You are very welcome Banaz! Good to hear from you. Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
It should 've been mentioned that the upper side of red square is tangent to the circle. By saying tangent to two squares someone might think it touches the corners as in square Blue. What if the distance berween the sqares was bigger than 2 , AC would not be perpendicular to none of red square sides as it is AB for blue box
Furthermore it should be stated circle is tangent to the red square upper side AND to its left upper corner. In other words the circle center is aligned with the whole red square left side. If NOT you can draw infinite numbers of circles between 2 points...
If you know the rules for tangents then that is not necessary since a tangent will always create a 90 degree angle since it’s perpendicular to the radius of the circle. While it would’ve been nice for clarity it’s not necessary.
@@reynaldowify Exactly. This would have the impact of making angle E non-perpendicular (and perhaps D depending on interpretation). That would cause BCDE to not be a rectangle.
This circle is not tangent to two squares. Is just tanget to cicle of area 4, and touch one corner of circle are 9. With the information you give, there are infite solutions to the problem.
@@Nikioko But if circle is tangent to blue square, and touches red square in corner, you get another isosceles triangle with two legs r (a different r than shown in video) and chord of length √5. You also get a right triangle with legs 1 and 2-r, and hypotenuse r. Use Pythagorean theorem to get r = 5/4 I tried to post a link showing a diagram, but youtube keeps deleting it, so I'll try to explain. In this second case, centre of circle is directly to the right of top right corner of larger blue square. We drop a vertical line from centre of circle so that the other end of this line is directly to the left of top left corner of smaller red square. This vertical leg has length = 3-2 = 1. We now draw horizontal leg of right triangle by joining this line to top left corner of red square. This horizontal leg has length = 2-r (where 2 is distance between squares). Hypotenuse is line from centre of circle to top left corner of red square = r.
@@Nikioko it's not true because you assume that the circle is tangent to the upper side of the red square not the square, witch was not the requirement. so his solution applies only for this drawing.
@@Nikioko of course you assume that. you should re-read it yourself. I don't need to re-read it. however if you are not smart enough than it also doesn't matter how often you re-read it
The problem need to state clearly that the center of the circle is aligned to both upper and bottom left vertices of the red square, because if this information is not given, one would never be sure about how far to te right the point of tangency would be ("sliding" it over the upper side of the red square, making the circle bigger) or not sure either about how little the circle can be (by wondering if the circle is not tangent with any side of the red square, as happens to be the case with the blue square).
Locate the center of the circunference in (0,0) and the tangent points with the squares in (0,-r) and (-2,1-r). So the points who belong to the circunference satisfy the equation x^2 + y^2 = r^2 Replacing (-2,1-r) on the equation se have: (-2)^2 + (1-r)^2 = r^2 4. + 1 -2r +r^2 = r^2 2r = 5 r = 2.5 Congratulations, your channel is superb!
So nice of you Peter! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
The problem is under-specified as stated. Nothing in the problem definition rules out the circle of radius 1 that lies in the space between the two squares -- is it not tangent to both squares? One more constraint fixes the problem definition, e.g., area of the LARGEST circle that can be tangent to the two squares. It seems you are using this assumption in the sketch geometry, but fail to cite it in the givens.
Sorry, but no, your solution only works for the peculiar and unspecified condition when Point A is directly over Point C. In fact if the circle is smaller your problem conditions would still be met, but no solution is possible.
He said the circle was tangent to the square. The chord BC = SQRT 5. Put a point F midway between B and C. Triangles BEC and ACF are similar. Therefore, SQRT5/2r=1/SQRT5. Thence r=2.5 and area =19.635.
I think that demo / solution is flawed. You did assume AD is colinear with the left side of the small square. That is not an absolute fact, it is not said in the problem nor anything to make you conclude it is provable. That has an impact of the value for BD deduced from CE.
With all due respect, given the description, this problem is unsolvable. Your solution assumes that the vertices on the left side of the smaller square is virtically below the center of the circle. However, that condition is not provided in the problem description. Without this condition, the problem has infinitely many solitions.
