This is a trick question because if you go and see the actual Mona Lisa in person, there will be so many people in the way that there are only two reasonable viewing distances and that is from either the back wall or the front of the crowd after you wait for 3 hours.
... and I haven't gotten over the trick of is Mona Lisa smiling as happy to see her painter Da Vinci? Or is her gaze sincere of why am I posing for this painter Da Vinci general gaze? A mathematical symmetrical formula from the triangular eyes as the base and her eyes-nose-bottom lips is the height and bottom lip center to either eye ends all form an equilateral triangle DaVinci started with before starting to paint and the pull back of the tip of the lower lips center to each eye end place doesn't suggest smile or gaze answer to the painting viewer! 😊
The Mona Lisa is the classic over-rated painting. Before the famous theft, hardly anyone gave it a second glance. It’s fine Give me a good Van Gogh or Monet anyday… da Vinci’s sketches hold more charm (all personal opinion, of course…. If you think otherwise, fine)
Couldn't remember the formula for the tangent of difference of angles, but writing it as arctan(32/x) - arctan(18/x) and finding the derivative when x is 0 did the job easily
The problem can easily be solved using the inscribed angle theorem. Any point Q on a circle through A and B forms the same angle AQB, and this angle increases when the radius of the circle through A and B decreases. The center M of a circle through A and B is on the horizontal line halfway between A and B. In order for a point P on the x-axis to fit on such a circle, the radius cannot be smaller than 18 + 14/2 = 25. Then by Pythagoras, the distance of P (and M) to the y-axis equals sqrt(25^2-7^2) = 24, and that is the answer…
This is clever, because the smallest possible circle offers the largest possible inscribed angle, the minimum size of the circle through A and B and P on the x-axis is already the circle you are looking for, and all you need there is the missing leg of the right triangle 7/x /25. Not bad.
And I don't think this proves the maximum angle couldn't be to the left of the point below the crnterlf the cirlce..how would you know..this only shows the maximum angle is not ont to the right of the center of the circle..
@@leif1075 if the left intersection of the circle with the x-axis moves further to the left, the circle must become larger again. And as the circle gets bigger, the angle gets smaller again.
Personally, I prefer this geometric proof below: Circumscribe a circle "w" in the triangle ABP (radius R). Consider the law of sines: (where α is the angle APB) |AB|/sin(α) = 2R -> R*sin(α) = |AB|/2 = const. -> sin(α) is maximal if R is minimal -> R is minimal if the circle "w" is tangent to the line OP This is how we prove that the circle "w" needs to be tangent to OP at the point P Radius of the circle "w" is equal to |OB| + |BA|/2, because the center of our circle is on the perpendicular bisector of AB. R = 18 + 14/2 = 18 + 7 = 25 So the radius is equal to 25. To calculate the distance, we use the legendary Pythagorean theorem. First label midpoint of AB as M and center of the circle "w" as C. Angle BMC is, by definition, a right angle, so: |BM|^2 + |MC|^2 = |BC|^2 |BC| is the radius R, so it's equal to 25 |BM| is exactly half of |AB|, so it's equal to 14/2 = 7 |BM|^2 + |MC|^2 = |BC|^2 -> 7^2 + |MC|^2 = 25^2 -> |MC|^2 = 25^2 - 7^2 = 625 - 49 = 576 -> |MC|^2 = 576 -> |MC| = 24 !!! "And that's the answer" ... achieved using only the basic theorems :)
I did a similar geometry puzzle on my channel related to height, and developed a formula for it, which I have yet to find anywhere in the literature. Very cool!
Using vectors and inner product: write P = [x, 0], B = [0, 18] and A = [0, 32]. The inner product between PB and PA will tell us the angle between them: = |PB| |PA| cos(theta). Squaring both sides to simplify the lengths gives the equation cos^2(theta) = (x^2 + 18*32)^2 / (x^2 + 18^2) * (x^2 + 32^2). Theta is maximized when cos^2(theta) is minimized, and WolframAlpha says that the RHS is minimized when x = -24, 24.
Using WolframAlpha is a bit of a cop-out for a problem as simple as this (especially when you can just type "maximize atan(x/18)-atan(x/24)" into WolframAlpha in the first place) -- I'm curious how simple that quadric-divided-by-a-quadric is to find a derivative for by hand...
I went through the pain of doing a cosine law bash and getting the same function as you, but I decided to torture myself further by taking a derivative. I got the correct answer after about 5 minutes😅
Interesting. When I saw the setup, I thought, "something to do with a circle and inscribed angles". But didn't get any further than just a 'hunch'. Thanks for sharing.
I used calculus to solve this problem...consider the function arctan(32/x) - arctan(18/x) then set it's slope to be 0..u can solve easily using inverse trig nd calculus
This problem is similar to the decision a rugby kicker has to make when deciding how far to place the rugby ball away from the touch line when taking a conversion kick following a try. They need to open the angle up whilst also limiting the total distance to ensure the ball has sufficient height and accuracy.
A geometric construction (with circles and straight lines only) of the "best" point P, given O, A and B: Consider the point C which is the symetric of A in the symetry of center O. Then you draw the semi cercle which diameter is [B,C]. Its intersection with the Ox axis is P (meaning OP = sqrt (ab)).
i dont know calculus yet, i solved it using trig and the am-gm inequality, (which i found much more simpler as idk any of the derivative stuff, and it looked complicated, prolly cause idk what it means). i like competitive math and the am-gm inequality is basically a must know, and its really simple and effective in its application
While I knew the answer to your question I just want to point out that no one looks at a picture from the floor, and you can easily move the picture. By moving the picture down you decrease the size of the circle which increases the angle.
