That "well known fact" can be easily derived as well, so no need to memorize it: using the center point of the inscribed circle, the original triangle can be partitioned in 3 new triangles, and their 3 areas obviously sum up to the area A of the original triangle. Their areas are AB * r/2, AC * r/2 and BC * r/2, and solving this for r gives: r = 2 * A / (AB + AC + BC) = 2 * A / P.
It is normally known as A = sr, where s is the semi-perimeter. If sides are a,b,c then the triangle is made from 3 triangles of area .5ar, .5br, and .5cr.
mmm. That should be covered in planar Euclidian geometry, typically in year 8 or 9. That is if your middle school curriculum is worth its salt. Classical geometry is classic for a reason: it builds axiomatic reasoning from a young age, and is useful by itself in further studies. Not covering it makes students miss out a lot.
7:00 I didn't know that... but I think I arrived at that Call the center of the circle O Draw OA, OB, OC The triangle ABC can be almost partitioned[ignoring the dividing lines we made which have 0 area] into OAB, OBC, OAC... in particular, their combined area must be the same A=area(ABC)=area(OAB)+area(OBC)+area(OAC) But the triangles we created have a base equal to the side of the original triangle, and height equal to the radius of the circle.... area(OAB ) = r ⋅ length(AB) / 2 area(OBC) = r ⋅ length(BC) / 2 area(OAC) = r ⋅ length(AC) / 2 A=r ⋅ length(AB) / 2 + r ⋅ length(BC) / 2 + r ⋅ length(AC) / 2 = (r/2) ⋅ (length(AB) + length(BC)+length(AC)) = (r/2) p Solving for r: r=A/(2p)
I don't think 0AB, 0BC and 0AC have the same area, unless the triangle is equilateral, it becomes evident with a right triangle, as if the angles are 90, 45, 45 then the tirangle of the hypotenuse has the same height as the triangles of both cathethus but a larger base, but if the angles are 90, >45,
@@ByNatureEnemiesyeah, he even wrote the formulas for the areas, which clearly depend on the size of each side. Nonetheless, the "big" triangle is still the sum of its parts whether the triangle is equilateral or not, so his derivation still works
Nice problem. I took it for granted that your answer was correct and tried to draw it to scale. Unfortunately, I saw that your computed value of r was too large. I figured out that the right answer is 2/(3+sqrt(5)+2*sqrt(2)), approximately 0.248. In case you do not like radicals in the denominator, the expression just mentioned is equal to (1+sqrt(5)+sqrt(2)-sqrt(10))/6. I saw that I am not the only one who gave a correct answer in a comment. By the way, I wrote two books about similar geometry problems and published them with Amazon. Best wishes, Leen Ammeraal
Set up a right handed co-ord sys with A as origin and x-axis, AB. Then using parallel projections, we can show C has co-ords (2/3, 2/3). Side lengths of triangle ABC then follow from Pythag. The area of triangle ABC = 1/2 - Area triangle BEC = 1/2 - 1/6 = 1/3, then (proceeding like Penn) we're home. NOTES: The r = 2A/P result is easily obtained by cutting triangle ABC into three triangles of height r The alternate angles theorem (Euclid, book 1) is often a handy piece of kit if you want to prove two triangles are similar
Got the result too. I knew there was a formula for r. I had to look it up, but then the calculations were easy. I found the coordinates of point C (origin A) by intersecting the 2 lines AE and BD: xc=yc=2/3 I named H the point for height HC in triangle ABC. xH =xC = 2/3 From there, I derived the triangle lengths, the perimeter, the area, and done!
@@roberttelarket4934 His clone does all the actual calculations for him. He only describe the solutions of the problems. It saves a lot of time and energy.
For anyone interested in the formula used to find the radius of the incircle (2A/P), en.wikipedia.org/wiki/Heron%27s_formula#Trigonometric_proof_using_the_law_of_cotangents provides a nice explaination
My HS teacher would have marked your similar triangles wrong. She was a stickler that the ordering had to be the same. First letter of each had to have the same angle, second letter, etc so ABC ~ EDC and not ABC ~ CDE
I agree with Michael. Ordering does not change the triangle's similarity. If you are stating the triangles are congruent, even, the ordering isn't the distinction, it's the unique set of three sides/angles and any of the six possible orderings must describe the identical object.