There are only two tangents to the vertex of a square, each that is perpendicular to the adjacent sides of the vertex and that contains the vertex. That said, the blue square is only the intersection of the corner vertex and the circle.
your solution is wrong, because you assumed that AC( were A is the center point of the circle and C is the point of tangent) is vertical and colinear with the side of the smaller square without being mentioned in the requirement. Or the requirement is incomplet.
The real answer is any area greater than or equal to PI*(root(5)/2)^2. The solution given relies on the left edge of the red square being colinear with the center of the circle, but was never explicitly stated as part of the question.
Excellent explanation to a problem that seems difficult at first glance, but is extremely easy once you apply the right formulas to get your missing variables. Always enjoy watching your work sir, thank you for your hard work and dedication.
You did not mention that the center of the circle is directly above the corner of the red square. that is why your problem is incomplete and can have many solution as long as the diameter of the circle is >2.
Mmm, slightly confusing, but still great fun 😃👍🏻in your more recent videos you often give a brief explanation in the notes about the strategy, this I like, it encourages me to know the pathway to take, but without giving the answer away, keep up the great work. Its very generous of you to publish these problems
So nice of you Scott! I appreciate that. You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
It's at 2:26 in the video. The red square is tangent to the circle. The radius is perpendicular to the tangent point. Using Google translate: На видео это 2:26. Красный квадрат касается круга. Радиус перпендикулярен точке касания.
So nice of you Yasser! Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Enjoy every moment of your life 🌻
You failed to mention that the center of the circle is directly aligned with the vertical left side of the smaller square. I cannot just assume that. I'm quite disappointed with you.
It is not given that AC is vertical. The illustration suggests that C is at the bottom of the circle --- BUT IT IS NOT STATED ----- so, we CANNOT ASSUME THAT. I wanted to assume it to solve it, but I refrained.
So nice of you Taha! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
great quickly resolution ... i have used other one ... based on to find the angle of the triangle costructed among the 2 squares and from it i have "comed up" to the angle of the center of the circle .... next R= sqrt(5) / 2 * sin(half_center_angle) ....
Ubique el centro de la circunferencia en (0,0) y los puntos de tangencia con los cuadrados en (0,-r) y (-2,1-r). Cómo los puntos pertenecen a la circunferencia satisfacen la ecuación x^2 + y^2 = r^2 Remplazando (-2,1-r) en la ecuación tenemos: (-2)^2 + (1-r)^2 = r^2 4. + 1 -2r +r^2 = r^2 2r = 5 r = 2.5
Sir. I don't know english it's google translate The distance from point A to point B may be r-1 or more. Because there is no condition that the center of the circle is the same as the Y axis of the red square
If the radius touches one corner of the squar with side 3 since it is a square it touches the other opposite corner of the sguare straight down so i thich we can add up r and diagonal as a hypotenouus of the big right angle triangle
Data insufficient, sir. You should have specified that upper side of smaller square is tangent to yhe circle, but nowhere in the question it is written.
Another unbelievably complex solution to such a simple question using the chord x chord intersection method 2 x 2 = 1 x (2r - 1) r= 2.5. That took 15 seconds.
I thought you would explain like this. Extend BD to the right of the circle, say F. Extend the radius AC up and call it G. BD x DF =DC x DG. Difference of height between two squares is 3-2=1. BD = 2. So DF=2. 2 x 2 = 1 x DG. DG=4, DIA =4+1=5. r=2.5 etc etc (your regular style of solving) Regards.
The problem as given does not define a single circle, but a family of circles. Point C as drawn is on a vertex of the square 2. However, as given, C can be anywhere on the top surface of square 2 and still be tangent. Thus EC can be any value between 2 and 4.
Стороны квадратов 3 и 2. Тогда катеты треугольника: 2 и 1, а гипотенуза V5- хорда окружности. Проведя к ней перпендикуляр, получим подобный по перпендикулярности сторон треугольник. 1:V5= 0,5*V5:R. R=5/2. S=25п/4. Ответ: 25п/4.
Very wrong definition ! The circle in tangent only to the red circle and exactly in its left corner above ! There are an infinity of circles tangent to the red square !!!
