The Lunes of Alhazen

We look at a classic problem from Medieval Islamic Mathematics.
Playlist: • Geometry
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[@dieguenec]

I think this problem can be solved geometrically rather than algebraically. We know that the hypotenuse of the right triangle is the diameter of the circumscribed circunference. Also by Pythagorean theorem we know that the areas of the semicircles attached to the smaller sides of the triangle add up to the area of the semicircle attached to the hypotenuse (Prof Penn also proved it). So we see that the two moon shapes plus the two circular segments add up to the right triangle plus the two circular segments. Therefore the area of the right triangle is equal to the sum of the two moon shapes. It’s way easier to show graphically but I hope I explained myself well enough 😅

[@dramwertz4833]

Yep thought the same

[@idohalamit]

Yeah, the claim he spent half the video proving is a standard sentence in geometry

[@shivampandey2937]

Yes very easy geometrical proof

[@TheOneThreeSeven]

came here to say this

[@jasonwoolf]

kzbin.info/www/bejne/jp_JZaKdmr2Cg68

[@ayoubfenni2586]

رحمه الله تعالى هو عالم جليل وهو من افضل العلماء العرب في الرياضيات

[@propea6940]

When I saw ابن الهيثم I checked 3 times whether this is Michael Penn's channel or not ..

[@haithamelatrache1231]

I usually watch most of Michael's videos but this one was with a bit of a smile ;-)

[@propea6940]

@@haithamelatrache1231 Same here even though the problem isn't hard but I liked it 😂

[@waelatfah1903]

Same here 😂

[@lisandro73]

I read “lunes” and I thought about “monday”

[@haithamelatrache1231]

in french Monday is Lundi the day of moon which is also Moon Day thus monday

[@SlipperyTeeth]

Taking the scenic route: lunes → monday → moon-day → moon

[@cipher3966]

I finally understood one of these, and my approach was pretty close.

[@yanmich]

Βeing Greek, I have to mention that the Lunes of Alhazen is an extension of the Lune of Hippocrates, where a single lune matches the area of a 45-90-45 triangle inside a circular quadrant.

[@thezoz9476]

Being Greek doesn't have anything to do with that

[@Stelios2711]

@@thezoz9476 Seems like you didn't get why he said that.

[@thezoz9476]

@@Stelios2711 I think i don't , but I know hippocrates is Greek but there is no need to know that Yannis michos is greek

[@thezoz9476]

@@jakubskop73 ikr

[@Stelios2711]

@@thezoz9476 ah, OK, I see. Then you are right.

[@brettaspivey]

Oh this is way simpler: (area of semicircle AB) = (area of semicircle AC) + (area of semicircle CB) Then flip AB around, and it follows immediately

[@nakamakai5553]

A beautiful piece of work. Thanks. Just the way to start a Sunday

[@VibratorDefibrilator]

Wow, I was unfamiliar with this problem (very beautiful approach!). Gotta check out the history of mathematics once again. Thank you!

[@aliberro]

A big SALUT to my Muslim Brothers and Sisters around the world, to Michel Penn and to everybody who's reading this comment, you are amazing, don't forget to #Always_Smile :)

[@cernejr]

It is interesting that the area of the lunes have no Pi in it. Without doing the calculation that is not something I would expect.

[@mohamedahmedeltohfa5540]

Yes, there is always a Pi in circles.

[@Tentin.Quarantino]

Especially when pi has a knack for appearing in places it ostensibly doesn’t belong.

[@dimadima5298]

Thank you micheal شكرا مايكل Alkhawarizmi also is a Muslim Mathematician who solved 2nd degree equations

[@ZainAlAazizi]

His name is Al-Hassan ibn Al-Haythem, an Iraqi Muslim scholar from Basra. He was born in 965 and died in Cairo, Egypt in 1040. He was the father of Optics science and he was the first who mentioned the concept of “Camera”.

[@haithamelatrache1231]

and it happens I'm his father lol

[@nikosdemetriou4527]

Το πιο πάνω πρόβλημα λύεται πολύ πιο εύκολα με την βοήθεια μιας "εξίσωσης". Δηλ. Εμβ. ημικυκλίουABC+Εμβ. (a) + Εμβ.(b)=Εμβ. Τριγώνου ABC+Εμβ. ημικυκλίου AC+ Εμβ. ημικυκλίουBC. Εύκολα τα 2 μικρά ημικύκλια αποδυκνύεται ότι ισούνται με το μεγάλο ημικύκλιο. Διαγράφονται στην εξίσωση και αποδυκνύεται εύκολα.