Her point, is that without seeing the two triangles, if you said ABC ~ EDC then you can write the equations AB/BC = ED/DC and know the angles of A = E and so forth because ordering mattered (at least for her)
@@TedHoppI agree to a point. If your methodology is to expect that the ordering corresponds, such that if ABC is congruent to DEF, it is always true that a/d = b/e = c/f, yes ordering is critical. Apparently that is not Michael's methodology. He tends to order the vertices in alphabetical order and then ensure appropriate sides and angles are considered. The former may be more rigorous in its design, but the latter isn't wrong per se, though it could be disaster in the hands of someone not as good at geometry.
1:40 You could argue like you did... but you could also argue that, since AB is parallel to DE(because, as he said before, the outer figure is a square), then CAB ≅ DEC because they are two parallel lines being crossed by a third one;
2:11 Just a quick alternative variant step, you could notice that since DE is parallel to AB that angles EDB and DBA are congruent (i.e. interior angles along the transversal of parallel lines.) Like the same reasoning for angles EAB and EDB being congruent.
My approach was more angles/trig-based, and I think it's more elementary. We only need the fact that the incircle is found at the intersection of the three angle bisectors of the triangle. (I promise the solution is really simple, but describing the diagram in text makes it more of a mouthful than necessary hahaha) Draw the angle bisector at A and the angle bisector at B, and let their point of intersection be denoted by F. Then, the radius of the in-circle is the altitude of the triangle AFB. Project point F onto the line AB, and give its "shadow" the name G. This splits triangle AFB into the two right triangles AFG and BFG. Let t = |AG| and so |BG| = 1 - t be the lengths of the two segments that AB was split into. Let h = |FG| be the altitute of the triangle. Using trig, h = t*tan(GAF) and h = (1-t)*tan(GBF). Simple algebra then gives us that h = tan(GAF)tan(GBF) / (tan(GAF) + tan(GBF)) Both of tan(GAF) and tan(GBF) can be solved for using the half-angle formula (or double-angle formula, I guess) for tangents, since by SOHCAHTOA we know that tan(CAB) = 1 and tan(CBA) = 2. Specifically, the formula tells us that tan(GAF) = (-1 + sqrt(2)) and tan(GBF) = (-1 + sqrt(5))/2. Plugging these values in gives us the same result as the one you got in the video, after some simplifications! :)
Technically you can put it on a graph map the lines using the coordinate system find the point of intersections for the vertices and the y coords of the incenter using the formula which would be the radius
triangles above and below are proportional - above is 2 times smaller, so you can figure out the sides of a triangle and then there is a formula how to get radius of the circle inside the triangle by its sides
A good exercise for everyone! I think an equivalent result (without square roots in the denominator) could be: (1 + Sqrt[2] + Sqrt[5] - Sqrt[10])/6 (checked with Mathematica).
I did it with analytic geometry and it took me an hour 😂. I called the bottom left corner the origin and wrote the general equation of the circle. I noticed that k equals r, and so eliminated k. Then the two lines y=x and y=2-2x intersect the circle and are tangent to it. So with two equations for the circle and two equations for its derivative, I had a system with unknowns, r, h, x1, and x2. After solving for r (I rationalized the answer), there were two independent plus or minuses. So I had 4 cases to check. The lower bounds for 3 of the radii were greater than 1/2, so obviously extraneous. The last one was 1/3 - 1/6*(sqrt(2)-1)(sqrt(5)-1) = 0.2480006.
Michael, I would like to see you solve this limit: The limit when n goes to infinity of the ratio between the nth fibonacci number and the (n-a)th fibonacci number for a natural fixed number a. I found by calculations that is phi to the ath power but couldn't prove. Thanks
I didn't use the well known fact as it was not well known to me but ended up with (√2 + √5 - √10 +1) / 6 which evaluates to exactly the same thing as the given answer.
