Can you solve this Oxford admissions question?
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- Опубликовано: 1 май 2024
- The 2007 Oxford entrance exam had a delightful question: given equations of two circles, can you find the shortest distance and the points that are closest to each other? Special thanks this month to: Mike Robertson, Daniel Lewis, Kyle, Lee Redden. Thanks to all supporters on Patreon! / mindyourdecisions
Oxford 2007 test question 1, part D
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Oxford 2007 solutions question 1, part D
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The shortest distance between two disjoint circles always lies on the line between their centres.
That's what I thought, then you take the radii off that distance.
It also helps to find coordinates of P and Q by adding/subtracting a vector AB, normalized to corresponding radii, from coordinates of A and B
Oh, so just like the officials proposed it to be solved.
Yes... that is the rule we all memorise , but thanks to this video i understand now where this rule comes from
I almost got this one, so it wasn't that incredible.
Only if the two circles have no intersection together.
3:08 "No obvious way to do this"
Uh, I think it should be immediately obvious that the shortest distance between two circles would be along a straight line through their centers.
It's not obvious. Maybe intuitive, but not obvious.
took me 3 seconds to realize this
It's a fairly obvious idea to think of, sure. To a lot of people anyway. But there's a difference between coming up with the idea, and coming up with a way to prove that the idea is right
Sometimes things in math look pretty obvious, but actually turn out to be wrong, so mathematicians don't believe something just because it looks like it's true, even if it feels really really obvious. They still check whether they can actually prove it
And I don't think the proof is obvious. It's not all that hard to come up with a proof, but it's also not so obvious that you can just take it for granted
@@douglaswolfen7820 Thats a very nice thought! 😄😄
It's pretty obvious honestly
This can be solved within 1 - 2 minutes.
The distance between centers, that is the distance between (5,4) and (1,1) is simply sqrt((5-1)^2+(4-1)^2) = 5
Now subtract the two radii from 5, which are 2 and 1 from 5
We get 5-2-1=2
That was the easy bit tho.
@@kimba381The coord of points P and Q is far easier if you use the interception theorem
@@kimba381You also know that the right side of the circle equations is r². Thus the first circle has a radius of sqrt(4) = 2 and the second of sqrt(1) = 1, hence you have the distance between the perimeters of 5 - 2 - 1 = 2.
Yeah, but this wasn't an exam question, it was an interview question. You're there to have a conversation with the interviewers and explore the problem. You don't _want_ to solve it in two minutes. You want to use the opportunity to show your skills
Finding the solution is easy. If you want to impress, you need to be able to prove that it's the right solution
@douglaswolfen7820 I would rather show my ability to solve a problem in a fast and efficient way. To show my skills, first I will solve it by the same way (that is in 1-2 minutes), then I will spend some time in proving the distance formula, sqrt(...^2+....2). And will spend some time in proving that the shortest distance is indeed lies on the centres of the two circles. Spending time can be in some related things, not in solving the problem itself. (My opinion). Thanks for sharing your thoughts.
I actually did this in my head. Once it was clear that the centers of the two circles had a difference of 3 and 4, the resulting hypoteneuse had to be 5, which is the distance between the two centers. Subtract the two radii (1 and 2), and you get 2. Figuring out the coordinates of P and Q followed the same logic, since the triangles are all similar (3,4,5; 3/5, 4/5, 1; 6/5, 8/5, 2); so the coordinates of Q are (1.8, 1.6) and the coordinates of P are ( 3.4, 2.8).
Absolutely, correct approach, head math only.
You made this WAY too hard. 1. Find the distance between the two points and subtract the radii. 2. Find the equation on the line between the radii and determine where it intersects each circle.
Using parametric formula of the segment AB is easier than finding the intersection of a line and a circle.
Oh
Its also NOT CLEAR fro the wuestion ifnone cirlce isninside thebother or.not right?
Because he needs to make the video 8 minutes long.😉
I agree. In addition, the conclusion at the end of the video ('and that's the answer!') does not answer the original question, which is "what's the shortest distance".
