A simple geometry problem with a nice generalization.

Рет қаралды 48,596

3 жыл бұрын

We give a solve a nice simple geometry problem with a connection to the harmonic mean.
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Пікірлер: 153

@michaelempeigne3519 3 жыл бұрын

12 / ( 3 + 1 ) is equal to 3; not 4 ( as put in video )

@trelligan42 3 жыл бұрын

Miswrote AND misspoke; unusual for Michael.

@MichaelPennMath 3 жыл бұрын

Classic Michael Penn arithmetic!

@michaelempeigne3519 3 жыл бұрын

@@MichaelPennMath maybe you should redo the video ?

@kurax9115 3 жыл бұрын

@@michaelempeigne3519 why

@peterchan6082 3 жыл бұрын

@@kurax9115 Just to clean up the errors.

@user-en5vj6vr2u 3 жыл бұрын

7:24 michael.exe has stopped working

@crazy4hitman755 3 жыл бұрын

I wanted to comment this but I knew that someone would have done it before me

@petersievert6830 3 жыл бұрын

Seems like it was forgotten to be cut, right?

@demenion3521 3 жыл бұрын

for those who are wondering how 12/(3+1)=4, you have to notice that 3>>1 so that we can neglect the 1 compared to the 3 and we are left with 12/3=4 :D

@Lucas125K 3 жыл бұрын

@@abbeleon lmao 😂

@michaelgian2649 3 жыл бұрын

A wonderful lesson (by declining to do an edit cut) teaching the pursuit of rigour. Not one, but two pauses to regroup. 7:24 (with a "b-beep") & 7:30 , both after uttering true statements that were not precisely relevant to the argument. The latter facial expression is priceless. This lesson has as much value as the maths content.

@javizaragoza1463 3 жыл бұрын

7:21 when you are trying to explain the situation to your jealous girlfriend but she keeps interrupting

@eugenekim3012 3 жыл бұрын

I have to mention a simpler method... If the radius of the circle is 2, then we have OC = 2 and OA = 4. Provided that angle OCA is a right angle, we know that angle COA is 60 degrees. Then if we connect PC, we have triangle OCP being an equilateral triangle. If PC = OC, then point B is the midpoint of OP, so x is half of the radius, and that is 1.

@goodplacetostop2973 3 жыл бұрын

13:06 Not a good place to stop but that’s a nice result

@thayanithirk1784 3 жыл бұрын

Good Place To Stop 😹

@thayanithirk1784 3 жыл бұрын

May I know you nationality

@________6295 3 жыл бұрын

@@thayanithirk1784 he is michael

@goodplacetostop2973 3 жыл бұрын

@@________6295 I’m not

@JalebJay 3 жыл бұрын

I think that's a nice result.

@pieguski3924 3 жыл бұрын

In the picture we can notice more means!! AO is the arithmetic mean. AC is the geometric mean. and there is a hidden square mean. If we will draw a straight line perpendicular to AQ and If we will call the intersection of our drawn line and circle R, the length of AR is mean square. This is a great geometric example of means inequalities. ( Of course all means are means of 2 lengths AP and AQ).

@djvalentedochp 3 жыл бұрын

that's awesome

@lucashoffses9019 3 жыл бұрын

You didn’t specify where along the line we draw the perpendicular.

@pieguski3924 3 жыл бұрын

@@lucashoffses9019 oh yeah. drawing perpendicular line from point O.

@latarnik1 3 жыл бұрын

I love math problems especially geometry and this channel is perfect for me. Good job, man!

@mathismind 3 жыл бұрын

Love your videos!

@mrflibble5717 2 жыл бұрын

I really liked this interesting Geometry result, and as usual, nice presentation. Thank you Michael.

@dzakytamir3048 3 жыл бұрын

Me : *look at the thumnail Also me : *pointing the "x"

@timurpryadilin8830 3 жыл бұрын

You could also easily prove here that the harmonic mean is always smaller than the arithmetic mean: point B is surely on the left from the point O.

@tasnimmahfuznafis8892 3 жыл бұрын

Thanks, I did not notice it at first.

@OH-pc5jx 3 жыл бұрын

This also immediately gives you the AM-HM inequality as AO is the arithmetic mean of a and b

@user-fp8qp6fh2g 3 жыл бұрын

Great this problem continue your work

@alperenkoken 3 жыл бұрын

I was studying harmonic division in geometry and this video was loaded.