I solved it before watching the video, and I'm a little proud that I found the correct answer ;-) My approach: I used analysis to determine the coordinates of the point of intersection C as (2/3, 2/3). From there it was easy to find out that the area A of the original triangle is 1/3, AC is 2 * sqrt(2)/3, BC is sqrt(5)/3. And AB is obviously defined as 1. And given that r = 2 * A / (AB + AC + BC), this was all I needed to determine r.
ps: where "used analysis" just means that I trivially determined the functions for the 2 lines as f(x) = x and g(x) = 2 - 2x, and used f(x) = g(x) to find the intersection point.
my method was to set up a coordinate system with the origin at A, then get two line equations, get the coords of point C as the intersection of the two lines, then use the same formula for r. your method was basically the same, but slightly easier & more geometric
Isn't this just about solving equation r/tan(CAB/2) + r/tan(ABC/2) = 1? To express tan(a/2) using tan(a) we can use known trigonometric identities. We know tan(CAB) and tan(ABC) from ratios sides of right triangles in the picture.
My first thought from the thumbnail and not seeing the video yet: the inradius of a triangle is two times the area divided by the perimeter. I get nice values if I say the base is 3, the left side is 2sqrt(2) and the right side is sqrt(5). And the area is 3 via various methods. So I got to r = 6/(sqrt(5) + 2sqrt(2) + 3) and with a lot of trouble I got (1 + sqrt(2) + sqrt(5) - sqrt(10))/2, and that's for a side length of 3. If it's a unit square, then that /2 should be a /6. Now I will watch and see where I went wrong.... ;) EDIT: Hmm, pretty much the same process. I got lucky this time hehehe. (I never get these right.)
Solved it the boring way: YELLOW: y = x GREEN: y = 2(1-x) PINK: (x-k)(x-k) + (y-R)(y-R) = RR YELLOW x PINK: 2xx - 2(k+R)x + kk = 0 which has only one root, therefore: RR + 2kR - kk = 0 Take the positive root, as the negative root is nonsense: k = R(sqrt(2) + 1) GREEN x PINK: (5)xx + (4R-2k-8)x + (kk+4-4R) = 0 which has only one root, therefore: 3(1+sqrt2)RR - (3+2sqrt2)R + 1 = 0 Take the negative root, because we want the smaller of the two circles that satisfy the above properties: R = (1/6)(sqrt2 + 1 - (sqrt2 - 1) sqrt5)
Wasn't it actually superfluous to include the lengths of the smaller similar triangle? While I know all of those trivial facts about similar triangles from what I learned from my first year high school geometry teachings, none of that seemed pertinent to solving the given problem re. the length of the radius of the main triangle's inscribed circle -- unless the problem were to find the radius measure of an inscribed circle inside of that smaller triangle rather than one inside of the bigger of the two similar triangles
Hello sir. Can you recommend me some sources for understanding the evolution of mathematics through the interaction between humans and animals. E.g counting evolved through the meed to count sheeps by shepherds who started designating a stone to each sheep...
How do you know counting evolved through the need to count sheep? It could have as well evolved through the need to count how many berries were gathered and how to split them evenly among all the people of a tribe.
@@bjornfeuerbacher5514 Ya, it could be but I read this in George Simon's topology. That can be contested and can even be distinct for different civilisations, but this premise is not improbabal given the hunter gatherer state of early societies.
@@bjornfeuerbacher5514 Ya, you're right, My appologies for a bewildered reply. The way I should have stated my point is that one of the historiographical perspectives on development of counting attributes its evolution to the need to keep track of sheeps by shepherds which is possible for agriculture and livestock rearing had been a part of initial human settlement period. Your argument is also acceptable and can be one perspective to look at this. But what I am requiring is a study of how animal-human interaction could have produced/evolved mathematics. Counting was a possible example.
Nice problem, Michael. However, from my perspective, the geometry problems that you solve in your channel are by far simpler than the algebra/calculus ones.
Every school classroom used to have those. (We also had slate tablets and ink wells, and in high school there was a giant slide rule up front, so I may be older than you.) 🙂 The compasses were made from hardwood and had a rubber tip, like an eraser, to form the point. Probably about a foot long.
@@TomFarrell-p9z: No ink wells for us but giant compasses I vaguely remember from high school in the 1960's. Ink wells? Hell you must be from the late 1930's!
@@roberttelarket4934 LOL! I exaggerated a little. No slate tablets, and the ink wells (and stains) were there, but no ink bottles. In 4th grade we transferred to a newly opened elementary school (in 1971) and--I found out much later--our principal spent a day at the school district''s warehouse picking out the best of the old desks and chairs to fill the classrooms after their new furniture budget ran out. We had one room schools in that district until the late 1940's and I'm sure some of those desks came from when they were closed. Not sure how much older than that the desks actually were. Wonderful teachers and a great math education though. Perfectly good prep for college and careers in the maths, sciences, and engineering.
