A quick geometry problem.

Рет қаралды 651,368

3 жыл бұрын

We solve a nice and quick geometry problem.
Playlist: • Geometry
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Пікірлер: 641

@manucitomx 3 жыл бұрын

“Super easy, barely an inconvenience”, what a great call out. Dr. Penn, you rock.

@josda1000 3 жыл бұрын

It made me laugh and do a spit-take. Awesome.

@fahad-xi-a8260 3 жыл бұрын

Hey could you explain what does this mean. I really didn't understand. I just thought he just meant it because it was easy.

@ThePiotrekpecet 3 жыл бұрын

@@fahad-xi-a8260 It's from screen rant pich meeting series

@themathhatter5290 3 жыл бұрын

@@ThePiotrekpecet It's also been turned into a song! kzfaq.info/get/bejne/iMidaKxilrHRZps.html

@ZiRR0 3 жыл бұрын

@@fahad-xi-a8260 Just search up george ryan or screen rant pitch meetings. Many people in the comments use it as a joke.

@jesusthroughmary 3 жыл бұрын

For something that is "super easy, barely an inconvenience", I think it's funny that you took 50 seconds to explain that the radius of a circle with diameter 1 is 1/2.

@jesusthroughmary 3 жыл бұрын

@Flewt Puns I just did.

@brixn2302 3 жыл бұрын

He is trying to catch as many "normal people" as he can. If he takes less time maybe regular people wouldn't even try to understand what's going on here.

@wwoods66 3 жыл бұрын

Then it takes 50 seconds to figure out that the radius of the small circle bisects the bottom of the equilateral triangle. Rather than simply saying, "By symmetry...".

@MansMan42069 3 жыл бұрын

I wouldn't think too much about it considering "super easy..." is a joke in the first place.

@kylezindel314 3 жыл бұрын

@Hand Solo there’s nothing wrong with explaining givens to make students become more familiar with cosine uses.

@psioniC_MS 3 жыл бұрын

I simply calculated the height of both points (1/2 and 1/3 of the height of △, which is √3/2), then subtracted the right one from the left one, and then used Pythagoras to calculate x.

@csababekesi-marton2393 3 жыл бұрын

Same.

@iplanes1 3 жыл бұрын

@@csababekesi-marton2393 so did I much less complicated

@dr_bean 3 жыл бұрын

Same thing: saw the thumbnail, went to calculator, came back with the answer like that

@AndreasStanglPlus 3 жыл бұрын

Then you would have to argue why you take 1/3 of the height. Don't get me wrong, this is totally valid, one can see that by symmetry, but it is not something one can just assume.

@psioniC_MS 3 жыл бұрын

@@AndreasStanglPlus I see what you mean and ofc it would be required in a formal proof, but for me that's just basic geometry, like using Pythagoras without proving it

@joshhickman77 3 жыл бұрын

Am I crazy or is this way easier without any circles at all and just using coordinate geometry?

@ramakrishnasen4386 3 жыл бұрын

Coordinates are usually considered bashes , and not elegant solutions. That's why synthetic proofs are better.

@AritraDaddy 3 жыл бұрын

@joshhickman You're absolutely right, there are a shit ton of people in the comments going crazy about how he should have solved it using only anytical geometry (and not trigonometry) whereas the simplest way to do it is just coordinate geometry. Of course you'd need to know the relation between the radius of a circle inscribed in an equilateral triangle and its side, but that's common knowledge.

@edal7066 3 жыл бұрын

you are totally right. the circeles are there to confuse you.

@ashgoku6966 3 жыл бұрын

Yeah I thought that too . But who tf needs an easy solution . We mathematicians and computer science engineers are known to make shit complicated and easier at the same time lol . It's more fun solving hard problems or alternate solutions

@mutammimirtiza2454 3 жыл бұрын

Basically, Coordinate geometry is what’s done here. Coz the distance formula comes from pythagorean theorem as such

@peetiegonzalez1845 3 жыл бұрын

I love how fully half of this video is him explaining in a roundabout way, with unnecessary calculations, that the horizontal distances were all 1/2 on the bottom line.

@crosshairs007 3 жыл бұрын

Na, half the video was him PROVING that the bottom horizontal distances were all 1/2. Sure you and I know that just from looking at it, but now he has actual mathematical proof of that from which he can make further factual statements with no possibility of error.