@andreivila7607 3 жыл бұрын

Wow! Too easy... I would like you to check out this problem from the JBMO 2014. This is its statement: Determine all the triplets of prime numbers (p,q,r) such that 3p^3 - 5q^4 - 4r^2 = 26. You can probably find the solution on AOPs, but I would really like you to showcase this problem on your channel :)

@matniet43 3 жыл бұрын

Hey, I did some calculations: if the circle has a radius r then the segment b is 2r+a and working on the harmonic mean of a and b gives a simpler relationship (1/x = [1/r] + [1/a])

@badremathsbadro7642 3 жыл бұрын

Thank you

@jhkst4751 3 жыл бұрын

Well this looks like circle inversion in analytical way. I like it especially for solving geometric problems. Eg. construct touching circle to 3 oher circles etc.

@badbud804 2 жыл бұрын

I propose one, in my eyes, quite elegant first-sight solution. As the angle OCA is a right angle, we know by Tales that C lies on a circle trough A and O with the center P and r=2. Therefore, the intersection of both circles must project on the middle of PO, which makes x half the radius, i.e. 1.

@dneary 3 жыл бұрын

It's a nice opportunity to use Power of the Point too - |AC|^2=|AQ|.|AP| = 12 - so we have a 2:1:sqrt(3) triangle, or a 30-60-90 triangle.

@lucho2868 3 жыл бұрын

AOC is congruent to COB thus 2×BO=OC so POC is equilateral and x=1.

@tomatrix7525 3 жыл бұрын

Lovely addition to an otherwise somewhat basic and boring problem. Very nice

@georgecaplin9075 3 жыл бұрын

He uses the phrase “a radii “ a lot...alright twice, but it’s two times too many for a good mathematician like him.

@michaelgian2649 3 жыл бұрын

Spoken more than once then makes them plural?

@EspyLacopa2 3 жыл бұрын

Looked at the initial drawing for about 30 seconds, figured out x = 1 becuase it's a right triangle with hypotenuse of 4 and a leg of 2. . .means it's a 30-60-90 right triangle. Very helpful. since that height drawn that seperates the triangle into two creates similar, that means the short leg is half the length of its hypotenuse (which we see is 2). 2-1 = 1.

@gamedepths4792 3 жыл бұрын

Me at 5:06 - 3=6/2 ?

@djvalentedochp 3 жыл бұрын

same lol

@The_Math_Enthusiast 3 жыл бұрын

He: What's the relation b/w 3, 2, 6. Me: 3*2=6

@bourhinorc1421 3 жыл бұрын

me too but that was too obvious to be true

@jofx4051 3 жыл бұрын

Others: 6=2*3; 6/2=3; 6/2=3

@jofx4051 3 жыл бұрын

2^3-2=6; 3^2-3=6; (3-2)^6=(2-3)^6; 3^2-2^3=6/6

@The_Math_Enthusiast 3 жыл бұрын

@@jofx4051 thanks man

@The_Math_Enthusiast 3 жыл бұрын

@@jofx4051 you open my eyes

@kaypee9187 3 жыл бұрын

Sir, another interesting property: If we set the distance AP as y, and the distance OB as x, then we get the result: x= r.r/ (r+y), which means that x and y share an inverse relationship. Thus as y goes to infinity, the distance x reduces to 0 and when y is 0 the distance x=r. It is as if the point B is an image of the point A.

@copyrightfreemusicforall5563 3 жыл бұрын

Thanks sir

@zsigmondtelek1612 3 жыл бұрын

Funny observations as well: AC gives geometric mean and AO the arithmetic mean.

@math-4-science32 3 жыл бұрын

I am now searching if the quadratic mean is somewhere on this figure :)

@zsigmondtelek1612 3 жыл бұрын

@@math-4-science32 hint: ((b-a)/2)^2+((a+b)/2)^2=(a^2+b^2)/2

@math-4-science32 3 жыл бұрын

So we take a radius OD which is perpendicular to AO, and the quadratic mean is the hypotenuse in the right triangle AOD ?

@zsigmondtelek1612 3 жыл бұрын

@@math-4-science32 yes exactly

@abdelilahsalimchatar4653 3 жыл бұрын

Thanks for this! Now I can show off in school.