@@TomFarrell-p9z: Well then you were born about 1962. Further you wrote maths not math implying you're not from the U.S. or was that just a spelling error?
@@roberttelarket4934 Spot on with the date. I grew up in Upstate New York. Not sure about "maths" in that context--I thought we say it that way in the US (when referring to the field). However, I was stationed in England for a few years, picked up some of their languistic details, and ever since my writing, and speaking, have been a bit confused.
So Mike that's why you’re doubly brilliant you have two brains in two bodies the second in your identical twin! Two brains are infinitely better than one!
I guess there is an unspoken 'well known method' that requires simple straight-lines, pythagoras and so forth, but NOT trigonometry, right? Because the trigonometric solution is also pretty darn direct, too. I'm going to use intersecting line functions to find the cross-point C. [1.1] f(𝒙) = 1 𝒙 ⊕ 0 (line AE) [1.2] g(𝒙) = -2 𝒙 ⊕ 2 (line DB) Solve for intersection [2.1] 𝒙 ⊕ 0 = -2𝒙 ⊕ 2 [2.2] 3𝒙 = 2 [2.3] 𝒙 = ⅔ … and pasting into [1.1] [2.4] 𝒚 = ⅔ This however is NOT the vertical line crossing the origin of the in-circle. the in-circle has a radius [𝒓]. Marking the center of the circle as 'O', then we note that the △AOQ (Q is base) will have left θ as ½ the angle of the slope 1 AE diagonal. [3.1] tan( θ = ½ arctan( ¹⁄₁ ) ) = 0.41421 Likewise, the △BOQ will have a φ as ½ the angle of the DB diagonal [4.1] tan( φ = ½ arctan( -²⁄₁ ) ) = -0.61803 These are the slopes then of the two corner-to-incircle-origin lines which similarly can be set as equal equations and solved [5.1] z(𝒙) = 0.41421 𝒙 ⊕ 0 … and [5.2] y(𝒙) = -0.61803 𝒙 ⊕ 0.61803 … so now make equal and solve [5.3] 𝒙 = 0.61803 / (0.41421 ⊕ 0.61803) [5.4] 𝒙 = 0.59873 and using [5.1] and the 𝒙, find [6.1] 𝒚 = 0.41421 × 0.59873 ⊕ 0.0 [6.2] 𝒚 = 0.24800 //• And that [𝒚] turns out to be the radius of the incircle. Tada. Never really needed to know the magic formula for radius, perimeter and triangle area, per your chalkboard diagram. I definitely grant however ... knowing that magic formula helps tons by comparison. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
That "well known fact" can be easily derived as well, so no need to memorize it: using the center point of the inscribed circle, the original triangle can be partitioned in 3 new triangles, and their 3 areas obviously sum up to the area A of the original triangle. Their areas are AB * r/2, AC * r/2 and BC * r/2, and solving this for r gives: r = 2 * A / (AB + AC + BC) = 2 * A / P.
I think that's way more interesting than the puzzle. It's easier, but it's elegant.
It would've been worth taking the 30 extra seconds to show this "wkf".
@7:00 - yet another "well-known fact" I'd never heard of :).
It is indeed well known. Also it is very easy to prove - for your homework.
It is normally known as A = sr, where s is the semi-perimeter. If sides are a,b,c then the triangle is made from 3 triangles of area .5ar, .5br, and .5cr.
mmm. That should be covered in planar Euclidian geometry, typically in year 8 or 9. That is if your middle school curriculum is worth its salt. Classical geometry is classic for a reason: it builds axiomatic reasoning from a young age, and is useful by itself in further studies. Not covering it makes students miss out a lot.
The final answer can also be written
[1+sqrt(2)+sqrt(5)-sqrt(10)]/6.