@kelly4187 3 жыл бұрын

@@crosshairs007 Just saying "By symmetry this line segment has length 0.5" is also a proof given it is a regular triangle. His method was just way more complicated than it needed to be.

@paulh4828 3 жыл бұрын

There is another way to solve the problem with coordinates: Let's consider the coordinate system having as origin the vertex at the bottom left of the square and as unit vectors the two sides resulting from this same vertex. Then the coordinates of the center of the great circle are those of the center of the square, i.e. (1/2, 1/2). Moreover, noting that the height of the triangle is √3/2 (in the video) and knowing that the orthocenter of the equilateral triangle is at a distance from its base equal to one third of its height, the coordinates of the center of the small circle are therefore (3/2, √3/6). The distance between the centers of the two circles is therefore according to Pythagoras: x = √((3/2-1/2)² + (√3/6-1/2)²) = √(1 + 1/12 + 1/4 - √3/6) = √((8-√3)/6) ◼️

@BCSEbadulIslam 3 жыл бұрын

Coordinate Geometry: "I also exist. Why no one uses me? 😭"

@BCSEbadulIslam 3 жыл бұрын

@DELTA en some people avoid Coordinate Geometry to make questions harder.

@jesusthroughmary 3 жыл бұрын

Coordinate geometry: "Am I a joke to you?"

@lialos 3 жыл бұрын

It's how I approached this one. Found the same distances, just viewed them as points on a plane, and then used the distance formula between them.

@WarpFactor999 3 жыл бұрын

@Enrique Mtz LoL! And what would be the fun in that?!? Why not try hard to confuse the hell out of most folks?

@tmtnyc1124 13 күн бұрын

yes, it's so quick with coordinate geometry.

@nickcannata5402 3 жыл бұрын

You made this much more complicated than it needed to be.

@mathadventuress 3 жыл бұрын

He does that for a reason. Haven’t you seen his overkill series lol

@GaryTugan 3 жыл бұрын

true of many of these vids. Which makes it interesting to pause the vid and do the problem first (to see how easy it REALLY) is first, and then watch and laugh at how complex he makes it look :D

@donaldbesong8853 Жыл бұрын

Not at all. I guess that's the best way to go about it.

@hellosquirrel7271 3 жыл бұрын

Super easy, barely an inconvenience.

@thomashughes4859 3 жыл бұрын

The centroid of an ET is 1/3 from any side. That side is 1/3 root 3. Subtract 0.5. Use old Pythagy's Theory and Bob's your uncle! Neat puzzle, thanks, Doc! :D

@tortillajoe9942 3 жыл бұрын

Easier way of calculating r: The height of an equilateral triangle is also its angle bisector. The intersection of all three angle bisectors is the incenter (the center of the inscribed circle). The length from the incenter to a vertex is twice the length of the incenter to the edge, thus r is 1/3 the height (which we know to be √3/2). Therefore r=√3/6

@nahblue 2 жыл бұрын

it would be a more frustrating youtube video that way: it would be a series of facts, and people who don't know them feel like "huh, so I'm no good if I don't know that". And the video would be less free standing. I think he knows what he's doing, and he's doing it well :)

@SzanyiAtti 3 жыл бұрын

I used the fact that rs=A, so r=A/s, and since the triangle is equilateral with a sidelength of 1, it was easy to put values in the places of A and s

@nicholasrhodes1729 3 жыл бұрын

Loving these geometry problems. Been giving these questions to my classes of pupils. Good work.

@ffggddss 3 жыл бұрын

Having picked up the problem from the thumbnail, before watching the vid, what I'd do is: 1. Drop an apothem from each polygon's center to its base 2. Draw a horizontal (i.e., parallel to the bases) through the ∆ center to the square's apothem. You now have a rt ∆ whose: • hypotenuse is x, • horizontal leg is a segment of the line we drew, of length b=1 (sum of half of each polygon's side); • and whose vertical leg is the difference in the heights of the two polygons' centers (that is, the difference between their incircle radii), a = ½ - 1/(2√3) = (√3 - 1)/(2√3) Enter Pythagoras, stage left: x² = a² + b² = (√3 - 1)²/12 + 1 = 1 + (2 - √3)/6 = (8 - √3)/6 x = √(8 - √3) / √6 = √(4/3 - 1/(2√3)) = 1.02208522087863126046709003932782... There may be a simpler symbolic form of the answer than this, idk... Fred

@nakamakai5553 3 жыл бұрын

Beautiful work. My new favorite channel. Thank you.