[@geosalatast5715]

Your solution is much easier and quicker than mine!! well done!! sometimes you give us the shortcut showing that we dont need to make our heads hurt, and sometimes you give us a complex solution to show us interesting properties and ways to handle a problem!! I went to solve all the areas with respect to the radius of the big circle hahahaha

[@davidchung1697]

What a nice problem. I was actually about to calculate the areas of the small and large lunes, but then saw the video!

[@suhailawm]

You r always great

[@manucitomx]

I loved this.

[@jackhandma1011]

Good to see Medieval Islamic math on this channel. It's kinda underrated to be honest.

[@caladbolg8666]

On that note, it'd be nice to get some videos on the lunes of Hypocrates, and in general on some math history (like squaring the circle and other classical problems).

And trisecting an angle.

[@ethanyap8680]

This also became a Western Australian Junior Olympiad problem, was pleasantly surprised when I saw this.

[@iabervon]

I think it's just as easy to show that, if two points on a circle are connected to first any point on the circle and second the center, the second angle is twice the first (measuring them both in the same direction). A right angle then gives a straight line as a special case. The proof of this has the advantage of not having a diagram which tempts you to assume the result.

[@purusotamasahoo9687]

Who can go to 1000 years back to get a (excellent) maths question? Obviously- Michael xD

[@sidimohamedbenelmalih7133]

That's interesting

[@curedent6086]

This problem is a great classic in French mathematics textbooks. I don't think any French high school student could have escaped it.

[@ibrahimalmetwale3282]

Greeaaat!💜

[@FireSwordOfMagic]

Very nice solution

[@JobBouwman]

Just add the area of the two black lenses to both sides of the equation. Then you must prove that the summed area of the two smaller semicircles (with diameter a and b) is equal to the area of the large semicircle (with diameter c). And this follows directly from Pythagoras.

[@jackhandma1011]

Dude, are you the guy on Quora? I think it's more likely than having two different Job Bouwmans who are both good at math.

[@JobBouwman]

@@jackhandma1011 Yes indeed! And are you by any chance the famous Jack Ma?

[@JobBouwman]

PS: what's your profile on Quora?

[@jackhandma1011]

@@JobBouwman No lol. I'm not someone that rich and famous. I prefer not to tell my profile. I mostly just read answers anyway, but I do spend a lot of time on quora.

[@viniciusmoretti]

I know this problem as "Lunes of Hippocrates".

[@user-ht1vg5we2p]

yeah, michael penn just gave all the credit to a guy who found this problem and republished it a thousand years later

[@prammar1951]

@@user-ht1vg5we2p Lunes of alhazen is an extension to the lines of Hippocrates.

[@xspeedox]

The area of the big circle = sum of the two circles by Pythagoras (and pi*R^2 formula for area). The desired result follows.

[@tgx3529]

We can use that the triangle has array K=a*b/2, blue part P , red part N, the part L and M... Then N+M=pi b^2/8 P+L=pi a^2/8 K+L+M=pi c^2/8 L+M=pi c^2/8-ab/2 So N+M=pi*(b^2+a^2-c^2)/8+ab/2

[@liab-qc5sk]

The next lesson can be at-tosi circles

[@goodplacetostop2973]

10:37

[@study5133]

@@nullsharp4610 best books for practice math is ????

[@goodplacetostop2973]

@@nullsharp4610 People keep me asking about that so I’m trying to post homework every other day. Actually, I think I should take a real break from it. I should rack up a lot of homeworks in advance and then post every day.

[@mushfikaikfat]

@@goodplacetostop2973 Can you kindly tell me that who are you ?? Pls 🥺 I've a great desire to be known about you

[@KatariaGujjar]

Imagine doing all these calculations without a calculator

[@gderghi]

You relax me

[@buxeessingh2571]

If memory serves me right, if you create another lune by creating a pi/2 radian arc centered at the midpoint of the semi-circle and joining the endpoints of the hypotenuse, then the area of the third lune matches the area of the triangle. Allegedly, this was the origin of the attempt to "square a circle".

[@tomatrix7525]

This one was actually really interesting

[@user-ui7mr5vh6f]

Great work, I enjoy your channel a lot! I would be interested to see more of those old "classic" geometry problems, you know, which have been famous for hundreds of years. I've heard there are a lot of them in Japan, it would be nice to hear more about them. Thank you for your work!