7:00 I didn't know that... but I think I arrived at that
Call the center of the circle O
Draw OA, OB, OC
The triangle ABC can be almost partitioned[ignoring the dividing lines we made which have 0 area] into OAB, OBC, OAC... in particular, their combined area must be the same
A=area(ABC)=area(OAB)+area(OBC)+area(OAC)
But the triangles we created have a base equal to the side of the original triangle, and height equal to the radius of the circle....
area(OAB ) = r ⋅ length(AB) / 2
area(OBC) = r ⋅ length(BC) / 2
area(OAC) = r ⋅ length(AC) / 2
A=r ⋅ length(AB) / 2 + r ⋅ length(BC) / 2 + r ⋅ length(AC) / 2 = (r/2) ⋅ (length(AB) + length(BC)+length(AC)) = (r/2) p
Solving for r:
r=A/(2p)
I don't think 0AB, 0BC and 0AC have the same area, unless the triangle is equilateral, it becomes evident with a right triangle, as if the angles are 90, 45, 45 then the tirangle of the hypotenuse has the same height as the triangles of both cathethus but a larger base, but if the angles are 90, >45,
@@ByNatureEnemiesyeah, he even wrote the formulas for the areas, which clearly depend on the size of each side. Nonetheless, the "big" triangle is still the sum of its parts whether the triangle is equilateral or not, so his derivation still works
@@Kaneeren Yes it does, that's the reason i only clarified that part
@@ByNatureEnemies good catch... I meant their *combined* area is the same as the area of the triangle;
Nice problem. I took it for granted that your answer was correct and tried to draw it to scale. Unfortunately, I saw that your computed value of r was too large. I figured out that the right answer is 2/(3+sqrt(5)+2*sqrt(2)), approximately 0.248. In case you do not like radicals in the denominator, the expression just mentioned is equal to (1+sqrt(5)+sqrt(2)-sqrt(10))/6. I saw that I am not the only one who gave a correct answer in a comment. By the way, I wrote two books about similar geometry problems and published them with Amazon. Best wishes, Leen Ammeraal
That twin drawing animation was really cool!!!
Set up a right handed co-ord sys with A as origin and x-axis, AB. Then using parallel projections, we can show C has co-ords (2/3, 2/3). Side lengths of triangle ABC then follow from Pythag. The area of triangle ABC = 1/2 - Area triangle BEC = 1/2 - 1/6 = 1/3, then (proceeding like Penn) we're home.
NOTES:
The r = 2A/P result is easily obtained by cutting triangle ABC into three triangles of height r
The alternate angles theorem (Euclid, book 1) is often a handy piece of kit if you want to prove two triangles are similar
Got the result too.
I knew there was a formula for r.
I had to look it up, but then the calculations were easy.
I found the coordinates of point C (origin A) by intersecting the 2 lines AE and BD: xc=yc=2/3
I named H the point for height HC in triangle ABC.
xH =xC = 2/3
From there, I derived the triangle lengths, the perimeter, the area, and done!
I didn't know you had an identical twin?
It's a clone
@@chrisriess1298: Very clever! Damn I should have thought of that!
@@roberttelarket4934 His clone does all the actual calculations for him. He only describe the solutions of the problems. It saves a lot of time and energy.
His name is Mitchell Penn
@@chrisriess1298 its satire 💀
I *_love_* these geometry problems! I've missed them lately.
I set x =-2x + 2 as those are the equations of the 2 lines, got x=y=2/3 which allows lengths via pythag and the r = 2A/p.
For anyone interested in the formula used to find the radius of the incircle (2A/P), en.wikipedia.org/wiki/Heron%27s_formula#Trigonometric_proof_using_the_law_of_cotangents
provides a nice explaination
Someone gave a 2 lines proof in the comments. See above.
Sangaku problems are some of my favorites. They are surprisingly difficult.
My HS teacher would have marked your similar triangles wrong. She was a stickler that the ordering had to be the same. First letter of each had to have the same angle, second letter, etc so ABC ~ EDC and not ABC ~ CDE
I agree with Michael. Ordering does not change the triangle's similarity. If you are stating the triangles are congruent, even, the ordering isn't the distinction, it's the unique set of three sides/angles and any of the six possible orderings must describe the identical object.
Her point, is that without seeing the two triangles, if you said ABC ~ EDC then you can write the equations AB/BC = ED/DC and know the angles of A = E and so forth because ordering mattered (at least for her)
@@Qermaq Ordering is most certainly important. Corresponding lengths are in the same ratio and ordering defines which lengths are in correspondence.