@rcabr31 3 жыл бұрын

It's very easy to find r, just need a bit of pythagorean theorem, if h is the height of the equilateral triangle: (h-r)^2 = r^2 + (1/2)^2 and set h = sqrt(3)/2

@ExaltedDuck 3 жыл бұрын

My answer was way different. I had "X is the 24th letter of the modern English language, having etymology dating back to the ancient greek chi and is often used as a stand-in for the primary unknown in the analysis of mathematical systems."

@deepjyoti5610 3 жыл бұрын

Quick geometry problem series are good basic builders or introduction to olympiad geometry

@manuelgonzales2570 Жыл бұрын

Excellent! Thank you!

@dsodragon8152 2 жыл бұрын

i don't know how you make this seem so easy... when you explain ... it's like tetris block falling together in my head perfectly!!!

@taufiqutomo 3 жыл бұрын

I tell my students that whenever a circle touches anything, draw a line from the center of the circle to the point of tangency. In this case, this resulted in five extra unnecessary lines.

@jamesmnolan 3 жыл бұрын

Actually, only 3 unnecesary lines (the 2 dropped perpendiculars from the centres of the circles are indeed necessary to show/calculate the difference in height between the centres)

@trelligan42 3 жыл бұрын

Center of an inscribed circle IS the center of a regular polygon (the circles are distracters). Horizontal distance is easily seen as 1 (half of each figure's width). Vertical distance is 1/2 for the square, and scaled for the triangle by the altitude versus 1. Then you use difference of altitudes versus horizontal difference, which leads to the Pythagorean solve he ends with. Same answer, fewer extraneous lines.

@Nicolas-zf3pv 3 жыл бұрын

Everyone is saying that “coordinate geometry would be easier”, but they end up doing exactly what he just did, saying that they would put a coordinate system in the left corner of the square, getting the positions of both centers and calculating the distance. However in the video he just did the same thing, got the horizontal distance between each point (x1-x2), got the vertical distance between each point (y1-y2) and apply Pythagoras, exactly the same. In some comments they calculate the height of the center of the circle inscribed in the triangle (y2) by the well know formula 1/3 of height of triangle (a * sqrt(3)/2), that really would be easier than using tryg, but he decided to use the most explicit method which was easily proved and didn’t require this formula memorization or any other kind of proof. And again, remembering and using this formula has nothing to do with coordinate systems being easier.

@stormlight131 3 жыл бұрын

Glad someone said it. I thought I might be crazy for thinking the coordinate geometry method is identical to the given method.

@RexxSchneider 2 жыл бұрын

@@stormlight131 Using coordinate geometry (or 2D-vectors) will always result in the same calculations, but finding the distance between (0, 1/2) and (1, sqrt(3)/6) is a standard procedure (which is actually applied Pythagoras). You don't have to construct extra lines to make a rectangle and transpose the y-coordinate of the triangle's incentre in order to find a side of right-triangle, just to then use Pythagoras; coordinate geometry does that for you. Similarly, finding the modulus of the difference between two vectors will require the same calculations, without having to construct a right triangle to apply Pythagoras to. These tools are useful conveniences, but they can't simplify the arithmetic.

@geoninja8971 3 жыл бұрын

I always like to see a 'real-world' value, like 1.022, more comforting you haven't screwed something along the way! Nice problem though, I enjoyed it.

@DubioserKerl 3 жыл бұрын

01:31 did he just...?

@thenuka9954 3 жыл бұрын

Yh he fucken did, could not be prouder

@JeremyKramer7 3 жыл бұрын

we know that phrase from the first movie!!

@nmmm2000 3 жыл бұрын

Good job! For some reason I like the video very much. On the square, you can divide the diameter by 2 and get the radius directly. On the triangle - use the fact that height and mediana are one and same. Then you can use the medicenter divideds the mediana in ratio 1:2

@wolo0048 3 жыл бұрын

I tried a Law of Cosines approach and found it pretty straight forward, using the diagonal of the square from the center, the angle bisector of the triangle to the incenter, and the given segment as side C across the angle of 105 degrees. This is known since their is a 30 degree angle between the square and the triangle, as well as the 90 degree angle of the square, and the 30 degree angle of the angle bisector.