Thank u so much for this video and take care and stay safe and God bless u
You are very welcome Banaz! Good to hear from you.
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Which God would you recommend? I have yet to find an effective one.
@@japeking1 yourself
@@japeking1 no recommendation, each person must do it for himself
@@Gnome5555 Good. I'll continue to do without then ;-)
It should 've been mentioned that the upper side of red square is tangent to the circle. By saying tangent to two squares someone might think it touches the corners as in square Blue. What if the distance berween the sqares was bigger than 2 , AC would not be perpendicular to none of red square sides as it is AB for blue box
Agreed. This question lacks sufficient information. The solution assumes this tangency which is unclear given circle is "tangent to two squares".
Furthermore it should be stated circle is tangent to the red square upper side AND to its left upper corner. In other words the circle center is aligned with the whole red square left side. If NOT you can draw infinite numbers of circles between 2 points...
If you know the rules for tangents then that is not necessary since a tangent will always create a 90 degree angle since it’s perpendicular to the radius of the circle. While it would’ve been nice for clarity it’s not necessary.
@@st3althyone The center of the circle could not be on the vertical from the rightmost square, and still being tangent ro the circle at the vertex
@@reynaldowify Exactly. This would have the impact of making angle E non-perpendicular (and perhaps D depending on interpretation). That would cause BCDE to not be a rectangle.
This circle is not tangent to two squares. Is just tanget to cicle of area 4, and touch one corner of circle are 9. With the information you give, there are infite solutions to the problem.
I have the same thoughts too..
@@Nikioko But if circle is tangent to blue square, and touches red square in corner, you get another isosceles triangle with two legs r (a different r than shown in video) and chord of length √5. You also get a right triangle with legs 1 and 2-r, and hypotenuse r. Use Pythagorean theorem to get r = 5/4
I tried to post a link showing a diagram, but youtube keeps deleting it, so I'll try to explain. In this second case, centre of circle is directly to the right of top right corner of larger blue square. We drop a vertical line from centre of circle so that the other end of this line is directly to the left of top left corner of smaller red square. This vertical leg has length = 3-2 = 1. We now draw horizontal leg of right triangle by joining this line to top left corner of red square. This horizontal leg has length = 2-r (where 2 is distance between squares). Hypotenuse is line from centre of circle to top left corner of red square = r.
I expected to see a general solution to the infinit variations
@@Nikioko it's not true because you assume that the circle is tangent to the upper side of the red square not the square, witch was not the requirement. so his solution applies only for this drawing.
@@Nikioko of course you assume that. you should re-read it yourself. I don't need to re-read it. however if you are not smart enough than it also doesn't matter how often you re-read it
The two intersecting chords theorem works nicely also to derive the radius
The problem need to state clearly that the center of the circle is aligned to both upper and bottom left vertices of the red square, because if this information is not given, one would never be sure about how far to te right the point of tangency would be ("sliding" it over the upper side of the red square, making the circle bigger) or not sure either about how little the circle can be (by wondering if the circle is not tangent with any side of the red square, as happens to be the case with the blue square).
Locate the center of the circunference in (0,0) and the tangent points with the squares in (0,-r) and (-2,1-r).
So the points who belong to the circunference satisfy the equation x^2 + y^2 = r^2
Replacing (-2,1-r) on the equation se have:
(-2)^2 + (1-r)^2 = r^2
4. + 1 -2r +r^2 = r^2
2r = 5
r = 2.5
Congratulations, your channel is superb!
Does it need to prove the line AC is parallel/perpendicular to the red square first!?
You have to make the assumption that EC is tangent to circle, The problem doesn't give that info so it can't be solved
Which square is tangent? Red ? Blue ? or both?
There are 2 squares so i guess both.
Best math channel on youtube
Thank you for such thorough explainations. I enjoy these videos very much.
So nice of you Peter! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
The problem is under-specified as stated. Nothing in the problem definition rules out the circle of radius 1 that lies in the space between the two squares -- is it not tangent to both squares? One more constraint fixes the problem definition, e.g., area of the LARGEST circle that can be tangent to the two squares. It seems you are using this assumption in the sketch geometry, but fail to cite it in the givens.