@@TedHoppI agree to a point. If your methodology is to expect that the ordering corresponds, such that if ABC is congruent to DEF, it is always true that a/d = b/e = c/f, yes ordering is critical. Apparently that is not Michael's methodology. He tends to order the vertices in alphabetical order and then ensure appropriate sides and angles are considered. The former may be more rigorous in its design, but the latter isn't wrong per se, though it could be disaster in the hands of someone not as good at geometry.
Oh silly 😭
Nice problem. I decided to solve along with the video, but by generalizing by letting DE=x (0
1:40 You could argue like you did... but you could also argue that, since AB is parallel to DE(because, as he said before, the outer figure is a square), then CAB ≅ DEC because they are two parallel lines being crossed by a third one;
2:11 Just a quick alternative variant step, you could notice that since DE is parallel to AB that angles EDB and DBA are congruent (i.e. interior angles along the transversal of parallel lines.) Like the same reasoning for angles EAB and EDB being congruent.
My approach was more angles/trig-based, and I think it's more elementary. We only need the fact that the incircle is found at the intersection of the three angle bisectors of the triangle. (I promise the solution is really simple, but describing the diagram in text makes it more of a mouthful than necessary hahaha)
Draw the angle bisector at A and the angle bisector at B, and let their point of intersection be denoted by F. Then, the radius of the in-circle is the altitude of the triangle AFB.
Project point F onto the line AB, and give its "shadow" the name G. This splits triangle AFB into the two right triangles AFG and BFG.
Let t = |AG| and so |BG| = 1 - t be the lengths of the two segments that AB was split into. Let h = |FG| be the altitute of the triangle.
Using trig, h = t*tan(GAF) and h = (1-t)*tan(GBF). Simple algebra then gives us that h = tan(GAF)tan(GBF) / (tan(GAF) + tan(GBF))
Both of tan(GAF) and tan(GBF) can be solved for using the half-angle formula (or double-angle formula, I guess) for tangents, since by SOHCAHTOA we know that tan(CAB) = 1 and tan(CBA) = 2. Specifically, the formula tells us that tan(GAF) = (-1 + sqrt(2)) and tan(GBF) = (-1 + sqrt(5))/2.
Plugging these values in gives us the same result as the one you got in the video, after some simplifications! :)
Thanks, I'd love to see more geometry problems.
Technically you can put it on a graph map the lines using the coordinate system find the point of intersections for the vertices and the y coords of the incenter using the formula which would be the radius
triangles above and below are proportional - above is 2 times smaller,
so you can figure out the sides of a triangle and then there is a formula how to get radius of the circle inside the triangle by its sides
A good exercise for everyone! I think an equivalent result (without square roots in the denominator) could be: (1 + Sqrt[2] + Sqrt[5] - Sqrt[10])/6 (checked with Mathematica).
I did it with analytic geometry and it took me an hour 😂. I called the bottom left corner the origin and wrote the general equation of the circle. I noticed that k equals r, and so eliminated k. Then the two lines y=x and y=2-2x intersect the circle and are tangent to it. So with two equations for the circle and two equations for its derivative, I had a system with unknowns, r, h, x1, and x2.
After solving for r (I rationalized the answer), there were two independent plus or minuses. So I had 4 cases to check. The lower bounds for 3 of the radii were greater than 1/2, so obviously extraneous. The last one was 1/3 - 1/6*(sqrt(2)-1)(sqrt(5)-1) = 0.2480006.
I thought he was going to use Heron's formula to obtain the area of the triangle. Using the similarity of triangles is much easier.
Michael, I would like to see you solve this limit:
The limit when n goes to infinity of the ratio between the nth fibonacci number and the (n-a)th fibonacci number for a natural fixed number a. I found by calculations that is phi to the ath power but couldn't prove.
Thanks
I didn't use the well known fact as it was not well known to me but ended up with (√2 + √5 - √10 +1) / 6 which evaluates to exactly the same thing as the given answer.