Another stating of the problem, maybe more catching, is like this: « Your usual standing (or sitting, for that matter) point is 2,4 meters from the wall, eyes at height 1,7 m. At which height should you hang your painting if its length is 1,4 m? » The answer is not 1,7+3,2=4,9 m high; it is placing the center at eye level. Meaning that find the best P for a fixed H is not find the best H for'fixed P. counterintuitive. That is because we prefer our eyes watching horizontally, not straining them or our neck upwards. The difference in viewing angle is sufficiently small to not matter: 30,25° against 23,77°.
The locus of P for a given angle APB is a circle, You want the smallest possible circle to maximise angle APB, that is the circle through A,B tangent to the base line. The rest follows from Pythagoras. Easy.
I don't understand this problem. When I hang pictures on my walls, point B is typically BELOW my height, i. e. below point O, making OB a length on the negative y axis. Why would I place paintings so I have to tilt my head backwards, looking up, instead of just looking straight at the painting? Obviously I also consider the aesthetics of the placement in relation to other stuff on the wall, as well as furniture etc. This problem makes no sense at all to me...
If you notice you can inscribe a circle that contains the points A B (on the wall) and a point on an x axis (eye level) you can use the Tangent-Secant theorem x² = AB, so x =√ AB. All angles on the circle are the same. All angles outside the circle are smaller. All angles inside the circle are larger. Since you are constrained on the x axis, the point of the circle that coincides with the x axis is the answer for the maximum viewing angle. √ (18 units (14 units + 18 units)) = 24 units.
@@leif1075 the inscribed angle theorem says it ... any angle that has a vertex on the circle is between one that has a point inside the circle and outside the circle. if it is on the line then the angle that is in the center is twice the size. The special case is when it is at the center it is 2 times the angle. caddellprep.com/wp-content/uploads/2013/07/Central-And-Inscribed-Angles.jpg So if it is ANYWHERE outside the circle it is smaller.
At the geometric solution it's possible to do it without any measurement, just with a straight edge/ruler and a compass. Find the midpoint (M) of A and B by bisecting the line between with the compass. Than draw a circle with distance OM around M. With this same radius starting from A or B, you will get point E (where it intersect the perpendicular line from the first step or just draw two arcs from A and B to get E) Now draw another circle with the same radius (OM) from point E. Take the radius ME to get point P from the origin O
I have a simpler solution. To maximize the angle, both triangles should be the same (by same, I mean having the same angles, didn't learn mathematics in English) and in this case, it is easy to see that both of them are 3-4-5 triangles. The small one is 18-24-30 whereas the big one is 24-32-40. Hope this helps.
I saw this problem for rugby conversion kicks where you choose how far downfield to kick from, but must be at the same left/right position where the try crossed the goal line.
Sir, please solve this problem and tag me . Question:- Take a square PQRS. Draw two arcs PR and QS such that they are centered at Q and P respectively. Now, draw circle touching the arcs PR and QS externally as well as the side RS. If side length of your square is 16 cm, what is the diameter of the small circle formed?
Interesting geometric proof at the end! I wouldn't say it's a practical solution since it's not at all obvious, but I imagine mathematicians before Newton and Leibniz needed to resort to methods like these to solve this kind of problem.
How I solved it: We need to find the line OP for which angle APB is the greatest. We can write an function to find angle APB in terms of OP. APB is equal to APO-BPO. Tan(APO)=32/OP (trig ratio) => APO=arctan(32/OP). Tan(BPO)=18/OP => BPO=arctan(18/OP) APB=arctan(32/OP)-arctan(18/OP) From there, if Pre-Calc/Trig taught me anything, plug it into Desmos. Solution is (24,16.26), or 24 ft away, which gives a view angle of 16.26.
@@suspended3785 if you want to find 24, then its basic algebra and can be done on pen and paper as shown in proof 1, after that the viewing angle can be found by simple division
@@suspended3785 I don't know a whole bunch about calc, however, one way that would work would be to find when the derivative equals 0 (if you don't know what a derivative is, it is essentially the slope at any given point on a function. I highly recommend 3blue1brown's channel if you want to learn more about derivatives.) You would take the derivative of the function arctan(32/x)-arctan(18-x) (with x representing OP), which I don't know how to do, otherwise I would explain it, but per Google you get -32/((x^2)+1024)-(-18/((x^2)+324). From there, you need to find the zeroes of that function, so you need to rewrite the function with common denominators, which is -32((x^2)+324)/((x^2)+1024)((x^2)+324) - (-18)((x^2)+1024)/((x^2)+1024)((x^2)+324). Now we simplify by distributing the -32 and -18 in the numerator. Since we are subtracting a negative value, I also changed the sign to addition. -32x^2-10368/((x^2)+1024)((x^2)+324) + 18x^2+18432/((x^2)+1024)((x^2)+324) Since there are common denominators, we can combine the numerators. 14x^2+8064/((x^2)+1024)((x^2)+324) Now, since we want to find the zeros, we can ignore the denominators, because all we want to find is when the numerator equals 0, because that makes the function equal 0. So, we have 14x^2+8064=0 Divide both sides by 14, x^2+576=0 Subtract 576 to the other side, x^2=-576 **I made a sign error somewhere , but I really don't feel like doing error analysis, so someone else can correct my mistake if they want to** I should have gotten x^2=576 Square root both sides, x=+-24 Since you cant see through the wall the painting is hanging on, -24 does not work. Going back to when we disregarded the denominators, you have to make sure 24 doesn't cause you to divide by 0, because that would mean the answer is undefined. Therefore, 24 is the answer.