Also AB is the hypotenuse of a 3-4-5 right triangle. We know the two sides from the graph are length 3 and 4, so AB is 5.
And for part 2, an equivalent approach to the vector stuff is to draw vertical lines through P and Q, then work with smaller triangles that are similar to the 3-4-5.
I love how you go into agonizing detail to show a the shortest distance between two points is a straight line (ugh), and then totally gloss over how to add vectors.
It's actually the inverse:
* The shortest distance between the centers is a straight line, yup.
* The shortest distance between two points on the edges is also a straight line, yup.
* But unlike the centers, we have a choice of edge points. Which ones lead to the shortest straight line? The ones that lie along the straight line between the centers. That may be intuitively obvious, but the video actually spells out the algebra underlying it (minimizing X + a couple constants is equivalent to minimizing X).
He doesn't actually show the shortest distance between 2 points is a straight line. He uses that fact, however, to prove the minimum distance between 2 points on disjoint circles lie on such a line (since the radii are constant and therefore the distance between the points on the circles must also be minimum).
That's not what he shows. Jesus, you people.
@@markkennedy9767thank you. You seem to be one of the few here who understands his proof. It's quite good.
It's hilarious that none of these 'math geniuses' answering your comment have managed to understand your point 😂
Shortest path is the line segment connecting the two centers. Subtract the radius of both circles from that line segment to get the distance. Don't even need to plot anything.
Easiest question ever LOL
Is it though, what about 1+1
Shift the axis by (1,1) so the small circle is at the origin, then find the minimum of the larger circle.
What exactly is the "minimum of the larger circle".
🤨
Minimum of the magnitude makes sense
You would prob need some kind of differential equation involving the magnitude of a parametric
I'd take the equation for the line, and tried to find a px and py thats both on the line and the circke equation. the vector variant was proberly easier to calculate.
I did the same. Far easier to add the vectors, and I had not considered it. Also, when he said vectors, I was going to make it more difficult by changing the vector into polar coordinates and then back to cartesian coordinates.
And I complain that my students make problems harder than they need be.
This isn't even a circle problem. It's finding the distance between two desired points on a line between two given points, knowing the distance from the given points to their nearest desired point. This is just subtraction on a hypotenuse.
Nice! I found the equation o0f the line and then the intersection of the curves. The vector solution is elegant.
And what an incredible explanation.
I like how you can use pythagerous for this.
So process is.
1) Draw two circles on a graph at the stated centres and radii.
(x-h)sq +(y-k)sq=rsq
centre=(h,k)
radius=(r)
Circle 1 is origin 5,4 radius 2
Circle 2 is origin 1,1 radius 1
2)Draw a straight line point to point from the centre one circle to the other. Because this is the shortest point between PQ where we know the fixed lengths of both R.
3)Find the length of the line by using pythagerous theorum.
4)Subtract the radii of each circle from the line length.
5)This is the shortest point between the circumfrence of both circles.
6) Lables
H= Hypotenuse, X= horizontal, Y=vetical, rSubscript1= first circle (larger one), rSubscript2=second circle (smaller circle)
P= co-ordinate large circle
Q= co-ordinate small circle
A= Centre of Large Circle
B= Centre of small Circle
7) Vectors which I am not going to simplify as it doesnt lend itself to long writing.
Both are going to use point B as the vector origin.
Vector BA=(4,3) [(BQ)/(BA)=1/5]
Q= B+[1/5 of the vector BA]
Q= 1,1+[1/5 of (4,3)]
Q= 1.8,1.6
Vector BA=(4,3) [BP/BA)=3/5]
Note H minus Rsubscript 1 makes point P the location on the hypotenues length from vector start point B
P= B+[3/5 of vector BA]
P=1,1 +[3/5 of 4,3]
P=3.4, 2,8
Comment added for my own working through the process as a learning tool.
I am terrible at math. However, I watch just in case something gains access to that rock I call my 'math' brain... still blows me away. I am highly proficient in EVERYTHING but math. But damn committed to learning.❤
Start solving 10th grade problems of geometry or algebra by giving it some time, once you will start getting correct answers, your confidence will skyrocket which will increase your interest and, interest and peace of mind are the only things you need for problem solving.
