; We get c=b-a. on putting value of c in terms of a and b in expression of OA^2; We get OA^2=a^2+(b-a)^2+a(b-a)=25; that is a^2+b^2-ab=25; So AB^2=25(as AB^2=a^2+b^2-ab); Hence x=AB=5.
Assume Chord as A and B and centre as O, draw AO ane BO, we know that inscribed angles subtended by the same arc (AB) are equal, hence angle AOB will be 60 degeee, and traingle AOB will be equilateral, hence AB would be 5.
The solution discussed by Mathbooster is awesome! One very interesting property of circle that we can conclude from this problem is that: If A and B are two points on circle such they subtend same angle at any point C on diameter as in figure. Let image of point A and B with respect to Diameter be respectively A' and B'. Then 1.A' and B' lie on Same Circle. 2.Points A',C,B and Points B',C,A will be Collinear.
Your solution using circle geometry is ideal for this one. Here is an alternative using trigonometry. Let _|AC| = y, |BC| = z , |OC| = t_ Using the cosine rule in: _ΔACB: x² = y² + z² - 2yzcos(60°) ⇒ _*_x² = y² + z² - yz_* ... ① _ΔACO: 5² = y² + t² - 2ytcos(120°) ⇒ _*_25 = y² + t² + yt_* ... ② _ΔBCO: 5² = z² + t² - 2ztcos(60°) ⇒ _*_25 = z² + t² - zt_* ... ③ Subtracting ② - ③ _y² - z² + (y + z)t = 0_ ⇒ _(y + z)(y - z + t) = 0_ But _y + z ≠ 0_ ∴ *_t = z - y_* ... ④ Adding ② + ③ _y² + z² + 2t² - (z - y)t = 50_ Substituting for *_t_* from ④: _y² + z² + (z - y)² = 50_ ⇒ _y² + z² - yz = 25_ ⇒ _x² = 25_ from ① ⇒ *_x = 5_*
@@syed3344 In ΔACB side AB has length _x_ and is opposite ∠ACB. Angles ∠ACP and ∠BCQ are each given as 60°, and together with ∠ACB they form half a circle, i.e. 180°. Therefore ∠ACB = 60°. In ΔACO side AO is a radius so has length _5_ and is opposite ∠ACO (O is the centre of the circle and lies on PQ). Angle ∠ACO = ∠ACB + ∠BCQ = 60° + 60° = 120°. In ΔBCO side BO is a radius so has length _5_ and is opposite ∠BCO which is given as 60°. Hope this helps.
@@imetroangola4943 I don't use a program - these are all Unicode characters. I get most of mine by using the online keyboard on my Windows 10 tablet. Pressing the "&123" key and then the "Ω" key gives many mathematical symbols, fractions, Greek letters etc. Otherwise if you do a Web search for what you want, e g. "angle symbol", "square root sign" you will get sites that use it and you can copy it from there (often straight from the search results) and paste it into your work. Also if you search for "superscript generator" and "subscript generator" you will find useful sites that convert letters and numbers to Unicode equivalents that you can use for exponents and subscripts. Finally I put an underscore _ at the beginning and end of each line of algebra that I write. In a UA-cam comment this converts the text to italics which is how algebra is normally presented in text books. Hope this helps.
extending line BC under the diameter until intersects the circle in D, we get an isosceles triangle ACD, being angle PCD = 60° because opposite to BCQ and then for simmetry reasons DC = AC. This means that angle ADB is 30°, then its angle in the center AOB is 60°. But AOB is also isosceles being AO = OB = radius. All this lend us to the conclusion that AOB is equilateral and X = radius = 5
Another example of easier than it looks. I wonder this combination of three circle theorems and defn of vertical angles would work if the subtended angle were not 60 degrees. And looks for some problems practicing and memorizing basic circle theorems should be considered as basic as multplication.
А, нельзя было сразу провести ОА и ОВ сразу получили бы равносторонний треугольник ОАВ, где х=5? И это не зависит от точки на диаметре, из которой проведены два отрезка под углом 60 градусов от линии диаметра! Так строится шестиугольник вписанный в окружность под разными углами!
The usual methods of Pythagoras theorem, sine law, cosine law and corresponding sides of similar or congruent triangles for length problems of triangle do not help in this problem. (I am not sure they can solve the problem in a complicated way) The two 60 degree angles given should be a hint that theorems about angles are the key to solution.
