Can you calculate the X value? | (Pitot's Theorem with Proof) | #math #maths | #geometry

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Can you calculate the X value? | (Pitot's Theorem with Proof) | #math #maths | #geometry

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[shaozheang]

You can solve using simultaneous equations. Thanks for the new method!

[Birol]

My way of solution ▶
By considering the Deltoids:
D(EAFO), D(HDEO), D(GCHO) and D(FBGO)
we can see that:
[FA]= [AE]
[ED] =[DH]
[HC] = [CG]
[GB] = [BF]

Let's say [AE]=x
[FA]= x
[ED]= 21-x
[DH]= 21-x
[HC]= 18-(21-x)
[HC]= x-3
[CG]= x-3
[GB]= 12-(x-3)
[GB]= 15-x
[BF]= 15-x
⇒
[AB]= [FA]+[BF]
[AB]= x+15-x
[AB]= 15

[stanbest]

Thanks, I enjoyed that. What a neat theorem.

[hortalextf]

Every day I'm looking more than 1 of your videos. You explain so good, sometimes too slow, but better slow than too fast. I try to resolve the problems by myself, but in many cases I'm not sucessful.
And there are so many new theorem. I never heard of Pitot's theorem, so I learnd something new today.
Google tells me that Henri Pitot is a French engineer from the 19th century.
Thanks for your awesome lessons.

[stephenbrand]

Thankyou for this video. I had never heard of Pitots theorem before and would be interested in seeing a proof of the general case rather than this specific scenario. I’m sure it will be on the internet somewhere.

[RAG]

I just did it the way you proved the theorem. Never heard of it.

[unknownidentity]

Let's find x:
.
..
...
....


According to the two tangent theorem we know:

AE = AF
BF = BG
CG = CH
DH = DE

Therefore we can conclude:

AB + CD
= AF + BF + CH + DH
= AE + BG + CG + DE
= AE + DE + BG + CG
= AD + BC

AB + CD = AD + BC
x + 18 = 21 + 12 = 33
⇒ x = 33 − 18 = 15

Best regards from Germany

[michaeldoerr]

The x = 15. I glad that have learned that some proofs are so simple that they require the use of one circle theorem!!! Please make a playlist of geometric proofs slash problems with or without circle theorems!!!

[LuisdeBritoCamacho]

The Pitot Theorem in Geometry states that in a Tangential Quadrilateral the two pairs of opposite sides have the same Total Length.
01) (AD + BC) = (AB +CD)

02) (AD + BC) = (21 + 12) = 33

03) (AB + CD) = (18 + X) = 33

04) 18 + X = 33 ; X = 33 - 18 ; X = 15

ANSWER : X = 15

IMPORTANT NOTE : I am always willing to go further, to learn more, in the Mathematics Field. Thank you for this opportunity.

[hortalextf]

It's crazy how many theorems exist, it is endless!
Looking for "Pitot's theorem", I found the "Japanese theorem for cyclic quadrilaterals" and the "Carnot's theorem".

[MrPaulc]

Looking up Pitot's theorem, it appears that x = 15 due to 21 + 12 = 18 + x, but it's the proof that is of interest due to its employment of two-tangent theorem .I looked up the history and was surprised that Pitot's theorem was not until 1725, but I imagine ancient mathematicians figured it out much earlier as it seems closely allied with two-tangent.

[ВалерийЛещук-ъд]

a and a ; 18-a and 18-a ; a-6 and a-6 ; 21-a and 21 -a ; 21-a + a-6 =15

[gelbkehlchen]

Solution:
Due to the equality of the tangent segments, the following applies:
(1) FB = BG
(2) CG = CH
(3) DH = DE
(4) AE = AF ⟹
AF+FB+CH+DH = AE+BG+CG+DE ⟹
AB+CD = AE+DE+BG+CG = AD+BC ⟹
x+18 = 21+12 |-18 ⟹
x = 21+12-18 = 15

[santiagoarosam]

Perímetro =12+18+21+x 3+a+12-a=15=x.
Gracias y saludos.

[himo]

AF=AE=j DE=DH=k CH=CG=l BF=BG=m
j+k=21 k+l=18 l+m=12 j+m=x
j+k+l+m=33 j+k+l+m=18+x 33=18+x x=15

[antoinegrassi]

Une petite suggestion. Plutôt que de faire la démonstration avec des 21 - a, des a - 3, n'était-il pas plus simple et plus parlant de faire deux choses:
1) sur le dessin de mettre des petites encoches ( l, ll, lll, lV) sur les tangentes qui ont la même longueur (DE = DH...), on voit tout de suite que les sommes des côtés opposés vont être egales, et cette image à elle seule suffira à retenir la démonstration.
2) de le démontrer en déroulant simplement le calcul en remplaçant successivement les longueurs égales . C'est-à-dire:
AD + CB = (AE + ED) + ( CG + GB) = AF + HD + CH + FB
= (AF + FB) + ( CH + HD) = AB + CD. CQFD.
Les démonstrations lourdes n'aiguisent pas les esprits et sont mal memorisées.
Vive les maths. 👍

[sergeyvinns]

RTSSIA! Это легко, если обозначить, к примеру, CG=y, DG=FB=12-y, CH=CG=y, DH=18-y=DE, AE=AF=21-18+y, =x-12+y,
3=x-12, x=15. Забыл добавить. что привёл доказательство того. что сумма противоположных сторон четырёхугольника. в который вписана окружнось. величина дпостоянная, АD+ВС=CD+AB. Формула для х, выглядит следующим образом: х=AD+BC-DC=21+12-18=15.

[sandhyaprabhu]

If you are the teacher there won't be a single student who can fail.

[wackojacko]

@ 7:19 Frick'n a! ...That's French for Right on Jack! Henri Pitot would know what I'm talkin about . 🙂

[marcgriselhubert]

Let a = AF = AE; b = BF = BG; c = CG = CH; d = DH = DE; then AD + BC = a + b + c + d = AB + DC (Pitot's theorem). So here 21 + 12 = x + 18, and so x = 15.