That would effectively make OA negative and OB positive, which makes the suggested formula the square root of a negative number, i.e., an imaginary number. I'm not sure if it's mentioned in the video, but in general, the derivative has three roots: x = 0, x = sqrt(OA * OB), and x = infinity. It appears that once A moves below the x-axis, sqrt(OA * OB) becomes imaginary, and the x = 0 root goes from being a minimum to a maximum. TL;DR, the maths works out just fine!
The problem can be solved from a symmetry perspective. If the viewing angle is cut in halves, to maximize it, there should not be any preference for either halve, even if the picture is seen upside down. For that reason, the viewing angle should be centered around a 45° angle. Therefore Tan ø = a/x = x/b Then x= sqrt(a*b)
Quite a nice brain teaser. I solved it by imagining a person walking to the right from O at a constant velocity v. Angle APB is maximum when the rate of change of angles Alpha and Beta is the same. If v*t is the distance OP at any time t, then solve for v*t by equating the derivatives of rates of change of angles Alpha and Beta w.r.t. t. You will end up with vt=24. Yay!
I could only remember the law of cosines, plugged everything in, took derivative to find minimum of cosine of viewing angle, and after some simplifications I too arrived at sqrt((18^4-18^2*32^2+14*18^3+32^2*18*14)/(32^2-18^2-(18*28)))=24
I'm happy to say that I solved it in a minute. You just have to realize that the angle APB is maximum when the sine of it's measure is maximum. Then by the law of sines we have 2R = a / sin(alpha), so you just need to find a Point P such that the circumcircle of the triangle APB is minimal. The midpoint of the circumcircle has to be on the perpendicular bisector of AB. So we can use the pytharorean theorem (or ghogus theorem or whatever you call it) and get (|OP|² + (14/2)²) = (14 / 2 + 18)² which results in |OP| = 24.
Why does the midpoint of the circumcirlce have to be on the perpendicular bosectpr of AB..that's not true if by circumcise you mean a circle inscribed by the top smaller triangle APB..only if you meant a bigger circle in the complete larger right triangle AB whatever the 3rd point was...or did I misunderstsnd you?
Why is the painting so damn high? Is the observer a child? I suppose if an adult were viewing the painting, the maximum angle would be reached if they put their eye on the painting.
The units are completely arbitrary. It isn't necessarily a realistic dataset for this problem. Those could be decimeters, and the result would still be the same, except in units of decimeters.
@@carultch The painting is above the person's head. The units are indeed arbitrary, but in order for the maximization problem to have a non zero solution, the painting must be above the person's head! What art gallery does that?
Just use the fact OP/a=cot(x+cot^-1(OP/b)) Solving for OP gives a quadratic equation with discrriminant (a-b)^2cot^2(x)-4(ab) since discrminant is always postive and cotx is a decreasing function. You can set equal zero to get max value of x cot(x)=√(4ab/(a-b)^2)) Put this back in the equation (a-b)cotx/2+discriminant(0)= Simpliyng gives √ab It also gives x=cot^-1(√(4ab/(a-b)^2)
Why would anyone think to duplicate the bottom half of the line AB like in Preshs second method??why not duplicate both the lines not just the bottom half..that would make more sense and be less contrived right? But I don't see why anyone would think to inscribe a circle in the triangle like your method..what made you think of that? Why?? And how would you know whether to inscribe it in the "lower" triangle versus the bigger one? Only trial and error I supppse right? But even then how eould you know what to do with the circle??
Why would anyone think to duplicate the bottom half of the line AB..why not duplicate both the lines...that would make more sense and be less contrived right?
Um, I have a problem with this problem. The point "P" represents some hypothetical viewer's line-of-sight with wall-OA, but it's positioned at floor-level where no one's natural line-of-sight would realistically be. Since the average person is a little more than five and a half feet tall (I don't know exactly what that is in meters for the metric fetishists out there, but I do know that it's between 1.5 and 2m.), then average eye-level is around 5'4"; so, why isn't eye-level for this diagram adjusted for that?
The line OP represents eye level, not the floor. All we're interested in is that point B is 18 inches/centimeters above eye level, regardless of the viewer's eye level above the floor.
Where did the negative go in front of (a-b)? And where did the 2x^2 go when the (a-b) was factored? Just confused on the jump from one to the next. Thanks in advanced.
So if a 14" picture was hanging 18" off of the ground (to the bottom), and you laid your face down on the floor and wanted to look up at it-- you would only need to be 2ft away. It sure seems like that wouldn't work out very well. lol.. Math is weird. But I do get how the angle will flip and decrease as you get further away. You should have made a graphic to show that. At the "maximum point, the angle decreases in both directions as you move from there.
As usual, I paused the video once you stated the problem and solved it. I ended up using calculus, since this is the most obvious method for optimizing functions of continuous variables. However, instead of maximizing the tangent of the angle, I maximized the angle itself, which can be expressed as the difference of two arctangents, specifically arctan(32/x) - arctan(18/x). Solving it this way turned out to be somewhat simpler than your derivation. However, I did find your use of the AM-GM inequality quite clever!
In geometric method you have taken point p' away from op to prove angle AP'B is lesser than APB , what if p' inside op it also can be done but problem is very good trigonometric solution is more methodical
The same argument still works. The key point is that if P' is inside OP, it's still OUTSIDE the circle ABP. And the rest of the argument shown in the video still flows from there.