@ffggddss 3 жыл бұрын

"What is the harmonic mean?" It's one of a class of transformed means whose transforming function is f(x) = 1/x [or a/x, a ≠ 0]. A transformed mean is M(x[1] ... x[n]) = f⁻¹( [1/n] ∑f(x[i]) ) And the corresponding transformed sum is S(x[1] ... x[n]) = f⁻¹( ∑f(x[i]) ) Basically, it's done in 3 steps: 1. Transform the x's 2. Take the average (or sum) 3. Untransform the result Other examples: When f(x) = ax + b, with a ≠ 0, M(x...x) is just the ordinary average; S(x...x) is the ordinary sum. In a sense, these are "null transformations." When f(x) = a ln(x), with a ≠ 0, M(x...x) is the mean proportional; S(x...x) is the product. When f(x) = ax², with a ≠ 0, M(x...x) is the RMS (root mean square); S(x...x) is the RSS (root sum of squares) Generally, you can construct a transformed sum and mean from any function, restricted to a domain where it is (strictly) monotonic. Fred

@udic01 3 жыл бұрын

For the specific problem in the beginning just connect C to P.. Since ACO is a right triangle and AP=PO => CP=AP=2. or another way triangle ACO is 30-60-90 plus CO=OP meaning triangle CPO is equilaterral. Either way x is half of PO thus x=1

@djvalentedochp 3 жыл бұрын

superb

@patrickpablo217 3 жыл бұрын

I always think of the harmonic mean equation as: ( (a^-1 + b^-1) /2 )^-1 which i like since it looks a lot like (a+b)/2

@isaacchen23 Жыл бұрын

By projecting through C, we can obtain -1 = (A, B; P, Q). This result trivializes the problem.

@xl000 3 жыл бұрын

Dude, can you close a Captains of Crush #2 ? I'm 2 mm away from closing a #2, both on the left and right hand

@JalebJay 3 жыл бұрын

5:58 12/4 = 4 now guys ;)

@tonyhaddad1394 3 жыл бұрын

Yes he make a little mistake

@MrRyanroberson1 3 жыл бұрын

well if you recall e = 3 but also e ~ 2 so 12/(3+1) = 12/(e+1) = 12/(2+1) = 12/3 = 4

@m4riel 3 жыл бұрын

I think this would be harder for general values, but I thought of a faster way to do it. If we divide everything by 2 (so that r = 1): •r=1 •AO can be interpreted as sec(þ) •AP=1 •AC as tan(þ) •BO as cos(þ) •PO as r=1 That would imply that x = PB = PO-BO = 1-cos(þ), but also that sec(þ) = AO = AP+PO = 1+1 = 2 => sec(þ) = 2 => cos(þ) = 1/2 => x = 1-cos(þ) = 1-1/2 = 1/2 Multiply it by 2 to get x=1.

@tonyhaddad1394 3 жыл бұрын

Very easy probleme but beautiful

@SlidellRobotics 3 жыл бұрын

I admit that I had to look up the method to find a tangent to a circle through a point outside the circle, and I don't recall having seen it before (though I knew it was possible as this video showed that it only involved multiplication and division). Anyway, putting the pieces together: To find the harmonic mean of lengths x and y, draw an arbitrary line and select an arbitrary point A on it. Find points P and Q at distances x and y from point A (both on the same side of A). Bisect PQ to find point O. Draw the circle with origin at O and radius OP (or equivalently OQ). Bisect OA, finding point R. Draw a circle centered on R with length OR (or equivalently AR). Draw a segment where the circles centered on O and R intersect, calling the intersection of that segment with the original line point B, and either of the circle intersections C. Then the length of AB is the harmonic mean of AP and AQ (and therefore x and y). And OBTW, by the "power of a point" theorem, AC is the geometric mean of AP and AQ. Obviously AO is the arithmetic mean of AP and AQ. So three means constructed!

@CM63_France 3 жыл бұрын

Hi, 5:59 : 3 instead of 4 For fun: 1:57 : with moderation. 13:09 : nice result No good place to stop. 1 "then next what we are going to do", 1 "another thing we are going to do", 1 "the first thing we are going to do", 1 "now what I want to do", 1 "what I want to notice here", 5 "let's go ahead and", including 4 "so let's go ahead and".