Have you tried "3 Blue 1 Brown" videos? The topics are usually bit more advanced, but I think they do a much better job of explaining _why_ mathematicians would do the things they're doing
Presh isn't always great at explaining what assumptions he's starting from, or why he's taking a particular approach
Me too. Its fun being baffled though and knowing someone understands it.
might I recommend you cut out a bunch of little squares of paper and treat these as number units..
It can really help people who don't grasp math as it is typically written down.
but it will become visually obvious when you see, addition, multiplication, etc...
it will also make it obvious why we call a number times itself 'squared'.
If you do it with lego - cubed will become obvious too.
It really helps non mathsy people grasp whats actually going on.
I know it sounds quite childish.. but honestly its a REALLY good exercise for learning.
I surprised myself by figuring the first part out in my head. I tackled part 2 differently and couldn’t make it all the way through without writing some of it down.
I make a linear equation connecting the centers of the circles, solved for y, and plugged the resulting expression into the two circle equations. Solving for x then yields four different x possibilities (two for each circle) and the closest two will be the desired points on the circumferences. Then you plug the x numbers back into the linear equation to solve for the two y coordinates.
The vector solution here feels more intuitive, but I’m not sure I would have thought of that on my own.
Any approach that works, works. And this sounds like a totally valid approach, even if it was a bit more work
Did you have a diagram to look at? Sketching it out on paper can sometimes really help you find the more intuitive solutions
I spent a couple of seconds thinking I was going to need trigonometry to figure out the coordinates of P and Q. Then my brain kind of short-circuited when I tried to figure out the details
"But what's the angle? Oh wait… basic trig is about starting with an angle, and then using it to find the gradient. I don't actually know the angle, and I _already_ have the gradient. So I can skip that part"
@@douglaswolfen7820 I had a mental diagram going, and that was enough for part 1. I figured out the length of the line between the centers and subtracted off both radii to get the distance between the circumferences. But in part 2, I was tracking too many numbers and equations to have it stay in my head. That’s when the paper came out. 🙂
Straight to the point. Good.
The only "incredible" thing is that a low-level, straightforward, technically undemanding analytic geometry exercise is also an Oxford admission problem.
The shortest distance lye on line joining centers.Line joining centers is of length √4^2 + 3^2 =5. Shortest distance is 5 - (sum of radii of circles) = 5 -3=2.By forming and solving quadratic equations we get coordinates of P and Q are:(17/5 , 14/5) and (9/5, 8/5,)
Step 1: You have to consider the line between the centers of the circles, where one must be at (5,4) and the other at (1,1) given a simple visual inspection of the formulae provided.
Step 2: Delta y and delta x between the two points are 4 and 3, respectively, so the length between the centers is simple since it represents a 3,4,5 right triangle, i.e., the total length is 5.
Step 3: Immediately upon inspection again of the formulae you know that the radius of one circle is 1 and the other is 2 so you subtract the sum of these two radii from the total leading to a final answer of length = 2.
Great question and great video SO A BIG THANKS AS ALWAYS! but I would skip the circle diagrams. Instead just look at it analytically this way as it saves a ton of time and avoids unnecessary complexity.
Does this even deserve a video? i just read the question and its done like literally in my head within like 5 secs.
Depends. Do you want to learn more about constructing a mathematical proof? Figuring out the answer in 5 seconds is good, but proving it takes more work. I think that's what the video is about
This is a way more complicated solution that I had. Here’s what I did:
1)
Take the circles center points and find the slope. This ends up being 3/4.
2)
This slope created a special right triangle of side lengths of 3, 4 and a hypotenuse of 5.
3)
Create a triangle scaled 1/5 and 2/5 the size. These will have a hypotenuse equal to the radii.
4)
Accordingly add/subtract the x and y components of the triangles from the center points to get the answer.
This isn't a hard problem, just multiple easy problems
for the second question, I made a function for the hypotenuse (the line of the shortest distance between the two centers of the circles) and just used abit of trig to get the x values of q and p
I guess the lenghty reasoning to solve Q1 is to demonstrate that the shortest distance is on a straight line between centers, which is intuitively obvious.