I solved it before watching the video, and I'm a little proud that I found the correct answer ;-) My approach: I used analysis to determine the coordinates of the point of intersection C as (2/3, 2/3). From there it was easy to find out that the area A of the original triangle is 1/3, AC is 2 * sqrt(2)/3, BC is sqrt(5)/3. And AB is obviously defined as 1. And given that r = 2 * A / (AB + AC + BC), this was all I needed to determine r.
ps: where "used analysis" just means that I trivially determined the functions for the 2 lines as f(x) = x and g(x) = 2 - 2x, and used f(x) = g(x) to find the intersection point.
my method was to set up a coordinate system with the origin at A, then get two line equations, get the coords of point C as the intersection of the two lines, then use the same formula for r. your method was basically the same, but slightly easier & more geometric
Michael used to solve almost every geometry problem using coordinate geometry. I think the style he used in this video is more elegant.
@@TedHopp agree. i was a little disappointed at myself for missing the similarity of the triangles
Am I the only one amazed by how good Penn’s free hand drawing is? The computer-generated lines and the circle fit so well over his initial sketch.
It was not done by free hand.
So, if I did my math(s) correctly, that radius measure is a smidgen less than 1/4 of the length of one of the edges of the unit square
2 Michael's for the price of 1 is always a good deal.
Isn't this just about solving equation r/tan(CAB/2) + r/tan(ABC/2) = 1? To express tan(a/2) using tan(a) we can use known trigonometric identities. We know tan(CAB) and tan(ABC) from ratios sides of right triangles in the picture.
2:09 2 angles is enough
Finally a geometry problem!!
My first thought from the thumbnail and not seeing the video yet: the inradius of a triangle is two times the area divided by the perimeter. I get nice values if I say the base is 3, the left side is 2sqrt(2) and the right side is sqrt(5). And the area is 3 via various methods. So I got to r = 6/(sqrt(5) + 2sqrt(2) + 3) and with a lot of trouble I got (1 + sqrt(2) + sqrt(5) - sqrt(10))/2, and that's for a side length of 3. If it's a unit square, then that /2 should be a /6. Now I will watch and see where I went wrong.... ;) EDIT: Hmm, pretty much the same process. I got lucky this time hehehe. (I never get these right.)
MIchael finally cloned himself.
This becomes very easy if you use analytic or coordinate geometry
That one was not that difficult. It is nice to be able to solve the problem on my own before watching the video in full!
Thank you as usual.
Solved it the boring way:
YELLOW:
y = x
GREEN:
y = 2(1-x)
PINK:
(x-k)(x-k) + (y-R)(y-R) = RR
YELLOW x PINK:
2xx - 2(k+R)x + kk = 0
which has only one root, therefore:
RR + 2kR - kk = 0
Take the positive root, as the negative root is nonsense:
k = R(sqrt(2) + 1)
GREEN x PINK:
(5)xx + (4R-2k-8)x + (kk+4-4R) = 0
which has only one root, therefore:
3(1+sqrt2)RR - (3+2sqrt2)R + 1 = 0
Take the negative root, because we want the smaller of the two circles that satisfy the above properties:
R = (1/6)(sqrt2 + 1 - (sqrt2 - 1) sqrt5)
I just used coor bashing, pythag theorem and herons formula :D
Wasn't it actually superfluous to include the lengths of the smaller similar triangle? While I know all of those trivial facts about similar triangles from what I learned from my first year high school geometry teachings, none of that seemed pertinent to solving the given problem re. the length of the radius of the main triangle's inscribed circle -- unless the problem were to find the radius measure of an inscribed circle inside of that smaller triangle rather than one inside of the bigger of the two similar triangles
Hello sir. Can you recommend me some sources for understanding the evolution of mathematics through the interaction between humans and animals. E.g counting evolved through the meed to count sheeps by shepherds who started designating a stone to each sheep...
How do you know counting evolved through the need to count sheep? It could have as well evolved through the need to count how many berries were gathered and how to split them evenly among all the people of a tribe.
@@bjornfeuerbacher5514 Ya, it could be but I read this in George Simon's topology. That can be contested and can even be distinct for different civilisations, but this premise is not improbabal given the hunter gatherer state of early societies.
@@sukritisuri1137 Hunters and gatherers did not herd sheep.
@@bjornfeuerbacher5514 Ya, you're right, My appologies for a bewildered reply. The way I should have stated my point is that one of the historiographical perspectives on development of counting attributes its evolution to the need to keep track of sheeps by shepherds which is possible for agriculture and livestock rearing had been a part of initial human settlement period. Your argument is also acceptable and can be one perspective to look at this. But what I am requiring is a study of how animal-human interaction could have produced/evolved mathematics. Counting was a possible example.