[@Elouard]

From Wikipedia on the problem: (Alhazen explored what is now known as the Euclidean parallel postulate, the fifth postulate in Euclid's Elements, using a proof by contradiction,and in effect introducing the concept of motion into geometry. He formulated the Lambert quadrilateral, which Boris Abramovich Rozenfeld names the "Ibn al-Haytham-Lambert quadrilateral". In elementary geometry, Alhazen attempted to solve the problem of squaring the circle using the area of lunes (crescent shapes), but later gave up on the impossible task.The two lunes formed from a right triangle by erecting a semicircle on each of the triangle's sides, inward for the hypotenuse and outward for the other two sides, are known as the lunes of Alhazen; they have the same total area as the triangle itself.) Thank you Michael and please dig deep into such problem

[@doingmath5870]

Mr. Good vídeo. Maybe one day You solve and exercises relates to tensors. José from Nicaragua.

[@hybmnzz2658]

This way of doing geometry problems is better than uneccesary trigonometry and integration

[@lucastellez2558]

Could you prove the first claim with the inscribed angle theorem? If the inscribed angle is 90 degrees, then the central angle must be twice that, or 180, meaning the hypotenuse is a straight line passing through the center. Idk if this is logically sound or if I'm making some sort of mistake?

[@extensionsorbit7727]

This is fantastic. Thanks so much for all that you do.

[@SlipperyTeeth]

This is actually equivalent to the Pythagorean Theorem. If you add back on those portions bordering the legs of the triangle to both sides of the desired equation, then you have the statement that the sums of the areas of the smaller semicircles is the area of the larger semicircle - which is effectively the Pythagorean Theorem. This video goes into further detail kzbin.info/www/bejne/lYHcgYKOjbV8gLM

[@shatishankaryadav8428]

Area of shaded region is always Equal to base area

[@user-fp8qp6fh2g]

أنا من المغرب اتابعك واحب فيديوهاتك

[@JalebJay]

Through PIE this pie creates equal areas with the triangle and semi-pies. QED :)

[@GreenMeansGOF]

There is a theorem that says that it a triangle has one side as the diameter and the oposite corner on the circle, then that corner is a right angle. Does this mean that the converse is true in the sense that if we have a right angle, then the oposite side is a diameter?

[@carlosjuarez2686]

Function analysis from the second phase of OBM 2019 (question 4)?

[@vedants.vispute77]

Since angle BCA is 90°, ACB are collinear. Think of if BC isn't with A, it would not make 90°

[@txikitofandango]

I wouldn't have tried to show AOB is a straight angle. I would've proven that AB is a diameter. This can be achieved with the Inscribed Angle Theorem, which doesn't rely on measurement of angle, and thus more in line with Euclid, for what that's worth. I think it's more elegant, since angles are a problematic idea in pure geometry.

[@engrvarsi3774]

Great

[@DavidSavinainen]

Let’s call the area between the triangle and the blue lune M, and the area between the triangle and the red lune N. The area of the triangle is the same as half of the large circle, except for M and N, so in other words π/2 c² - (M+N) The areas of the lunes is πa²/2+πb²/2 - (M+N) = = π/2 (a²+b²) - (M+N) But since a and b are the sides of a right triangle, using Pythagoras, π/2 (a²+b²) - (M+N) = π/2 c² - (M+N) which is the same as the triangle’s area. It’s essentially the same approach as Mr Penn, but less adding-subtracting back and forth.

[@lawlietl1370]

What's the area of lunes construct on hypotenuse?

[@TheMrbrayn]

Maybe a little pedantic but wouldn't you also have to show that O lies on the line AB in order for the colinear argument to be sound?

[@CharIie83]

its like a pythagoras but with circles

[@MrRyanroberson1]

You know... shouldn't this therefore translate into some kind of lune theorem where you can convert lunes into right triangles?

[@user-A168]

good

[@barbietripping]

Big circle + lunes = 3 half circles + triangle, right?

[@athysw.e.9562]

I don't often comment your videos but I find this problem very pretty. It's even more as soon as you realize it's 1000 years old.