@muffinthesoulstealer 3 жыл бұрын

Haha didn't know you liked pitch meetings too, Dr. Penn! Loved the problem, also.

@pdeteresa 3 жыл бұрын

can also solve it by knowing that centroid of triangle is h/3 and the distance from centroid to the midpoint of any side is r = h/3 (for equilateral triangles)

@Ahldor 3 жыл бұрын

I love how mathmatical answers are so untelling.

@BboyKeny 2 жыл бұрын

It's not about the destination but about the theory's that we made friends with along the way.

@jansustar4565 2 жыл бұрын

Another way to get the triangle radius without trigonometry; The radius of the circle is equal to its Y coordinate (assuming the bottom side is at Y 0). And that also happens to be the center of the triangle. And to get the center, just add up all of the vertexes and divide by 3. Since X is already known, we only need Y. And it also turns out that two vertexes are at Y 0. So (0 + 0 + h) / 3 = h / 3. And h, in equilateral triangle is sqrt(3) / 2. And (sqrt(3) / 2) / 3 = sqrt(3) / 6. And it is the same answer.

@louisreinitz5642 3 жыл бұрын

super easy barely an inconvenience! I see what you did there!

@fahad-xi-a8260 3 жыл бұрын

Could you explain what did he do there.

@Ch1pp007 3 жыл бұрын

@@fahad-xi-a8260 It's a reference to Ryan George's genius Pitch Meeting series over on the Screenrant KZfaq channel.

@zhangruoran 3 жыл бұрын

Don't forget something called the "law of cosine". Connect the centers of the two circles to the right down corner of the square. Then solve the triangle.

@andreimazilu2339 3 жыл бұрын

Nice. That's what I was about to comment as well. We get a triangle with two easily to find sides (one is half of the diagonal of the square: √2/2, another is the distance between one corner of the equilateral triangle and the center of it's circle: √3/3 and the angle is 105 degrees). According to the rule of cosine, the distance we're looking for is the square root of the summ of the squares of these two sides minus the duble of the product of the sides multiplied by the cosine of the angle between them.

@AlphaCurveMath 3 жыл бұрын

Everybody asks what is x, but no one asks how is x :(

@jesusthroughmary 3 жыл бұрын

I'll do you one better. Why is x?

@notjoshidk1107 3 жыл бұрын

This comment was expected, but it is funny ngl...

@themaverick1891 3 жыл бұрын

How do you know that no one asks that? Maybe many people do ask.

@lpjdrummer12295 3 жыл бұрын

@@jesusthroughmary I'll do you one better. Y = X.

@fintan. 3 жыл бұрын

@@jesusthroughmary when is x

@asklar 3 жыл бұрын

The length of the segment can be obtained by pythagoras. The horizontal side of the triangle is ½+½. The vertical is the difference in height of both circles, so 1-√3/2 (the circumcenter of an equilateral triangle is at √3/2). So x²=1+(1-√3/2)²

@denischarushev8737 3 жыл бұрын

Excellent solution and explanation

@iTriguy1 3 жыл бұрын

And this is why so many kids in school say "I hate math". You used trig when simple coordinate geometry would do.

@ethanwilliamson782 3 жыл бұрын

You don’t understand teaching then. The point of this problem is for it to be simple enough for them to understand the solution and idea of the problem, and apply a concept that is new to them, therefore allowing what is a new and likely complicated process to be easily understood in terms they already understand

@ethanwilliamson782 3 жыл бұрын

@@xa1257 you don’t understand my point; you likely can’t solve the complicated variant without previously understanding simpler math, so by applying new solutions to previously understood circumstances, a student isn’t thrown into the deep end. If you don’t know the simpler solutions, click off the video and leave, you don’t understand enough math.

@jonathanstudentkit 3 жыл бұрын

what you say is hateful nonsense

@KamalSharma-uq1jb 3 жыл бұрын

If a equilateral triangle we can't we just directly use inradius formula a/2root3.. ???