I'm ashamed to say that I'm math illiterate, but when it comes to paintings not hung ridiculously high, the best viewing distance is usually the diagonal of the painting.
My attempt : APB = APO - BPO = tan-¹(32/OP) - tan-¹(18/OP) So we want to maximum f(x) = tan-¹(32x) - tan-¹(18x) for x positive. (x = 1/OP) f'(x) = 32/(1+(32x)²) - 18/(1+(18x)²) This has the same sign as : 32(1+18²x²) - 18(1+32²x²) = 14 - 32×18×14x² This has the same sign as : 1 - 32×18x² = 1 - 24²x² = (1+24x)(1-24x) This has the same sign as : 1-24x Therefore, f(x) has a maximum for x = 1/24 Therefore, OP = 24
Idk, I just made an equation for theta in terms of x and then found the stationary point via differentiation. Got the same answer. I can't imagine the torment ancient mathematicians had to go through to find the answer to a simple equation without using calculus. 🤔
In fact, for some paintings, I dare say that the best viewing angle actually minimizes the field of view rather than maximizing, since this reduces the risk of actually seeing the picture.
am-gm inequality. the minimum value occurs when they are equal to each other. its kind of intuitive as well when u think of it, but u should also understand the proof of am-gm. its pretty cool, and a hella useful inequality in competitive math.
Wait and why did you draw a second circle at 11:58..that makes no sense and also comes out of nowhere. Sorry.i dont think this second solution makes much sense..
A Ben Sparks numberphile video about a rugby kick (ua-cam.com/video/rHdYv62F5fs/v-deo.html) is essentially the same problem. In addition to solving the problem in a similar fashion (with the cricle), he concludes that a close approximation to the ideail angle is to choose x to be the same distance as the distance to the midpoint of AB. This gives a result of 25, pretty close to the actual result of 24.
@@henrik3141That doesn't answer the question of why. I could come up with plenty of adversarial combinations of 18 and 14 that have nothing to do with this situation, and still get to 24. It doesn't mean it is the correct method, just because it gets to 24 the end. It could just be a coincidence that it gets to 24. In other words, what specifically do the operations you did in your method, have to do with this situation, and how did you determine them?
An interesting application! But rugby players might want to take a slightly smaller value of OP, to gain a substantial reduction in the length of the kick in return for a negligible reduction in apparent angular size of the goal. Whereas the solution in this video is completely obsessed with the apparent angle, and ignores distance.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
This is a trick question because if you go and see the actual Mona Lisa in person, there will be so many people in the way that there are only two reasonable viewing distances and that is from either the back wall or the front of the crowd after you wait for 3 hours.
... and I haven't gotten over the trick of is Mona Lisa smiling as happy to see her painter Da Vinci? Or is her gaze sincere of why am I posing for this painter Da Vinci general gaze? A mathematical symmetrical formula from the triangular eyes as the base and her eyes-nose-bottom lips is the height and bottom lip center to either eye ends all form an equilateral triangle DaVinci started with before starting to paint and the pull back of the tip of the lower lips center to each eye end place doesn't suggest smile or gaze answer to the painting viewer! 😊
Was Mona Lisa a mona or a screema?
The Mona Lisa is the classic over-rated painting. Before the famous theft, hardly anyone gave it a second glance.
It’s fine
Give me a good Van Gogh or Monet anyday… da Vinci’s sketches hold more charm (all personal opinion, of course…. If you think otherwise, fine)
“Kwekt”
In short. The best angle is from a book📚📚
Couldn't remember the formula for the tangent of difference of angles, but writing it as arctan(32/x) - arctan(18/x) and finding the derivative when x is 0 did the job easily
did same, took only 5 lines
Same, using the formula is actually worse.
Yep did same thing here
it was an exam problem right after we learned the derivative of arctan(x). I think half of our class got it right:)
This video is a must see! I was astonished with the geometric proof. So beautiful! ❤❤❤
The problem can easily be solved using the inscribed angle theorem. Any point Q on a circle through A and B forms the same angle AQB, and this angle increases when the radius of the circle through A and B decreases. The center M of a circle through A and B is on the horizontal line halfway between A and B. In order for a point P on the x-axis to fit on such a circle, the radius cannot be smaller than 18 + 14/2 = 25. Then by Pythagoras, the distance of P (and M) to the y-axis equals sqrt(25^2-7^2) = 24, and that is the answer…
This is clever, because the smallest possible circle offers the largest possible inscribed angle, the minimum size of the circle through A and B and P on the x-axis is already the circle you are looking for, and all you need there is the missing leg of the right triangle 7/x /25.
Not bad.
But there are no circles here so why would you think of that and how does it make sense??
And I've never heard of the tangent secsnt theorem..is that well known..if you don't know that theorem can't you solve some other way at that point??
And I don't think this proves the maximum angle couldn't be to the left of the point below the crnterlf the cirlce..how would you know..this only shows the maximum angle is not ont to the right of the center of the circle..
@@leif1075 if the left intersection of the circle with the x-axis moves further to the left, the circle must become larger again. And as the circle gets bigger, the angle gets smaller again.
I typically don’t crawl around on my stomach in an art gallery 😂
Personally, I prefer this geometric proof below:
Circumscribe a circle "w" in the triangle ABP (radius R).
Consider the law of sines: (where α is the angle APB)
|AB|/sin(α) = 2R
-> R*sin(α) = |AB|/2 = const.