[@genalkali1779]

I think you don’t really need to calculate the area of the triangle at all for the answer

[@tonyhaddad1394]

That a very easy problem but very beutiful

[@mahmoudqaddoumi28]

Mashallah

[@titassamanta6885]

This is similar to lune of hippocrates. (Sorry if I have done any spelling mistake)

[@jkid1134]

Very pretty chalk

[@user-ht1vg5we2p]

Hippocrates: Works hard to prove this case. Michael Penn: gives the credit to a guy who lived a thousand years later and just found and republished the discovery. Hippocrates: Am I a joke to you?

[@prammar1951]

This is different from Hippocrates' version, you can check yourself.

[@jrkirby93]

3:11 woah, getting political now? That's a first.

[@goodplacetostop2973]

Actually, kzbin.info/www/bejne/Zn3WZYZunLtpsMU

[@frozenmoon998]

Lune=luna=moon.

[@tonyhaddad1394]

Lebanon adds .im from lebanon 😊

[@robert-skibelo]

I thought it really nice that you want to prove red and blue make purple. Because of course they do.

[@bluepine384]

👌

[@sidimohamedbenelmalih7133]

Can we find what represant the bleu area or the red area in the purple triangle?🤔🤔🤨

[@Gauteamus]

Interesting if we could find geometric representations of the red and blue lunes as partitions of the purple triangle. The partition is NOT divided by the altitude from the point C to the diameter c.

[@sidimohamedbenelmalih7133]

@@GauteamusSo what do you think 🤔 can we find it . I think there is a problem because the red area is a multiple of π so I don't think that we can turn it to a small triangle and that make the repartition of the red and the bleu area in the purple triangle is separed by a curve.

[@Gauteamus]

@@sidimohamedbenelmalih7133 Hmm, let's say we find some point D by construction with compass and ruler (putting some constraints on the possible x,y-coords of D?). Trace the line from C to D and let this line partition the right angle triangle ABC. Can a point D in the constructibles give rise to a "linear partition" of the triangle with partitioned areas dependent on pi? If not, can we neatly partition the triangle by arcs of circles?

[@jamirimaj6880]

10:17 You mean the GOUGO THEOREM, right?!

[@alejandrolagunes5697]

Oh no, Presh Talwalkar from MindYourDecisions did it again😂

[@user-ez4ol2ri4h]

I am arabic, and proud 🌝❤️❤️

[@haithamelatrache1231]

Hey !! that's my name here ! he he ;-)

[@SakiKnin83]

Thales

[@vedants.vispute77]

We want a 'Who Am I' video of Michael Penn.

[@mhmdsalhab8254]

🙏👍

[@mohamedbabe7324]

شكرا علة ذكره 🤣🤣

[@sidimohamedbenelmalih7133]

لو ذكر الإسم بالإنجليزية فقط لما عرفنا أنه ابن الهيثم

[@omarhassan2124]

@@sidimohamedbenelmalih7133 نعم

[@taransingh5026]

micheal 's video titles are way better then his problems...............

[@random-td7tf]

Pretty easy

[@sayansircar360]

Sir please make a lecture series on ancient Indian geometry and mathematics.

[@anubratasaha4367]

kzbin.info/www/bejne/aHnQfq2ej52tgMk

[@raystinger6261]

5:04 - That's circular logic IMO. I'd say don't assume the value of angle C (angle of the vertice C), but we know that it is equal to alpha+beta. We also know that: (180-2*alpha)+(180-2*beta) = 180 => 180 - 2*alpha - 2*beta = 0 => 2*alpha + 2*beta = 180 => alpha + beta = 90 Since angle C is alpha+beta, therefore angle C is 90°, thus triangle ABC is a right triangle. Not that Prof. Penn got it that wrong, but he presented the explanation wrong.

[@omarsamraxyz]

Wow thank you bro doing something from Islam! ❤️👏

[@Stelios2711]

Lines Of Hippocrates, 1500 years older than this problem itself?...

[@user-gx6jm2ik8b]

I found Michael Penn's youtube channel. And that's a good place to stop.

[@musaburakerdihan5148]

Proof, just in case :)

[@lucho2868]

Lunes means Monday in spanish. I hate lunes.

[@Shacontrol]

and it means Moon in arabic.

[@lucho2868]

@@Shacontrol In fact in spanish, Monday = Lunes from Luna (Moon) Tuesday = Martes from Marte (Mars) Wednesday = Miércoles from Mercurio (Mercury) Thursday = Jueves from Jupiter Friday = Viernes from Venus ---- Saturday and Sunday are called Sábado (from the Sabath) and Domingo (from latin Domine, Christian God). ---- IRONICALLY the english language completes our unfinished puzzle with the words derived from "Saturn day" and "Sun day".