Maybe it's obvious, maybe it's not. But as you say, his proof does demonstrate that fact.
It’s on a graph just create a right angle triangle and you see it has sides of 4 and 3 between the 2 centres with the hypothenuse being the min distance of 5.
The shortest distance between circles is on the line through both centers which crosses the circles at a right angle. It is the distance between the centers minus the radii of both circles.
From the formulas we know the origins (5,4) and (1,1) and the radii √4 (2) and √1 (1).
The offset between the origins is (4,3) like the sides of the known 3-4-5 triangle so the distance must be the hypotenusa 5. Subtracting the radii 2 and 1 gives 2 as the distance between the circles.
this is the beauty of math, such that you can even do the simplest problems in different ways, the hard and simple problems in simple and hard ways🤷♂
Yes, I got same:
circle formula, so:
Circle A is of radius 2 (square root of 4) and centered at (5,4).
Circle B is of radius 1 (square root of 1) and centered at (1,1).
Shortest distance D between the circles is
distance from P to Q,
which is distance between the centers of the two circles,
less the radius of each
D=((5-1)^2+(4-1)^2)^.5-2-1
D=(25)^.5-2-1
D=5-2-1
D=2
Draw line L through the radius of both circles,
it also passes through points P and Q.
L: y=Sx+C
4=5S+C
1=S+C
3=4S
S=3/4
Slope of line is S=3/4
right triangle, 3*3+4*4=5*5, so sides 3, 4, 5
So, for distance 5 along line PQ, x goes up by 4, y goes up by 3,
so, for distance 1 along line PQ, x goes up by 4/5, y goes up by 3/5,
Q is distance radius of B (1) away from center of B (1,1)
P is distance radius of B + D = 3 away from center of B (1,1)
Q: (1+4/5,1+3/5)
P: (1+3*4/5,1+3*3/5)
Q: (1.8,1.6)
P: (3.4,2.8)
A lot of people are missing the point. It's visually "obvious" that the shortest distance between the two circles should lie on the line connecting their centers. But *why*? The insight that the solution is obtained by translating one minimization problem to another is the satisfying one. Your response to this problem may indicate whether you are an engineer or a mathematician at heart.
For the line between the circles. Connect the circle centers with a line. Determine the length of that line. Subtract from that line the radius of both circles. The result is the length of the shortest line connecting the circles. In this case the line is the hypotenuse of a 3x4x5 triangle. 5 minus 1 minus 2 is 2. The line connecting the circles is length 2.
You can figure the centre and radius out by the equation itself and then calculate the distance from the centres then subtract the sum of both radii
That'll give you the shortest distance
I'm definitely not the sharpest tool in the shed. But I still think that the approach you took to solve the first part of the problem is unnecessarily complicated.
Yes, there is no need to go into the optimization digression. APQB is a quadrilateral, so AB is smaller than the sum of the other three sides. But still an awesome video.
Well he was trying to prove the result, not just get the result.
Part 1:
Those equations make it very straightforward. The first circle has center (5,4) and the second (1,1). The distance between those two points is 5. The radii of the two circles are 2 and 1 respectively, so that means the shortest distance between two points on the two circles is 5-2-1=2.
Part 2:
To go from the center of A to B, you have to follow a line that goes 4 left and 3 down. If you follow 2/5th of that line, you land on point P, which is (5-4*2/5,4-3*2/5) = (3 + 2/5, 2 + 4/5). If you follow 4/5th of that same line, then you land on point Q, which is (5-4*4/5,4-3*4/5) = (1 + 4/5, 1 + 3/5).
This question was really easy. We just need to use distance formula and parametric coordinates of a circle
I used the slope of the line through their midpoints a tiny touch of trig to add/subtract to/from the coords of the midpoints. It was made trivial by the fact that it was a 3,4,5 triangle. Your way was a little cumbersome.
For the distance to the circles, just make a line using the centers and then subtract the radii.