@JilariousJamesGaming 3 жыл бұрын

Nice, easy to understand

@randybell101190 3 жыл бұрын

Inscribed circles are Tight!!!

@m.h.6470 3 жыл бұрын

You don't need the circles AT ALL. X is just the line between the centerpoint of the square of the triangle. As such, if we put a coordinate grid at the bottom left square, the centerpoint for the square is at 0.5/0.5 and the center point of the triangle is defined by a right triangle starting at 0.5/0 with the right angle at 1/0 and a 30° (Pi/6) angle at the starting point. Use cotangents to get the y-value of the centerpoint of the triangle. Having both the start and end point of x, you simply need to use a² + b² = c², a being the difference of the x-values, b being the difference of the y-values.

@tychophotiou6962 2 жыл бұрын

I have watched a few of this mans videos and he always makes a meal of an easy problem. I did this in about 1 minute by noticing that the equilateral triangle when you drop a perpendicular is a 30, 60, 90 triangle, ie 1,2, root 3 triangle cut in half. therefore height is root 3/2. Therefore r = root 3 /6. Then apply pythag... simple!

@turnandfacethedragon 3 жыл бұрын

We can use trig to find that 1/2 of 1 is 1/2? That's brilliant! Have you never heard of special triangles? 30-60-90=1-2-3^1/2

@bekhaddaderrar2111 3 жыл бұрын

It's very very good . this is incredible bro

@methodiconion8523 3 жыл бұрын

Since getting back into higher-ish maths, I forgot that rationalizing denominators was convention since we're always told we don't have to do stuff like that, or even simplifying for that matter. Incidentally, it took me a second after finishing the video to figure out why the radius mattered until I realized it was the y-component of the right endpoint.

@fredfrancium 3 жыл бұрын

I like how he is teaching. Amazing

@OldHumbleDistillingCompany 3 жыл бұрын

Great Ryan George reference, Dr. Penn.

@mccloysong 3 жыл бұрын

1:45 doesn't that automatically divide the triangle base in half?

@user-vh1cu4iq8d 3 жыл бұрын

Easy-peasy. Solved it in my mind.

@kelly4187 3 жыл бұрын

Man, this was total overkill. Good job!

@kiran.i 3 жыл бұрын

Excellent explanation....

@rahulronaldo6813 3 жыл бұрын

Fantastic...

@balazsgyekiczki1140 3 жыл бұрын

Ooooh, nice Ryan George reference you've got there

@c-bass9968 3 жыл бұрын

When he busted out the trig to prove an altitude from a vertex to a side of an equilateral Δ bisects the side.

@jacobcrihfield1092 2 жыл бұрын

How do you know it’s an equilateral?

@denispiralic2669 3 жыл бұрын

you can also do the second part by letting the bottem left of the square be an Origin. Then use vectors and find the modulus between the centre of the two circles

@goodplacetostop2973 3 жыл бұрын

8:18 Homework... A triangle ABC is to be drawn with AB = 10, BC = 7 and the angle at A equal to θ, where θ is a certain defined angle. Of the two possible triangles that could be drawn, the larger triangle has three times the area of the smaller one. What is the value of cosθ?

@deepjyoti5610 3 жыл бұрын

Nyc one

@srijanbhowmick9570 3 жыл бұрын

Is it sqrt(17)/5 ?

@goodplacetostop2973 3 жыл бұрын

@@srijanbhowmick9570 I dunno... Did you just throw this number randomly? 😛

@mouseberry5257 3 жыл бұрын

@@goodplacetostop2973 his ans is right try using cosine rule

@abhaykumar-xo3rs 3 жыл бұрын

is it sqrt of (51/75)

@germanarestrepov518 3 жыл бұрын

Super chévere el ejercicio, da buenas ideas para nosotros los gomosos de echarle cabeza a este tipo de retos.

@fedyarudkovskiy6178 3 жыл бұрын

Would love to notice the fact that in equal sided triangle center of inscribed cirle is in intersection of bisectrices which are also heights and medians. So that pieces on the down side of triangle could be calculated without cosine function.