[@Shacontrol]

@@lucho2868 THANK YOU

[@zlatkodekanic6784]

Why is this made to look so complicated? Small semicircle + medium semicircle = big semicircle (Pythagorean theorem) Two smaller lunes = triangle + small semicircle + medium semicircle - big semicircle = triangle = ab/2. 15 seconds.

[@abezworld9435]

I think that’s what he did. But correct n me if I’m wrong he first he has to prove that ab is a diameter not a cord of the big circle.

[@zlatkodekanic6784]

Valid comment. My thinking is that people following this channel know that the center of the big circle is in the middle of the hypotenuse. Also, wanted to point out that no calculation was necessary.

[@lovingphysics5865]

#ignitedphysics

[@robertlynch7520]

I, not being very good at geometry, but not too bad with trigonometry, and PERL ... did it this way: #!/usr/bin/perl; # include your favorite trig library if needed, such as 'use Math::Trig;' my $a = 12; my $b = 5; my $c = sqrt( $a ** 2 + $b ** 2 ); pp( 'a', $a, "" ); pp( 'b', $b, "" ); pp( 'c', $c, "" ); my $zeta = asin( $b / $c ); # this is the angle at the TOP of the triangle. my $phi = 2 * $zeta; # this is at the bottom... my $theta = pi - $phi; # and this is the one of the left, the bisecting angle. pp( 'zeta', $zeta, "rad" ); pp( 'phi = 2 zeta', $phi, "rad" ); pp( 'theta = pi - phi', $theta, "rad" ); my $oa = (1/2 * $a) * $b/$a; my $ob = (1/2 * $a); pp( 'oa', $oa ); pp( 'ob', $ob ); my $lensa = 1/2 * $theta * ($c/2)**2 - (1/2 * $a * $oa); my $lensb = 1/2 * $phi * ($c/2)**2 - (1/2 * $b * $ob); pp( 'lens a', $lensa ); pp( 'lens b', $lensb ); my $la = 1/2 * pi * ($a / 2)**2 - $lensa; my $lb = 1/2 * pi * ($b / 2)**2 - $lensb; my $lunes = $la + $lb; my $tri = 1/2 * $a * $b; pp( 'lunes', $lunes ); pp( 'triangle', $tri ); Which GIVES the output... (I have a 'pp()' function that does 'pretty printing') ---------------------------------------------------------------------------------------------------- a = 12.000000 b = 5.000000 c = 13.000000 zeta = 0.394791 rad 22.62 deg phi = 2 zeta = 0.789582 rad 45.24 deg theta = pi - phi = 2.352010 rad 134.76 deg oa = 2.500000 ob = 6.000000 lens a = 34.686220 lens b = 1.679925 lunes = 30.000000 triangle = 30.000000 A HAH! They're the same. So, yah... it works by straight computation asl well (without any surprise). However, I liked the challenge of working this out trigonometrically BEFORE watching your video. By comparison, I worked pretty stupidly to find the result, without your clever math. Ah, well ... professional mathematicians are just wicked-cool crazy dudes! GoatGuy

[@ramtindorostkar765]

islamic? kidding me? Ebn_el_heisam was an iranian mathematician as he said in his books.... پشمام

[@manjumanl5279]

This is a 522 year old maths lesson .

[@user-jk4qb2nc3s]

من هم وطن ابن هیثم هستم

[@HamzaBaqoushi]

كفى من نشر التخلف بأسم الإسلام. الإسلام بريء من الحركات الإسلامية، مثل تلك التي تحكم في إيران. العلم للبشرية جمعاء و العلماء ينتمون إلى الإنسانية و ليس إلى الحكومات المتخلفة.

[@bemusedindian8571]

The person after whom this is named is famous for insisting on the scientific method to explain natural phenomena in 1000 AD. Father of modern optics. One of the old greats.

[@samharper5881]

1:10 Shoutout to my fellow colour-blind friends who wouldn't have seen what he did there even given another 1000 years of looking. Especially the ones who aren't triggered by the u in colour.

[@loneranger4282]

12 dislikes from arab haters

[@sumukhshankarhegde2853]

10:37 is good place to stop

[@suhailawm]

You r always great