If you know how circle equations work and know the 3-4-5 triangle, you can literally solve this in seconds. I did!
It’s the distance between the centers, which is 5 ( 3 and 4 are the other sides of a rectangle triangles), minus the 2 radius, so 5-(2+1)=2
You can also easily find the coordinates of points P and Q by first determining the equation of the straight line through the centers of circle A and circle B which is y=0.75x +.25. Then all you have to do is solve for the intersection of that straight line and each circle which gives you point P (9/5, 8/5) and Q (17/5, 14/5) 👍
2:07 The equations would give you the respective radius, but what gives you the coordinates of their centres? Can they not be tangent?
I only watched up to 2m20s. The 2 centres are corners of a triangle and we know the vertical and horizontal lengths so we know the diagonal length ie distance between centres. Then subtract the radius of each circle
Also you could just use the triangle inequality to say |AB|
What an ungainly solution to a fairly simple problem?
Part A: Line between centres of circles must give the shortest distance because radii cut at 90 degrees. Shortest is simply pythagoras - total radii
Part B is just similar triangles
I saw from the very beginning that the shortest distance is the straight line that goes through both centers. Would I be wrong, would my answer be incomplete if I don't prove that and start the answer from the straight line?
Nice problem! Managed to solve both parts with a pencil and paper, no calculator.
If the circles are not intersecting, then those two points lies on the line that goes through their centres. We can create an equation for such a line and that gives us two points for each circle. We then pick the points that are nearest to each other. These two points also answer the first question, because their distance is also the shortest distance between those two circles, so you can solve second question and get answer to the first, or you can solve the first by calculation of distance of their centers and then subtracting the radius.
Pitagoras 3,4 =>5 5-2-1= 2
The distance between 2 circles (or any shapes) depends on where they are placed relative to each other! If you place them 1cm apart, then that's the distance between them. If you place them touching, then the distance is zero. So zero is the shortest distance between 2 circles...
I just found the slope between the two centers of the circle, (1,1) and (5,4), using slope intercept form to get 4-1=m(5-1), 3=4m the slope is 3/4, plotting 3/4x we are not at the center, since plugging in 1 would get us 3/4, i add 1/4 to the line, ending up with 3/4x+1/4. Using this we can figure out the edges of the circle by substituting the y in each circle with the equation of line. Set each equation equal to 0, after solving for each solution, you get (0.2,0.4) and (1.8,1.6) for the smaller circle and (3.4,2.8) and (6.6,5.2). We can get rid of each extremes, as (1.8,1.6) and (3.4,2.8) are the closer points, the line connecting them is diagonal, which you could say is a hypotenuse of a triangle, subtracting the coordinates from each, you get a height of 1.6 and a length of 1.2, pythagorean theorem says a^2+b^2=c^2 or the square root of a^2+b^2=c, plugging in 1.6 and 1.2, you get 2.56+1.44, and that equals 4, and then you take the square root, which equals 2. The shortest distance between the two circles is 2.
It's 2 am and I saw this ... Solved it in under a min (yea)
Every circle equation is in the format (x-h)^2 + (y-k)^2 =r^2
Where the centre of the circle is (h,k) and radius is r
After finding centres of both circles use distance formula to find distance √{(x-a)^2 + (y-b)^2} where (a,b) is centre of other circle.. now subtract sum of radiii of both circles from the distance calculated above and tada...
All this is taught to 15 year olds in India who are preparing for IIT (a super difficult exam and I am preparing for it)
For part 2 of the exercise, you can take the equation for line AB, and simply calculate the coordinates where the line intersects each circle.
I got the answer (2) in my head in about four seconds. The centerpoints of the circles make a 3-4-5 right triangle, the radii are 1 and 2, so 5-1-2 = 2.
distance between two centers - sum of each circle's radius = 5 - (2 + 1) = 2
I suspect that this highlights a difference between mathematics education in the UK and in the US.
In the UK we are taught at secondary school level that the shortest distance between two circles lies on the line between their centres*. We are also taught that the line segment is normal to the edges of both circles, although you don't need that to solve the given problem.
Oxford University would expect applicants already to know that.