The problem could also be solved in this way: let point C be origin and length AC =a & BC =b and OC =c; Then Coordinates of Points A,B & O are respectively (-aCos60°,aSin60°),(bCos60°,bSin60°) & (c,0); So,AB^2=a^2+b^2-ab OA^2=a^2+c^2+ac=5^2 OB^2=b^2+c^2-bc=5^2 On equating OA^2=OB^2; We get c=b-a. on putting value of c in terms of a and b in expression of OA^2; We get OA^2=a^2+(b-a)^2+a(b-a)=25; that is a^2+b^2-ab=25; So AB^2=25(as AB^2=a^2+b^2-ab); Hence x=AB=5.
![ОРБАН приехал к ПУТИНУ и "отчитался" о поездке в Киев 😁 [Пародия]](http://i.ytimg.com/vi/_0ukDovOWq8/mqdefault.jpg)
![ОРБАН приехал к ПУТИНУ и "отчитался" о поездке в Киев 😁 [Пародия]](/img/tr.png)
The problem implies that the answer does not depend on the position of C on the diameter, so just move C to the center and it immediately follows that x = r
.
This is my first thought....
They should give some more distances :)
Keep it up sir 👍
Excellent job!
Extremely easy !!!
Position of point C is not defined, we can choose it , meeting the original conditions. ( Also is not defined position of points A and B ).
I chose to unify C with O center of semicircle, then AC and BC becomes the radius of semicircle, and knowing its angle is 60°, then
triangle ABC becomes an equilateral triangle, with segment AB becoming horizontal.
So X = R
X = 5 cm ( Solved √ )
Is not necessary any calculations
This result is valid for any position of point C between P and Q.
Other method, unifying C with P, then A, P ,and C are the same point, AB and CB become same segnent,
and being it angle 60° , we obtain an equilateral triangje OAB, then X = R= 5 cm.
Construct the circle with PQ diameter. O is the center of the circle.
Let BM⊥PQ (construction , M point of the circle and N=BM∩PQ )
Triangle OBM is isosceles (OM=OB=R) and ON height => ON is median => MN=BN.
Now let CM (construction).
You can prove easily that right triangles BCN=MCN => ∠QCM =60°.
Notice that ∠ACP=∠QCM=60° so points A,C,M belong to the same line.
Hence ∠AMB=30° => arc AB=60° => x=AB=λ₆=R=5
(side of an equilateral triangle inscribed in a circle with radius R)
The problem could also be solved in this way:
let point C be origin and length AC =a & BC =b and OC =c;
Then Coordinates of Points A,B & O are respectively (-aCos60°,aSin60°),(bCos60°,bSin60°) & (c,0);
So,AB^2=a^2+b^2-ab
OA^2=a^2+c^2+ac=5^2
OB^2=b^2+c^2-bc=5^2
On equating OA^2=OB^2;
We get c=b-a.
on putting value of c in terms of a and b in expression of OA^2;
We get OA^2=a^2+(b-a)^2+a(b-a)=25;
that is a^2+b^2-ab=25;
So AB^2=25(as AB^2=a^2+b^2-ab);
Hence x=AB=5.
Assume Chord as A and B and centre as O, draw AO ane BO, we know that inscribed angles subtended by the same arc (AB) are equal, hence angle AOB will be 60 degeee, and traingle AOB will be equilateral, hence AB would be 5.
Pls do IOQM level questions
The solution discussed by Mathbooster is awesome!
One very interesting property of circle that we can conclude from this problem is that:
If A and B are two points on circle such they subtend same angle at any point C on diameter as in figure.
Let image of point A and B with respect to Diameter be respectively A' and B'.
Then
1.A' and B' lie on Same Circle.
2.Points A',C,B and Points B',C,A will be Collinear.
Nice!
Your solution using circle geometry is ideal for this one.
Here is an alternative using trigonometry.