Sorry, but no, your solution only works for the peculiar and unspecified condition when Point A is directly over Point C. In fact if the circle is smaller your problem conditions would still be met, but no solution is possible.
Exactly !
He said the circle was tangent to the square. The chord BC = SQRT 5. Put a point F midway between B and C. Triangles BEC and ACF are similar. Therefore, SQRT5/2r=1/SQRT5. Thence r=2.5 and area =19.635.
I think that demo / solution is flawed. You did assume AD is colinear with the left side of the small square. That is not an absolute fact, it is not said in the problem nor anything to make you conclude it is provable. That has an impact of the value for BD deduced from CE.
With all due respect, given the description, this problem is unsolvable. Your solution assumes that the vertices on the left side of the smaller square is virtically below the center of the circle. However, that condition is not provided in the problem description. Without this condition, the problem has infinitely many solitions.
There are only two tangents to the vertex of a square, each that is perpendicular to the adjacent sides of the vertex and that contains the vertex. That said, the blue square is only the intersection of the corner vertex and the circle.
your solution is wrong, because you assumed that AC( were A is the center point of the circle and C is the point of tangent) is vertical and colinear with the side of the smaller square without being mentioned in the requirement. Or the requirement is incomplet.
The real answer is any area greater than or equal to PI*(root(5)/2)^2. The solution given relies on the left edge of the red square being colinear with the center of the circle, but was never explicitly stated as part of the question.
Excellent explanation to a problem that seems difficult at first glance, but is extremely easy once you apply the right formulas to get your missing variables. Always enjoy watching your work sir, thank you for your hard work and dedication.
Thank you!
You're welcome!
احسنتم وبارك الله فيكم وعليكم والله يحفظكم ويرعاكم جميعا
Great video. Love you ❤️
You did not mention that the center of the circle is directly above the corner of the red square. that is why your problem is incomplete and can have many solution as long as the diameter of the circle is >2.
Mmm, slightly confusing, but still great fun 😃👍🏻in your more recent videos you often give a brief explanation in the notes about the strategy, this I like, it encourages me to know the pathway to take, but without giving the answer away, keep up the great work. Its very generous of you to publish these problems
I enjoy your videos very much as I am a math/chemistry teacher and love to challenge my students with these great math puzzles. Thank you!
So nice of you Scott! I appreciate that. You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
As a math teacher, you should know that the circle is not tangent to the blue square.
Где в условии задачи сказано, что центр окружности находится на одной прямой со стороной квадрата площадью 4????
It's at 2:26 in the video. The red square is tangent to the circle. The radius is perpendicular to the tangent point. Using Google translate:
На видео это 2:26. Красный квадрат касается круга. Радиус перпендикулярен точке касания.
👍! You are wonderful!
Very useful and please keep uploading videos they really help us 👍👍
So nice of you Yasser!
Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Enjoy every moment of your life 🌻
You failed to mention that the center of the circle is directly aligned with the vertical left side of the smaller square. I cannot just assume that. I'm quite disappointed with you.
This assumes BDCE is a rectangle. There's no proof of that.
It is not given that AC is vertical. The illustration suggests that C is at the bottom of the circle ---
BUT IT IS NOT STATED ----- so, we CANNOT ASSUME THAT.
I wanted to assume it to solve it, but I refrained.
The figure in incomplete, does not have full information about 4u² square side is aligned with the vertical radius projection.
Thank for premath...
So nice of you Taha! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
great quickly resolution ... i have used other one ... based on to find the angle of the triangle costructed among the 2 squares and from it i have "comed up" to the angle of the center of the circle .... next R= sqrt(5) / 2 * sin(half_center_angle) ....
Thanks Roberto for the feedback. You are awesome 👍 Take care dear and stay blessed😃
I did this too! But I like PreMath's solution better, 'twas more elegant.
Good video!
Your volume is too faint. I can barely hear you.
Ubique el centro de la circunferencia en (0,0) y los puntos de tangencia con los cuadrados en (0,-r) y (-2,1-r).