Yan, tan, tethera, methera, pip, sethera, levera, hovera, dovera, dick - sheep counting numerals (1-10) in the Derbyshire Dales
Nice problem, Michael. However, from my perspective, the geometry problems that you solve in your channel are by far simpler than the algebra/calculus ones.
It's a good place to have some sake!
How do you make such nice chalk circles? Do you have a giant compass?
Every school classroom used to have those. (We also had slate tablets and ink wells, and in high school there was a giant slide rule up front, so I may be older than you.) 🙂 The compasses were made from hardwood and had a rubber tip, like an eraser, to form the point. Probably about a foot long.
@@TomFarrell-p9z: No ink wells for us but giant compasses I vaguely remember from high school in the 1960's.
Ink wells? Hell you must be from the late 1930's!
@@roberttelarket4934 LOL! I exaggerated a little. No slate tablets, and the ink wells (and stains) were there, but no ink bottles. In 4th grade we transferred to a newly opened elementary school (in 1971) and--I found out much later--our principal spent a day at the school district''s warehouse picking out the best of the old desks and chairs to fill the classrooms after their new furniture budget ran out. We had one room schools in that district until the late 1940's and I'm sure some of those desks came from when they were closed. Not sure how much older than that the desks actually were. Wonderful teachers and a great math education though. Perfectly good prep for college and careers in the maths, sciences, and engineering.
@@TomFarrell-p9z: Well then you were born about 1962.
Further you wrote maths not math implying you're not from the U.S. or was that just a spelling error?
@@roberttelarket4934 Spot on with the date. I grew up in Upstate New York. Not sure about "maths" in that context--I thought we say it that way in the US (when referring to the field). However, I was stationed in England for a few years, picked up some of their languistic details, and ever since my writing, and speaking, have been a bit confused.
「算額」が外国でも「Sangaku」と呼ばれているのに驚き
So Mike that's why you’re doubly brilliant you have two brains in two bodies the second in your identical twin!
Two brains are infinitely better than one!
nice
日本がだいすきだよ🇯🇵🇯🇵🇯🇵
Ah so!
I guess there is an unspoken 'well known method' that requires simple straight-lines, pythagoras and so forth, but NOT trigonometry, right? Because the trigonometric solution is also pretty darn direct, too.
I'm going to use intersecting line functions to find the cross-point C.
[1.1] f(𝒙) = 1 𝒙 ⊕ 0 (line AE)
[1.2] g(𝒙) = -2 𝒙 ⊕ 2 (line DB)
Solve for intersection
[2.1] 𝒙 ⊕ 0 = -2𝒙 ⊕ 2
[2.2] 3𝒙 = 2
[2.3] 𝒙 = ⅔ … and pasting into [1.1]
[2.4] 𝒚 = ⅔
This however is NOT the vertical line crossing the origin of the in-circle. the in-circle has a radius [𝒓]. Marking the center of the circle as 'O', then we note that the △AOQ (Q is base) will have left θ as ½ the angle of the slope 1 AE diagonal.
[3.1] tan( θ = ½ arctan( ¹⁄₁ ) ) = 0.41421
Likewise, the △BOQ will have a φ as ½ the angle of the DB diagonal
[4.1] tan( φ = ½ arctan( -²⁄₁ ) ) = -0.61803
These are the slopes then of the two corner-to-incircle-origin lines which similarly can be set as equal equations and solved
[5.1] z(𝒙) = 0.41421 𝒙 ⊕ 0 … and
[5.2] y(𝒙) = -0.61803 𝒙 ⊕ 0.61803 … so now make equal and solve
[5.3] 𝒙 = 0.61803 / (0.41421 ⊕ 0.61803)
[5.4] 𝒙 = 0.59873
and using [5.1] and the 𝒙, find
[6.1] 𝒚 = 0.41421 × 0.59873 ⊕ 0.0
[6.2] 𝒚 = 0.24800 //•
And that [𝒚] turns out to be the radius of the incircle.
Tada. Never really needed to know the magic formula for radius, perimeter and triangle area, per your chalkboard diagram. I definitely grant however ... knowing that magic formula helps tons by comparison.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Cloning at work.
Now let's have a geometry problem inspired by having eaten some Chun King Chow Mein.