Let _|AC| = y, |BC| = z , |OC| = t_
Using the cosine rule in:
_ΔACB: x² = y² + z² - 2yzcos(60°) ⇒ _*_x² = y² + z² - yz_* ... ①
_ΔACO: 5² = y² + t² - 2ytcos(120°) ⇒ _*_25 = y² + t² + yt_* ... ②
_ΔBCO: 5² = z² + t² - 2ztcos(60°) ⇒ _*_25 = z² + t² - zt_* ... ③
Subtracting ② - ③
_y² - z² + (y + z)t = 0_
⇒ _(y + z)(y - z + t) = 0_
But _y + z ≠ 0_
∴ *_t = z - y_* ... ④
Adding ② + ③
_y² + z² + 2t² - (z - y)t = 50_
Substituting for *_t_* from ④:
_y² + z² + (z - y)² = 50_
⇒ _y² + z² - yz = 25_
⇒ _x² = 25_ from ①
⇒ *_x = 5_*
Você faz onde todas essas nomenclaturas e símbolos matemáticos? Você usa seu computador com algum programa?
How did y find the angles ??
@@syed3344 In ΔACB side AB has length _x_ and is opposite ∠ACB. Angles ∠ACP and ∠BCQ are each given as 60°, and together with ∠ACB they form half a circle, i.e. 180°. Therefore ∠ACB = 60°.
In ΔACO side AO is a radius so has length _5_ and is opposite ∠ACO (O is the centre of the circle and lies on PQ). Angle ∠ACO = ∠ACB + ∠BCQ = 60° + 60° = 120°.
In ΔBCO side BO is a radius so has length _5_ and is opposite ∠BCO which is given as 60°.
Hope this helps.
@@imetroangola4943 I don't use a program - these are all Unicode characters. I get most of mine by using the online keyboard on my Windows 10 tablet. Pressing the "&123" key and then the "Ω" key gives many mathematical symbols, fractions, Greek letters etc. Otherwise if you do a Web search for what you want, e g. "angle symbol", "square root sign" you will get sites that use it and you can copy it from there (often straight from the search results) and paste it into your work. Also if you search for "superscript generator" and "subscript generator" you will find useful sites that convert letters and numbers to Unicode equivalents that you can use for exponents and subscripts.
Finally I put an underscore _ at the beginning and end of each line of algebra that I write. In a UA-cam comment this converts the text to italics which is how algebra is normally presented in text books.
Hope this helps.
extending line BC under the diameter until intersects the circle in D, we get an isosceles triangle ACD, being angle PCD = 60° because opposite to BCQ and then for simmetry reasons DC = AC. This means that angle ADB is 30°, then its angle in the center AOB is 60°. But AOB is also isosceles being AO = OB = radius. All this lend us to the conclusion that AOB is equilateral and X = radius = 5
Дуга опирающаяся на угол 60 градусов. равна пи/3, а хорда равна радиусу.
Another example of easier than it looks. I wonder this combination of three circle theorems and defn of vertical angles would work if the subtended angle were not 60 degrees. And looks for some problems practicing and memorizing basic circle theorems should be considered as basic as multplication.
А, нельзя было сразу провести ОА и ОВ сразу получили бы равносторонний треугольник ОАВ, где х=5? И это не зависит от точки на диаметре, из которой проведены два отрезка под углом 60 градусов от линии диаметра! Так строится шестиугольник вписанный в окружность под разными углами!
The usual methods of Pythagoras theorem, sine law, cosine law and corresponding sides of similar or congruent triangles for length problems of triangle do not help in this problem. (I am not sure they can solve the problem in a complicated way) The two 60 degree angles given should be a hint that theorems about angles are the key to solution.
The problem could also be solved in this way:
let point C be origin and length AC =a & BC =b and OC =c;
Then Coordinates of Points A,B & O are respectively (-aCos60°,aSin60°),(bCos60°,bSin60°) & (c,0);
So,AB^2=a^2+b^2-ab
OA^2=a^2+c^2+ac=5^2
OB^2=b^2+c^2-bc=5^2
On equating OA^2=OB^2;
We get c=b-a.
on putting value of c in terms of a and b in expression of OA^2;
We get OA^2=a^2+(b-a)^2+a(b-a)=25;
that is a^2+b^2-ab=25;
So AB^2=25(as AB^2=a^2+b^2-ab);
Hence x=AB=5.
@@RahulKumar-id5cq Thank you for your sharing. Your method is surprisingly not complicated. Probably I should learn more analytic geometry.
@@hongningsuen1348
I will keep doing in future 🙂.
asnwer=8 cm isit
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