* Except if they are concentric or if they intersect.
Given we know the radius of each circle Pythagoras tells us the answer based on circle centers being points on a right angle triangle with the triangle offset relative to the origin.
I used the parametric formula of the segment BA to find Q and P.
I completely overcomplicated this by missing the fact that the vector between their centers is composed of the legs of a 3-4-5 triangle. Ended up finding P and Q the hard way first with the angle of this vector and the x-axis (transformed from polar coordinates) then the distance between P and Q. Got the correct answer at least!
15 seconds to solve.
1) Center of the first circle (x; y = 5; 4)
Center of the second circle (x; y = 1; 1)
2) Distance between centers:
√((5-1)^2 + (4-1)^2) =
√(4^2 + 3^2) =
√(16 + 9) = √25 = 5
3) Subtract the radii of the circles from the resulting distance, and get the shortest distance: 5 - 2 - 1 = 2
Answer: shortest distance is 2
The distance between the two centres is 5, calculated from the coordinates. Taking away the two radii (1 and 2), the shortest distance is 2. That's it! There is no need to work out the coordinates of the points where the perimeters intercept the straight line connecting the two centres.
It took 4 minutes to explain that the distance between two points is 5, and the radii around those points are total length 3. The second solution was interesting; I used trig ratios, but I like how it was done here.
Can't we take the line equation and take the intersection of 2 circles? Intersection of 1st till intersection of 2nd gives us the distance between them.
Which software you use to make this video?
Circle A has a radius of 2 and circle B has a radius of 1. Circle A’s center is at (5,4) and circle B’s is at (1,1). The distance between these two points is sqrt((4-1)^2 + (5-1)^2) = 5 units. Subtract both radii from this value and you get 2.
The whole line drawing and point moving along the circle was harder to do than the whole math on this one. ;) Got the first part on looking at the question and the second one easily after. That said, i got a masters degree in maths and so this SHOULD be easy. 😅
I need you to solve this one .
Solve for integral value of x such that,
[(5^x )-1]/3 is prime number or perfect square numbers.
Ohh I used section formula for second one i.e. (mx2+ nx1)/m+n to get answer more easily
How I solved part 1: Shortest distance between 2 points is a straight line, so points P and Q will be along Segment AB. A line down from point A and a line right from point B will intersect at 5,1 and form a right angle. Segment AB then forms the hypotenuse and makes this a right triangle whose sides are 3 and 4...this is a 3-4-5 right triangle. Segment AB is 5 units long. Circle A's radius is sqrt(4), so it's 2. Circle B's radius is sqrt(1), so it's 1. 5 - 2 - 1 = 2, so Segment PQ will be 2 units.
For me, part 1 use 5-2-1=2 , and part 2 use point of intersection formula x-coor of P=(3*5+2*1)/(3+2) and so on...
The shortest distance will obviously be a straight line from a to b, i did a²+b²=c²
and got PQ=2
For the coordinates
i used trigonometry
With θ and β being angels of the same triangle in Q1
So it will be
Q ⇒ (1+cosθ,1+sinθ) ⇒(1+⅘,1+⅗)⇒(1.8 , 1.6)
P⇒(5-2sinβ,4-2cosβ) ⇒(5-2.⅘ ,4-2.⅗) ⇒(3.4 , 2.8)
Using similarity of the triangles
Managed to get this one, but I was a little uncertain. While it was not difficult to construct the line from (1,1) to (5,4) and check where it intersected the circles, and perpendicular lines to the first one through those intersection points certainly LOOKED like tangent lines. I ended up feeling decently confident but would have hated having to try and prove that those perpendicular lines were in fact tangent to the circle at the intersection points (even if it is probably doable via related rates).
This can be solved in head easily, one just needs to know how to read the centers and the radii off of the equations. The distance can also be calculated easily as 25 is a square number, then just subtract the sum of the two radii.
They are circles, so the distance betwen them is the distance between the centers minis the sum of the radii.
As a draftsman I had to find clearances between circles, whether they be holes or gears, a lot of times. I just assumed that the clearance, that is the closest they got, would always lie on a connector between the centers. I never thought it needed a proof. Was I doing something wrong?