-> sin(α) is maximal if R is minimal
-> R is minimal if the circle "w" is tangent to the line OP
This is how we prove that the circle "w" needs to be tangent to OP at the point P
Radius of the circle "w" is equal to |OB| + |BA|/2, because the center of our circle is on the perpendicular bisector of AB.
R = 18 + 14/2 = 18 + 7 = 25
So the radius is equal to 25.
To calculate the distance, we use the legendary Pythagorean theorem.
First label midpoint of AB as M and center of the circle "w" as C.
Angle BMC is, by definition, a right angle, so:
|BM|^2 + |MC|^2 = |BC|^2
|BC| is the radius R, so it's equal to 25
|BM| is exactly half of |AB|, so it's equal to 14/2 = 7
|BM|^2 + |MC|^2 = |BC|^2
-> 7^2 + |MC|^2 = 25^2
-> |MC|^2 = 25^2 - 7^2 = 625 - 49 = 576
-> |MC|^2 = 576
-> |MC| = 24 !!!
"And that's the answer"
... achieved using only the basic theorems :)
I did a similar geometry puzzle on my channel related to height, and developed a formula for it, which I have yet to find anywhere in the literature.
Very cool!
Can you give a link?
Using vectors and inner product: write P = [x, 0], B = [0, 18] and A = [0, 32]. The inner product between PB and PA will tell us the angle between them: = |PB| |PA| cos(theta). Squaring both sides to simplify the lengths gives the equation cos^2(theta) = (x^2 + 18*32)^2 / (x^2 + 18^2) * (x^2 + 32^2). Theta is maximized when cos^2(theta) is minimized, and WolframAlpha says that the RHS is minimized when x = -24, 24.
Using WolframAlpha is a bit of a cop-out for a problem as simple as this (especially when you can just type "maximize atan(x/18)-atan(x/24)" into WolframAlpha in the first place) -- I'm curious how simple that quadric-divided-by-a-quadric is to find a derivative for by hand...
@ Original poster -- You are missing required grouping symbols around that denominator. Your intended equation is of this form: A = B/(C*D).
I went through the pain of doing a cosine law bash and getting the same function as you, but I decided to torture myself further by taking a derivative. I got the correct answer after about 5 minutes😅
Interesting. When I saw the setup, I thought, "something to do with a circle and inscribed angles". But didn't get any further than just a 'hunch'. Thanks for sharing.
I used calculus to solve this problem...consider the function arctan(32/x) - arctan(18/x) then set it's slope to be 0..u can solve easily using inverse trig nd calculus
This problem is similar to the decision a rugby kicker has to make when deciding how far to place the rugby ball away from the touch line when taking a conversion kick following a try. They need to open the angle up whilst also limiting the total distance to ensure the ball has sufficient height and accuracy.
The geometric solution is beautiful. Thanks a lot for sharing it.
A geometric construction (with circles and straight lines only) of the "best" point P, given O, A and B:
Consider the point C which is the symetric of A in the symetry of center O.
Then you draw the semi cercle which diameter is [B,C]. Its intersection with the Ox axis is P (meaning OP = sqrt (ab)).
i dont know calculus yet, i solved it using trig and the am-gm inequality, (which i found much more simpler as idk any of the derivative stuff, and it looked complicated, prolly cause idk what it means). i like competitive math and the am-gm inequality is basically a must know, and its really simple and effective in its application
My solution, which is more calculation-heavy: Draw the triangle from 1:03. Let
While I knew the answer to your question I just want to point out that no one looks at a picture from the floor, and you can easily move the picture. By moving the picture down you decrease the size of the circle which increases the angle.
Beautifully explained through all methods
Another stating of the problem, maybe more catching, is like this:
« Your usual standing (or sitting, for that matter) point is 2,4 meters from the wall, eyes at height 1,7 m.
At which height should you hang your painting if its length is 1,4 m? »
The answer is not 1,7+3,2=4,9 m high; it is placing the center at eye level. Meaning that find the best P for a fixed H is not find the best H for'fixed P.
counterintuitive. That is because we prefer our eyes watching horizontally, not straining them or our neck upwards. The difference in viewing angle is sufficiently small to not matter: 30,25° against 23,77°.
The locus of P for a given angle APB is a circle, You want the smallest possible circle to maximise angle APB, that is the circle through A,B tangent to the base line. The rest follows from Pythagoras. Easy.
That's actually a lovely way of looking at it. Nice!
What do you mean by this and where can I also get that information? Thanks
Thanks! Now I can find the optimal position to place my TV for the start of NFL season.
I don't understand this problem.
When I hang pictures on my walls, point B is typically BELOW my height, i. e. below point O, making OB a length on the negative y axis.
Why would I place paintings so I have to tilt my head backwards, looking up, instead of just looking straight at the painting?
Obviously I also consider the aesthetics of the placement in relation to other stuff on the wall, as well as furniture etc.
This problem makes no sense at all to me...
If you notice you can inscribe a circle that contains the points A B (on the wall) and a point on an x axis (eye level) you can use the Tangent-Secant theorem x² = AB, so x =√ AB. All angles on the circle are the same. All angles outside the circle are smaller. All angles inside the circle are larger. Since you are constrained on the x axis, the point of the circle that coincides with the x axis is the answer for the maximum viewing angle. √ (18 units (14 units + 18 units)) = 24 units.
Why would anyone think of inscribed a circle..and how does that prove the an angle to te left of the circle x axis point could be greater..