@udic01 3 жыл бұрын

Why are you using trigonometry for a pure geometry problem? If you know that the line from the center of the small circle and perpendicular to the "base" is passing through the top vertex then you know you can use that it is an angle bisector and it is a median which cuts the base in to 2 equal segments. And the medians in every triangle interaect at the centroid and divide each other in to segments whose ratio is 2:1. There is also the 30-60-90 triangle with the edges ratio of 1:sqrt(3):2. I know that trigonometry is faster since you can do the calculations faster with calculators but then it should have been called a trigonometry problem and not a geometry problem.

@yashvardhan6521 3 жыл бұрын

Yes so true,I did it the same way

@txikitofandango 3 жыл бұрын

I've wanted to say this before too. Trig is less elegant if you're using 30-60-90 triangles

@peterklenner2563 3 жыл бұрын

Engineer here. The problem describes two points in the plane and asks for their Euclidean distance. How does casting this as a geometry problem or a trigonometry problem help or make a difference?

@udic01 3 жыл бұрын

@@peterklenner2563 you should watch his overkill videos first. This is a geometry problem and can be resolved with pure geometry. It's like solving 5+3 with geometric series and limits and taylor series and complex numbers. Of course we can solve with them but 5+3 is a simple algebric problem and we can solve it with simple tools. You can solve the problem also with 1. vectors 2. Cartesean axes (analytic geometry? I forgot the correct term) And probably with topology also. But my question was why do it that way? It is not more simpler then the one i suggested with pure geometry (he introduced the problem as a geometric(!) one and not as just a problem)

@peterklenner2563 3 жыл бұрын

@@udic01 Mr. Penn approaches the problem quite straightforwardly. Which shortcuts would be possible? More to the point, how could the last step of invoking the Pythagorean theorem be avoided or simplified?

@malignusvonbottershnike563 3 жыл бұрын

Nice, I could actually do this one in my head! Took a few minutes, but still worth it so that I could stay lying down and not get a piece of paper lol

@nilsastrup8907 3 жыл бұрын

Nice, did the same😊

@d314159 3 жыл бұрын

Thought the same and so did this one in my head to try and keep my aging brain in shape!

@NoName-jh7yj 3 жыл бұрын

@moisesggomes 3 жыл бұрын

Me too. I looked at the problem and solved in my head. My solution was wrong but I tried, and with this amazing feeling of not getting afraid of the question I went to sleep proud of myself.

@tgx3529 3 жыл бұрын

We can use that centre of circle lies at the "centr of gravity" of the triangel . Then we have r=1/3*(sin pi/3)=sqrt(3)/6

@JSSTyger 3 жыл бұрын

Its better to say "center of geometry".

@sublivion5024 3 жыл бұрын

Centre of gravity, centre of mass, and centre of geometry are different things

@1234567qwerification 3 жыл бұрын

Don't they coincide for an equilateral triangle?

@tamarpeer261 3 жыл бұрын

height of left point: 1/2 Height of right point: 1/3 Horizontal distance between the two points:1/6 Vertical distance between the two points:1/2+1/2=1 1+(1/6)^2=37/36 d=sqrt(37)/6

@Nikioko 3 жыл бұрын

Big secret: in an equilateral triangle, the heights are identical to the medians (and both angular and perpendicular bisectors) which intersect in the centroid at a ratio of 2:1. So the radius of the small circle is equal to 1/3 of the triangle's height which is √3/2. So the difference in x dimension is 1 (2x1/2) and in y dimension it is 1/2 - √3/6 (or 1/2 - 1/(2√3)). These are the legs of a rectangular triangle. And then use Pythagoras to solve for x which is the hypothenuse.

@michaeltieber3550 3 жыл бұрын

I really don't know what you were doing with angular functions there. This was fully solfable by Pythagoras. I did it in my head, without writing it down.

@kinekt2000 3 жыл бұрын

most difficult way to find radiuses can't be imagined

@happygimp0 3 жыл бұрын

2:22 Did you just calculate cos(pi/3) for dividing by 2? You know you could just divide by 2 to find the length to the middle?

@arkadiosvotessocialist8436 3 жыл бұрын

This was pretty easy to solve using coordinates. The center of the square is at (0.5, 0.5), and the center of the triangle is at (1.5, sqrt(3)/6. Just use the distance formula to take care of the rest.

@MrCoffeypaul 3 жыл бұрын

Sweet little problem!

@josephcoon5809 3 жыл бұрын

The center of the triangle circle is 60degrees between a vertex of the triangle and the center of a side. That creates a 30/60/90 triangle. Half the base is the long side making the short side half the base divided by root 3.