@Bestape 3 жыл бұрын

Reminds me of a mix of Thales theorem and Einstein's Pythagorean proof.

@udic01 3 жыл бұрын

6:00 typo 12/(3+1)=3... Which is the length of AB

@user-ir8rv7zf6y 3 жыл бұрын

now we use the same argument that we did bef... 7:29 kinda fun

@victory159 3 жыл бұрын

Triangle AOC is similar to Triangle COB (and not OBC) correct? Also triangle AOC is similar to triangle ACB

@Parodiaseharlemshake 3 жыл бұрын

Hey Michael, do you have an e-mail for contact? I want to send you a really good problem

@markdenversorino4136 3 жыл бұрын

A perfect circle

@frozenmoon998 3 жыл бұрын

Uncut, has fails, but is legit

@zilongdu3064 3 жыл бұрын

Easier way to prove the general case: 1/AP+1/AQ = (AP+AQ)/(AP*AQ) = 2AO/AC^2=2/AB, where 2nd step we used power of a point (AC^2=AP*AQ), 3rd step we used similar triangles (AO/AC=AC/AB).

@TwilightBrawl59 3 жыл бұрын

1:55 Have you? 😛

@peterchan6082 3 жыл бұрын

1:55 . . . It looks like the corresponding vertices of the pair of similar triangles (∆OBC ~ ∆OCA) had been misaligned when you wrote ∆OBC ~ ∆AOC (! - - go vote) and then carried on to write OA/OC = OC/OB, which can be confusing to many viewers. It is good practice to properly align the corresponding vertices of similar/congruent triangles, Thus if you choose ∆OBC to start with, then write ∆OBC ~ ∆OCA (rather than ∆OBC ~ ∆AOC)

@rumbaallday5575 3 жыл бұрын

Minor type error at 5:57 12/(3+1)= 3 not 4

@OH-pc5jx 3 жыл бұрын

Wait a damn minute.... have these videos been a single take this whole time???

@user-A168 3 жыл бұрын

Good

@petek1365 3 жыл бұрын

07:24 That was clearly a 'bad place' to stop ;)

@billtensus 3 жыл бұрын

haha nice one. brother

@pattystomper1 3 жыл бұрын

I saw a 30-60 right triangle, because the hypotenuse is twice the length of the short side. So the hypotenuse is the radius of the circle (2), plus the line length outside the circle (2). 2+2=4. The short side is the radius of the circle (2). So now that we know the angle AOC (60 degrees), and the small triangle is a right triangle, Then we know the small triangle is another 30-60 right triangle. The hypotenuse of that triangle is the radius of the circle (2) The hypotenuse of a 30-60 triangle is twice the length of the short side. So the short side of the small triangle is (1). Subtract (1) from the radius at PO (2) and X = 1.

@SlidellRobotics 3 жыл бұрын

1:06 and then you take a radii? Anyway, I recognized the half equilateral triangle and the similar triangles during the description and knew the answer to part 1 by the time he finished stating the problem. Easiest. Penn. problem. ever. (before the generalization)

@The_Math_Enthusiast 3 жыл бұрын

Every problem in this channel is either too easy or too tough!!

@tt27footie52 3 жыл бұрын

You didn't tell that's a good place to stop.

@ramaprasadghosh717 3 жыл бұрын

think simple length of tangent t =√( 6*6-4*4)=√20 ∆ of edges t, (2+4=6) and r is similar to the ∆ of edges r and √(r*r +(r-x)*(r-x)) and r -x so r-x / r = r/ t or x = r-(r-x) = r- r*r/t = r*(t-r)/t Now r= 4 and t So x = 4-16/√20 = 4- 2/√5 = (20-2√5)/5

@harish6787 3 жыл бұрын

🔥

@kristianwichmann9996 3 жыл бұрын

Go vote!

@nickpower2623 3 жыл бұрын

Even more easy solution: AP=PO=OQ=2 PB=X BO= 2-X AB= 2+X Euclid’s theorem: CB^2= ABxBO Pythagorean theorem: CB^2= CO^2-BO^2 System: ABxBO=CO^2-BO^2 Substitute: (2+x)(2-X) = 4 - (2-X)^2 so: 4-X^2 = 4 -(4 - 4X + X^2) So: 4-X^2 = 4-4 +4X-4X^2 so: 4=4X so: X=1 It proves also that PC = CO because PCO is equilateral triangle

@renatolabita867 3 жыл бұрын

teorema tangente / secante

@Dilip077 3 жыл бұрын

Please solve more OLYMPIAD geometry problems.