@@leif1075 the inscribed angle theorem says it ... any angle that has a vertex on the circle is between one that has a point inside the circle and outside the circle. if it is on the line then the angle that is in the center is twice the size. The special case is when it is at the center it is 2 times the angle. caddellprep.com/wp-content/uploads/2013/07/Central-And-Inscribed-Angles.jpg So if it is ANYWHERE outside the circle it is smaller.
At the geometric solution it's possible to do it without any measurement, just with a straight edge/ruler and a compass.
Find the midpoint (M) of A and B by bisecting the line between with the compass.
Than draw a circle with distance OM around M.
With this same radius starting from A or B, you will get point E (where it intersect the perpendicular line from the first step or just draw two arcs from A and B to get E)
Now draw another circle with the same radius (OM) from point E.
Take the radius ME to get point P from the origin O
The geometric solution is cool.
In the case of the trigonometric solution path, there is a somewhat simpler derivation:
a = 32
b = 18
artan(32/x) - artan(18/x) = max (should be maximum)
artan(32*x^-1) - artan(18*x^-1) = max
derivation rule: f(x) = artan(g(x)) --> f'(x) = 1 / ( 1 + (g(x))² ) * f'(g(x))
--> derivation:
1/(1+(32/x)²) * (-32/x²) - 1/(1+(18/x)²) * (-18/x²) = 0
1/(1+(32/x)²) * (-32/x²) = 1/(1+(18/x)²) * (-18/x²)
-32/(x²+32²) = -18/(x²+18²)
-(x²+32²)/32 = -(x²+18²)/18
1/32 x² + 32 = 1/18 x² + 18
(1/18-1/32)x² - 14 = 0
0.024x² - 14 = 0
x² - 576 = 0
pq:
x = 24
I have a simpler solution. To maximize the angle, both triangles should be the same (by same, I mean having the same angles, didn't learn mathematics in English) and in this case, it is easy to see that both of them are 3-4-5 triangles. The small one is 18-24-30 whereas the big one is 24-32-40. Hope this helps.
Cool. I believe that the word you're looking for is "similar triangles" (they don't have to be the same size, they have to have same angles).
Nice and easy problem,but the 2 methods of solution are very nice and creative.
I saw this problem for rugby conversion kicks where you choose how far downfield to kick from, but must be at the same left/right position where the try crossed the goal line.
That's a numberphile video by Ben Sparks right?
@@phoquenahol7245 Could be. Memory being what it is, I'm 95% on Numberphile, and 85% on Sparks.
Sir, please solve this problem and tag me .
Question:-
Take a square PQRS. Draw two arcs PR and QS such
that they are centered at Q and P respectively. Now,
draw circle touching the arcs PR and QS externally as
well as the side RS. If side length of your square is 16
cm, what is the diameter of the small circle formed?
16 cm?
Interesting geometric proof at the end! I wouldn't say it's a practical solution since it's not at all obvious, but I imagine mathematicians before Newton and Leibniz needed to resort to methods like these to solve this kind of problem.
How I solved it:
We need to find the line OP for which angle APB is the greatest. We can write an function to find angle APB in terms of OP.
APB is equal to APO-BPO.
Tan(APO)=32/OP (trig ratio) => APO=arctan(32/OP).
Tan(BPO)=18/OP => BPO=arctan(18/OP)
APB=arctan(32/OP)-arctan(18/OP)
From there, if Pre-Calc/Trig taught me anything, plug it into Desmos.
Solution is (24,16.26), or 24 ft away, which gives a view angle of 16.26.
How to do it without desmos or any calculator
@@suspended3785 if you want to find 24, then its basic algebra and can be done on pen and paper as shown in proof 1, after that the viewing angle can be found by simple division
@@suspended3785 I don't know a whole bunch about calc, however, one way that would work would be to find when the derivative equals 0 (if you don't know what a derivative is, it is essentially the slope at any given point on a function. I highly recommend 3blue1brown's channel if you want to learn more about derivatives.)
You would take the derivative of the function arctan(32/x)-arctan(18-x) (with x representing OP), which I don't know how to do, otherwise I would explain it, but per Google you get -32/((x^2)+1024)-(-18/((x^2)+324).
From there, you need to find the zeroes of that function, so you need to rewrite the function with common denominators, which is -32((x^2)+324)/((x^2)+1024)((x^2)+324) - (-18)((x^2)+1024)/((x^2)+1024)((x^2)+324).
Now we simplify by distributing the -32 and -18 in the numerator. Since we are subtracting a negative value, I also changed the sign to addition. -32x^2-10368/((x^2)+1024)((x^2)+324) + 18x^2+18432/((x^2)+1024)((x^2)+324)
Since there are common denominators, we can combine the numerators. 14x^2+8064/((x^2)+1024)((x^2)+324)
Now, since we want to find the zeros, we can ignore the denominators, because all we want to find is when the numerator equals 0, because that makes the function equal 0.
So, we have 14x^2+8064=0
Divide both sides by 14, x^2+576=0
Subtract 576 to the other side, x^2=-576
**I made a sign error somewhere , but I really don't feel like doing error analysis, so someone else can correct my mistake if they want to**
I should have gotten x^2=576
Square root both sides, x=+-24
Since you cant see through the wall the painting is hanging on, -24 does not work. Going back to when we disregarded the denominators, you have to make sure 24 doesn't cause you to divide by 0, because that would mean the answer is undefined.
Therefore, 24 is the answer.