@ABc-es4nv 3 жыл бұрын

Nice solution, but I think a far easier one is by placing them both in a coordinate system. You set the points of the square at (-1, 0), (0, 0), (0, 1) and (-1, 1), the center of the inscribed circle is at S (-1/2, 1/2) which you can easily work out. You set the points of the triangle at (0, 0), (1, 0) and (1/2, sqrt(3)/2), the radius of the inscribed circle is sqrt(3)/6 and from that follows that the center of the circle is at point S1 (1/2, sqrt(3)/6). And then you just calculate the distance SS1 between the centers.

@lyrimetacurl0 7 ай бұрын

There it is!

@jamesfortune243 3 жыл бұрын

Off topic here. Could you give your opinion on George Boole's book on finite differences, including the best results that would be applicable to contest problems?

@casper7707 3 жыл бұрын

Yoo English is not my primary language, I didn't learn about sin,cos, tan yet... but I still understood everything... That's a solid proof that you're a rly good teacher!

@shatishankaryadav8428 3 жыл бұрын

Why you not using smart board Do for tangent and circle related problem

@shzaizzhang4465 3 жыл бұрын

I tried to build a x-y croodinate at the bottom left corner of that square, and using croodinates are fast too!

@parkerraley2248 3 жыл бұрын

I was on the right track until you transposed r. What allows you to do that and guarantee it is the same length?

@joeym5243 3 жыл бұрын

Another way to solve is since you know that a line from a corner of an equilateral triangle to it's center will bisect it, and that all angles are 60°, and since the x-distance is .5, you can find the y value (from the base) from .5tan(30). After that, it's pretty straight forward.

@oozmakappa8895 3 жыл бұрын

that ain't math, it's witchcraft

@zoranocokoljic8927 3 жыл бұрын

In a equilateral triangle all 4 significant points - centers of inscribed circle, center of circle around, orthocenter and center of mass - are the same point and it is situated on the third of a height, and the height is h=(a/2)*SQRT(3). So the radius of the circle is (a/6)*SQRT(3). The distance between the centers of the circles on y axis is a/2-(a/6)*SQRT(3) or (a=1) (3-SQRT(3))/6. Square that we get (12-6SQRT(3))/36 or (2-SQRT(3)/6. The distance between centers on x axis is 1/2+1/2=1. We add that and get (8-SQRT(3))/6. So x= SQRT[(8-SQRT(3))/6] . This is basic 5th grade stuff, anyone who deals in mathematics ever a tad should know by heart. No trigonometry needed.

@devondevon4366 3 жыл бұрын

Dr. Penn really did make it more difficult than it was as his answer sqrt [8-sqrt3/6] also =1.022. This could be done in a minute. I did it by finding the radius of the circle (in the equilateral triangle to be 0.28867 using sqrt(s-a)(s-b)(s-c)/s) and the radius of the circle in the square is 0.5. the difference between the two is 0.211324 (or the length of the .perpendicular line formed). So now we have 2 sides and need to find x using the Pythagorean theorem. so 1^2 + 0.211324^2= x^2 1 + 0.04465 =x^2 1.04464 = x^2 1.022 = x Answer, and his answer "sqrt (8-sqrt3/6)" = 8-1.732/6 = 6.627/6 = 1.04465 = (1.04465) sqrt= 1.022 is equivalent to mine '1.022. (18:17)

@Avighna Жыл бұрын

This was the only problem of yours I could actually solve! I used analytic geometry along with a couple of lines to find x = sqrt((8 - sqrt(3))/6)

@michaelgolub2019 3 жыл бұрын

The solution seems straightforward. However, you could use properties of an equilateral (or even isosceles) triangle (median = height = bisectrix) and the radius of inscribed circle as well. So, no extra explanations would be needed. I like your videos very much and this one contrasts with them. Here is no zest that exist in your other videos: just a school problem that does not require some trick to be solved efficiently.

@mikeh283 3 жыл бұрын

What's the limit as you increase the polygons sides by one, start with the triangle and go right....?