I'd have constructed a parametric linear interpolator (lerp) but, Yes.
The use of the expression "Daunting Problem" had me spending far too long thinking I had misunderstood the question, but no! it really was a straight line between the centres. Once I realised that I was thinking too hard it was easy and so was the rest of it. Is this an indictment of the university system or does "Oxford admissions" refer to the McDonalds in Oxford?
Got it in about 20 seconds, this can't be an Oxford entrance exam. The centre of the two circles are 1,1 and 5,4 meaning the line between the centres are literally marking out a 3,4,5 triangle. The radiuses are root 4 and root 1 so 1, and 2.
Therefore it's 2.
When, having solved it in my head in about 10 seconds, I heard Presh describe this as a 'daunting problem', I felt like goddam Will Hunting. Then I read the comments and instantly felt like Chuckie.
I solved this in like 2 minutes (toke the distance from center of circle A to center of circle B (hypotenuse of a right angle triangle) and substracted radius a ad radius B.... (didn't bother with part 2, no fun)
I then watched the video and wondered why on earth the solution was strating out so complicated... honestly dropped of before the end
The best way to teach math is to keep it simple
I thought you made that explanation more difficult than it needed to be.
Oh, this is hard, I thought. Then I thought about it for a minute and then I solved it in my head in 10 seconds.
Does this problem become ridiculous if the circles overlap?
I was actually able to solve this in my head when I realized… (spoiler)
…that the centers of the circles are on a 345 triangle. Definitely one of those problems that looks harder than it is at first glance.
But that's how you would know the distance between the centers a 5, you would need to use a distance formula and points and stuff to find the distance of the soace between the 2 circles because that's the actual question.
@@maxhagenauer24 no need for a distance formula when known catheti are 3 and 4.
@@feedbackzaloop Catheti? What is that?
@@maxhagenauer24 plural of cathetus - sides of a right triangle, adjecent to a 90 degree corner.
Same here.
I imagined it as the tangents being parallel to each other where as the first pair is shorter than the other
I didn’t think about using vectors. I was got as far as working out the equation for the line and was going to sub it into the equations. I didn’t bother after that.
it's a 3,4,5 triangle. the hyp is 5-rA-rb = 2
This is so easy I did this in my head in 30 seconds, 5-1=4, 4-1=3 -> special right triangle 3,4,5 -> 5-2-1=2 -> answer is 2. I'm not a genius but sometimes problems on this channel are easy, and sometimes are really hard. This was the easy one with a misleading "Oxford" caption.
I got 2
Got it by the 17-sec mark of the video! Hope I’m right!!
wouldnt u just subtract the respective radiuses from the distance from each centre
Now I felt smart when I found the solution obvious. 😊
Solved in about 10 seconds. Not usual for Presh problems by any means!!
Wow,this was easy for me! Never been in college or university.
Are they on the same plane? Fold the plane until the circles lay on each other.
They are on the same plane as nothing references the Z axis, only X and Y axis’ are referenced.
Why not use the distance formula backwards to get coordinates?
I did the second part the hard way(solving for the intersection of the circles and the line connecting a and B) all the while knowing there must be a simpler approach. However, I had a little difficulty with the vector explanation. Maybe to me a clearer Solution invoked marking off units on the hypotenuse, of the 345 triangle, Drawing lines, parallel to the Y axis through these points and then doing similar triangle exercises.😂
I did not do any of this agonizing over part one. I just did it, like everyone else. We must all be lacking something.The vector way of finding the coords was good though.
Lots of people more interested in saying they know how to do this, and uh way easier akshully, than people recognizing the point he's trying to make (how to think through a math problem). Its a lot less about bragging on right answers and more about teaching people how to think. This was a GREAT video for explaining to my 6th grader how the lesson is knowing how to combine simple math facts to think through a math problem.
Man, solved this one in 15 seconds in my head :)
Me, who thinks in 4D, would fold the paper over itself and stab ahole through it representing a worm hole. Clearly the distance gets really short if you warp space.