@jabunapg1387 3 жыл бұрын

I just solved it with Pythagoras Theorem and now I feel bad after seeing your solution 😂

@PrashantSharma-nw6gc 3 жыл бұрын

I think that the best relation between 3,2 and 6 is 3*2=6

@pherriejamesbacolod6789 3 жыл бұрын

This is an application of geometric mean...

@nullplan01 3 жыл бұрын

Before watching: So, ACO is a right triangle with hypotenuse AO = 4, and leg CO = 2. From Pythagoras we therefore know AC = sqrt(12) = 2 sqrt(3). We know that the cosine of angle COA is 2/4 = 1/2, which means that the angle COA is 60°. But then angle OAC must be 30°. Changing views to triangle ABC, which is also right-angled, we see that cosine BAC = (2+x)/(2 sqrt(3)). Little algebra later and we see that x = 2 sqrt(3) cos(30°) - 2 = 2 sqrt(3) sqrt(3)/2 - 2 = 3-2 = 1.

@user-xs2sw7rd8n 3 жыл бұрын

12:50

@vanshaj5700 3 жыл бұрын

Hey Michael.... brother 12/4 isn't 4 man.. plz do correction

@Iomhar 2 жыл бұрын

And where is "And that's a good place to stop"?

@bourhinorc1421 3 жыл бұрын

what's happening, he was troubled and didn't finish with the "good place to stop"

@kenichimori8533 3 жыл бұрын

π = 3 = x Probability x equal 3 ratio half 2+1

@debdeepplaygamez4282 3 жыл бұрын

its Damn easy question.

@romar8665 3 жыл бұрын

АС^2=АР*AQ for any secant and tangent from the same point, try to prove it in your future video, good luck

@MichaelRothwell1 3 жыл бұрын

Similar triangles (again)

@grahck4391 3 жыл бұрын

He makes solving this overly complicated.

@LucaIlarioCarbonini 3 жыл бұрын

12 / (3 + 1) = GO VOTE!!!

@jessehammer123 3 жыл бұрын

AOC! Woo!

@benjaminbrat3922 3 жыл бұрын

Wait... there is something fishy. Isn't it a good place to stop? 12/4=3?

@faviomotta4202 3 жыл бұрын

there is a mistake in 5:59 in other situations those little mistakes make us flush the whole work down the toilet

@CTJ2619 Жыл бұрын

The harmonic mean should be 3 not 4 as written

@Polpaccio 3 жыл бұрын

13:05 that was not a good place to stop

@giuseppebassi7406 3 жыл бұрын

It was much easier: I solved this in very easy way just using the 1st euclid's theorem

@HDQuote 3 жыл бұрын

I found a much easier method for all you trigonometryheads :) (link to desmos graph: www.desmos.com/calculator/jq1h0kpdx7 ) 1. remember what the secant is: en.wikipedia.org/wiki/Trigonometric_functions#/media/File:Unit_Circle_Definitions_of_Six_Trigonometric_Functions.png 2. if we would zoom out so everything is half the size, we can use the unit circle and the point A is at (-2,0) 3. use w=arcsec(-2) to find the right angle for C 4. zoom in again by multiplying with 2. So C=2(cos(w),sin(w))

@serenapanicacci8935 3 жыл бұрын

But x is more easily equal to a/2

@stvp68 3 жыл бұрын

AOC!!!!!

@slimal1 3 жыл бұрын

You lost me at 11:59

@benheideveld4617 3 жыл бұрын

Radius, not radii

@copyrightfreemusicforall5563 3 жыл бұрын

Sir, plz give basics knowledge about IMO problems

@ttyagraj9554 3 жыл бұрын

Lets solve it as follows: Let AC=p, OC=2 (radius) ,BC= a ,OB=2-x Since tangents are perpendicular to the radius, angle ACO=90° By pythagoras theorem p^2 + 4=16 p=2root3 Again by pythagoras theorem, a^2 + (2+x)^2 = p^2 ----------(1) a^2 + (2--x)^2 = 4 ----------(2) On subtracting (1) and (2), 8x= 8 x=1 If I did something wrong then pls tell.