@valijoneshniyazov3859 3 жыл бұрын

put O(0;0) at the point where square and triangle joint! A(-R,+R) B(+a/2,r) find distance between A and B! d=sqrt( 9/4 + (0.5-0.5/sqrt(3))^2)

@kiduzi9507 3 жыл бұрын

It's so sad...I graduated high school in 2020 and didn't attend college because I cannot learn online; I must go physically otherwise I cannot REALLY learn. My lack of practicing math enough has made me forget how exactly to do the steps to solve this problem. I found a way of solving it which was actually hilariously "obvious" early on, but I lost the skills to rigorously solve it. I can't wait to go back to school.

@fajardohansalexander3949 3 жыл бұрын

You can still simplify the 8/6 in the end right???

@5t33v 3 жыл бұрын

Super easy, barely an inconvenience

@grahamjoss4643 3 жыл бұрын

Very cool

@davideciamacco3402 3 жыл бұрын

At 4:47 you draw a XY axes and the origin in the left bottom corner of the square and easily find the distance of two point knowing their coordinates.

@miradota347 3 жыл бұрын

Make 2 lines from the center of the triangle's bottom sides to the facing angles, the crossing point is the center of the triangle, give them length R, use R.cos(x) = 0.5 (half of the base). Then by finding R , use R.sin(x) = y ( the height of the center from the bottom). Now make a trangle connecting both centers with side length of (0.5 - y) and base of , the distance between the centers can be found using pythagoras ( X^2 = (0.5-y)^2 + (1^2) ). Same answer, but in a minute, and without using the fact that there are 2 circles, because they share the same centers anyways.

@stromboli183 3 жыл бұрын

Misinterpreted it from the thumbnail, I thought it was only about the part of the line between the square and the triangle. Perhaps consider that a bonus question ;)

@livinglogically8180 3 жыл бұрын

That is one time taking question... I also thought the same and ended up taking half an hour

@giuseppeficara4876 3 жыл бұрын

Using trigonometry, this is a trivial problem. A little bit less trivial is not using any trigonometric function. The key is a very basic concept that doesn't come into the description in the video: highs, bisectors and median are coincident in a equilateral triangle. The center of the inscribed circle in a triangle is where the bisectors do intersect each other. Once this is noted, Pythagoras's theorem all what you need to solve the problem. And indeed I was used to solve this kind of problems at age of 12.

@liamyaroschuk9940 3 жыл бұрын

I solved this problem using the area of the triangle. The hard part was solving for the radius of the circle inside the triangle. First, I need the area of the triangle. The width is 1, and the height is one leg of a 30,60,90 triangle where the hypotenuse is 1 and the other leg is 1/2. So the height is 1/2 * sqrt(3) = sqrt(3)/2. So, the area of the triangle is sqrt(3)/(2^2). Now is the fun part. I drop the angle bisectors from each corner until I reach the center of the circle. Now I have 3 smaller triangles each with an area of r/2, and their sum is 3r/2. Solving for r: 3r/2 = sqrt(3)/(2^2) 3r = sqrt(3)/2 r = sqrt(3)/2*3 r = 1/2*sqrt(3). I solved for the distance the same way as him. This was a fun problem! Thanks!

@user-fz3xc2sf5i 3 жыл бұрын

The point at which triangles and rectangles meet (0,0) The left center is (-0.5, 0.5) The right center is (0.5, 0.5×3*(1/2)÷3) Pythagorean theorem of two points.

@carlosgiovanardi8197 3 жыл бұрын

Mr Penn, how do we tackle problems like the following? Smallest positive integer that starts with 3016 and is divisible by 3017. Generalization: smallest positive integer that starts with "n" and is divisible by "n+1". Is it more easy when "n is odd and n+1 is even" or when "n+1 is odd and n is even"?

@JSSTyger 3 жыл бұрын

a = (n*10^b+c)/(n+1) where a, b, and c are positive integers. There, I started it for you :)

@carlosgiovanardi8197 3 жыл бұрын

@@JSSTyger it is an idea for Mr. Penn (if he likes and have time) to develop a video on number theory and play with numbers. Here the reference math.stackexchange.com/questions/3894122/ Another to play with is math.stackexchange.com/questions/1367371

@JSSTyger 3 жыл бұрын

@@carlosgiovanardi8197 I think the number that starts with 3016 is 30160949, which is 3017 times 9997. This would mean 9997 = (3016*10^4+949)/(3016+1).