@@suspended3785 calculus or am-gm inequality
What if the midpoint between A and B was on the x axis? Would not the greatest angle be 180° when x equals 0?
That would effectively make OA negative and OB positive, which makes the suggested formula the square root of a negative number, i.e., an imaginary number. I'm not sure if it's mentioned in the video, but in general, the derivative has three roots: x = 0, x = sqrt(OA * OB), and x = infinity. It appears that once A moves below the x-axis, sqrt(OA * OB) becomes imaginary, and the x = 0 root goes from being a minimum to a maximum. TL;DR, the maths works out just fine!
The problem can be solved from a symmetry perspective. If the viewing angle is cut in halves, to maximize it, there should not be any preference for either halve, even if the picture is seen upside down. For that reason, the viewing angle should be centered around a 45° angle. Therefore
Tan ø = a/x = x/b
Then x= sqrt(a*b)
Amazing problem, nice solution. Congratulations 👏👏👏
Quite a nice brain teaser. I solved it by imagining a person walking to the right from O at a constant velocity v. Angle APB is maximum when the rate of change of angles Alpha and Beta is the same. If v*t is the distance OP at any time t, then solve for v*t by equating the derivatives of rates of change of angles Alpha and Beta w.r.t. t. You will end up with vt=24. Yay!
I could only remember the law of cosines, plugged everything in, took derivative to find minimum of cosine of viewing angle, and after some simplifications I too arrived at sqrt((18^4-18^2*32^2+14*18^3+32^2*18*14)/(32^2-18^2-(18*28)))=24
geometric solutions are always the best way to solve geometric problems, including optimization problems
I'm happy to say that I solved it in a minute.
You just have to realize that the angle APB is maximum when the sine of it's measure is maximum. Then by the law of sines we have 2R = a / sin(alpha), so you just need to find a Point P such that the circumcircle of the triangle APB is minimal. The midpoint of the circumcircle has to be on the perpendicular bisector of AB. So we can use the pytharorean theorem (or ghogus theorem or whatever you call it) and get (|OP|² + (14/2)²) = (14 / 2 + 18)² which results in |OP| = 24.
Why does the midpoint of the circumcirlce have to be on the perpendicular bosectpr of AB..that's not true if by circumcise you mean a circle inscribed by the top smaller triangle APB..only if you meant a bigger circle in the complete larger right triangle AB whatever the 3rd point was...or did I misunderstsnd you?
the second solution is absolutely beautiful, that is amazing
Wow! That letter took 552 years to arrive! :)
Why is the painting so damn high? Is the observer a child? I suppose if an adult were viewing the painting, the maximum angle would be reached if they put their eye on the painting.
The units are completely arbitrary. It isn't necessarily a realistic dataset for this problem. Those could be decimeters, and the result would still be the same, except in units of decimeters.
@@carultch The painting is above the person's head. The units are indeed arbitrary, but in order for the maximization problem to have a non zero solution, the painting must be above the person's head! What art gallery does that?
@@hujackus its not real i think, just a nice math problem
Just use the fact
OP/a=cot(x+cot^-1(OP/b))
Solving for OP gives a quadratic equation with discrriminant (a-b)^2cot^2(x)-4(ab) since discrminant is always postive and cotx is a decreasing function. You can set equal zero to get max value of x cot(x)=√(4ab/(a-b)^2))
Put this back in the equation (a-b)cotx/2+discriminant(0)=
Simpliyng gives √ab
It also gives x=cot^-1(√(4ab/(a-b)^2)
Why would anyone think to duplicate the bottom half of the line AB like in Preshs second method??why not duplicate both the lines not just the bottom half..that would make more sense and be less contrived right? But I don't see why anyone would think to inscribe a circle in the triangle like your method..what made you think of that? Why?? And how would you know whether to inscribe it in the "lower" triangle versus the bigger one? Only trial and error I supppse right? But even then how eould you know what to do with the circle??
14:07 why angle QAP' is greater than zero for any other point P' ?
I tilted my head 267 deg. Cheers.
Why would anyone think to duplicate the bottom half of the line AB..why not duplicate both the lines...that would make more sense and be less contrived right?
finally, 1 inch tall people at museums can have the best viewing experience
just brilliant !
Um, I have a problem with this problem. The point "P" represents some hypothetical viewer's line-of-sight with wall-OA, but it's positioned at floor-level where no one's natural line-of-sight would realistically be. Since the average person is a little more than five and a half feet tall (I don't know exactly what that is in meters for the metric fetishists out there, but I do know that it's between 1.5 and 2m.), then average eye-level is around 5'4"; so, why isn't eye-level for this diagram adjusted for that?
The line OP represents eye level, not the floor. All we're interested in is that point B is 18 inches/centimeters above eye level, regardless of the viewer's eye level above the floor.
Who said P is positioned at floor level? It is eye level
its positioned at eye level
Where did the negative go in front of (a-b)? And where did the 2x^2 go when the (a-b) was factored? Just confused on the jump from one to the next. Thanks in advanced.
You're much more likely to get help if you provide a timestamp to the point of the video you're talking about.
Thanks 🙏
So if a 14" picture was hanging 18" off of the ground (to the bottom), and you laid your face down on the floor and wanted to look up at it-- you would only need to be 2ft away. It sure seems like that wouldn't work out very well. lol.. Math is weird. But I do get how the angle will flip and decrease as you get further away. You should have made a graphic to show that. At the "maximum point, the angle decreases in both directions as you move from there.
A more modern and practical application of this, is how far back you should sit in a movie theater to maximize the viewing angle of the big screen.