Cómo los puntos pertenecen a la circunferencia satisfacen la ecuación x^2 + y^2 = r^2
Remplazando (-2,1-r) en la ecuación tenemos:
(-2)^2 + (1-r)^2 = r^2
4. + 1 -2r +r^2 = r^2
2r = 5
r = 2.5
Sir. I don't know english it's google translate
The distance from point A to point B may be r-1 or more.
Because there is no condition that the center of the circle is the same as the Y axis of the red square
Thank you sir
You are very welcome Gowri! Good to hear from you.
Thank you so much for your continued love and support. Take care dear and stay blessed😃
What if the hypotneous of the big right angle triangle is r+the diagonal of the biger square and the adjuscent is 3+2 and the oposite side is r+2
If the radius touches one corner of the squar with side 3 since it is a square it touches the other opposite corner of the sguare straight down so i thich we can add up r and diagonal as a hypotenouus of the big right angle triangle
Assumption that the radios and side of red square are linear is probably incorrect
As through 2 points infinite circles pass
Like side of blue square is not colinear with radius, the red square side as well need not be.
ग्रेट (Great) !
ac at this case is 90 degree to red square .But you can't thinking 100% to red block is 90 degree
if we extend the chords AC and BD, we will get (2r-1)x 1=4
tangent to the blue square?
Data insufficient, sir. You should have specified that upper side of smaller square is tangent to yhe circle, but nowhere in the question it is written.
Another unbelievably complex solution to such a simple question using the chord x chord intersection method 2 x 2 = 1 x (2r - 1) r= 2.5. That took 15 seconds.
Good effort sir carry on.God bless u sir.
Thank you so much Safeer for your kind words.
You are awesome 👍Take care dear and stay blessed😃
I thought you would explain like this. Extend BD to the right of the circle, say F. Extend the radius AC up and call it G. BD x DF =DC x DG. Difference of height between two squares is 3-2=1. BD = 2. So DF=2. 2 x 2 = 1 x DG. DG=4, DIA =4+1=5. r=2.5 etc etc (your regular style of solving) Regards.
The problem as given does not define a single circle, but a family of circles. Point C as drawn is on a vertex of the square 2. However, as given, C can be anywhere on the top surface of square 2 and still be tangent. Thus EC can be any value between 2 and 4.
This is a bad problem. By your tangent definition, their are many different circles that can be *tangent* to those 2 squares.
thanks you
Стороны квадратов 3 и 2. Тогда катеты треугольника: 2 и 1, а гипотенуза V5- хорда окружности. Проведя к ней перпендикуляр, получим подобный по перпендикулярности сторон треугольник. 1:V5= 0,5*V5:R. R=5/2. S=25п/4. Ответ: 25п/4.
Why not use intersecting chords theorem. 2 x 2 = 1 x 4 D=5 r=2.5 much easier.
Red square is NOT tangent to circle, since it is not given.
Sorry but with the given information there's no way to tell that BEC and EBD are 90* only if the left line of the red square is continuation of AC
Bem bolado!
Thanks Jammal!
Obrigado querido pelo feedback. Você é incrível 👍 Cuide-se querido e continue abençoado😃
I got the result by using the chord theorem..
👌👍👏❤
9 is noT at All concerned with Circle PLacE 4 inSiDe Circle radius 4 area 4× 4 --- 16 units
Ojo!!! El círculo sólo es tangente al cuadrado pequeño.
Very wrong definition ! The circle in tangent only to the red circle and exactly in its left corner above ! There are an infinity of circles tangent to the red square !!!
Wrong definition, The circle is tangent only to the red square !!!
Sorry sir, the blue square is not tangent to circle, so this qustian is wrong.
Also i enjoyed you're all qustians,
Thanks
Area of the circle=25π/4
The two squares can't support the circle because of the square vision of 2D.
A red square justly can enter between blue and red squares!
A Circle that is Tangent to Two Squares...!
*This video should've been deleted.*
You can go fast. No need to go so slow and tell in detail the basic steps. Will take much less time.