I'm not convinced. If You want to maximize the viewing angle You should be placed in the center of the screen. No good place for watching.
@@popogast I mean anywhere there are seats, LOL
outdone!!
What if value of x is more then 24??????????
As usual, I paused the video once you stated the problem and solved it. I ended up using calculus, since this is the most obvious method for optimizing functions of continuous variables. However, instead of maximizing the tangent of the angle, I maximized the angle itself, which can be expressed as the difference of two arctangents, specifically arctan(32/x) - arctan(18/x). Solving it this way turned out to be somewhat simpler than your derivation. However, I did find your use of the AM-GM inequality quite clever!
Wonderful!
I don't need highest angle cause I can see at right angle by using the 'ladder stand' and view that painting!
This works only if the picture is above or below eye level, which is hardly ever the case. If the picture is within your eye level, ab
yea
In geometric method you have taken point p' away from op to prove angle AP'B is lesser than APB , what if p' inside op it also can be done but problem is very good trigonometric solution is more methodical
The same argument still works. The key point is that if P' is inside OP, it's still OUTSIDE the circle ABP. And the rest of the argument shown in the video still flows from there.
I should watch this math video again
I need to rewatch it again.
The geometric solution went over the head. From where did he get that 😅
I'm ashamed to say that I'm math illiterate, but when it comes to paintings not hung ridiculously high, the best viewing distance is usually the diagonal of the painting.
My attempt : APB = APO - BPO = tan-¹(32/OP) - tan-¹(18/OP)
So we want to maximum f(x) = tan-¹(32x) - tan-¹(18x) for x positive. (x = 1/OP)
f'(x) = 32/(1+(32x)²) - 18/(1+(18x)²)
This has the same sign as : 32(1+18²x²) - 18(1+32²x²) = 14 - 32×18×14x²
This has the same sign as : 1 - 32×18x² = 1 - 24²x² = (1+24x)(1-24x)
This has the same sign as : 1-24x
Therefore, f(x) has a maximum for x = 1/24
Therefore, OP = 24
That trigonometry solution is hugely over complicated. Similar to what others have done, I just graphed the difference of arctan 32/x and arctan 18/x.
its not if u just use am-gm inequality, i did the trig approach and the inequality and solved it super fast
Idk, I just made an equation for theta in terms of x and then found the stationary point via differentiation. Got the same answer. I can't imagine the torment ancient mathematicians had to go through to find the answer to a simple equation without using calculus. 🤔
Depends on the painting. They're not all best viewed from the same distance, even if they were all made the same size.
fr
In fact, for some paintings, I dare say that the best viewing angle actually minimizes the field of view rather than maximizing, since this reduces the risk of actually seeing the picture.
9:00 how the minimum occurs at x=ab/x ?
am-gm inequality. the minimum value occurs when they are equal to each other. its kind of intuitive as well when u think of it, but u should also understand the proof of am-gm. its pretty cool, and a hella useful inequality in competitive math.
With the differentiation of arctan, one quickly arrives at the equation 14x² = 8064.
Wow , I was also able to solve by Calculus method .
i did it with trigonometry by taking the maximum value theta can reach is 90 degrees .
According to every woman I’ve ever met, the correct height to hang something is exactly 3 inches lower than I have just hung it.
Wait and why did you draw a second circle at 11:58..that makes no sense and also comes out of nowhere. Sorry.i dont think this second solution makes much sense..
A Ben Sparks numberphile video about a rugby kick (ua-cam.com/video/rHdYv62F5fs/v-deo.html) is essentially the same problem. In addition to solving the problem in a similar fashion (with the cricle), he concludes that a close approximation to the ideail angle is to choose x to be the same distance as the distance to the midpoint of AB. This gives a result of 25, pretty close to the actual result of 24.
Used that kind of approximation as a lazy evaluation of the problem
neat!
Incredible this proof.
You lost me on the last two proofs.
Total Bouncer! 😂
The best viewing distance is when you are 2m tall and can see from above the crowd….😅
Complicated way of getting the direct answer. 18 + (18-14) + (18-14)/2 = 24
But WHY does your calculation work? What is the reasoning behind it?
@@carultch Because 24 is the correct answer
@@henrik3141That doesn't answer the question of why. I could come up with plenty of adversarial combinations of 18 and 14 that have nothing to do with this situation, and still get to 24. It doesn't mean it is the correct method, just because it gets to 24 the end. It could just be a coincidence that it gets to 24.
In other words, what specifically do the operations you did in your method, have to do with this situation, and how did you determine them?
@@carultch Yes but you can not just take random ways of combining 18 and 14.
@@henrik3141that's literally what you just did
And now my head hurts
You forget that humans have 2 eyes
11th comment? maybe?
🤔
Ngl, I used Pythagorean triplet of 3,4 and 5, solved it in a second and got my answer 24 as well XD
The Perfect Goal Kicking Angle - Numberphile
(ua-cam.com/video/rHdYv62F5fs/v-deo.html)
Rugby players use this all the time. Oh wait- no they don’t. They should
An interesting application! But rugby players might want to take a slightly smaller value of OP, to gain a substantial reduction in the length of the kick in return for a negligible reduction in apparent angular size of the goal. Whereas the solution in this video is completely obsessed with the apparent angle, and ignores distance.
You made it too complicated. Could be solved much easier.
Komen kedua, hadir Mr.
Trig sucks dude.
Just hang the painting at eye level.
the booba spot
I solved for 25 (allowed for high heels)😁
incredible content🤌🏻