- #1
Albert1
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QR Code Length: Examining the Fixed Size of QR Codes
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SUMMARY
The length of QR in a QR code is fixed and independent of the position of point P. This conclusion is derived from geometric principles involving a circle with center C and diameter OP, where the radius is defined as half of r. The relationship QR = r sin(α) demonstrates that the length solely depends on the radius r and angle α, confirming that QR remains constant regardless of P's location.
PREREQUISITES- Understanding of basic geometric concepts, including circles and angles.
- Familiarity with trigonometric functions, specifically sine.
- Knowledge of coordinate geometry, particularly the properties of midpoints.
- Ability to interpret mathematical proofs and geometric relationships.
- Explore the properties of circles in geometry, focusing on diameters and midpoints.
- Study trigonometric identities and their applications in geometric proofs.
- Research the implications of fixed lengths in geometric constructions.
- Investigate the mathematical foundations of QR codes and their encoding mechanisms.
Mathematicians, geometry enthusiasts, educators teaching geometric principles, and anyone interested in the mathematical foundations of QR codes.
- #2
Opalg
Gold Member
MHB
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I assume that $A$, $B$ and $O$ are meant to be fixed, but that $P$ varies, and the aim is to show that the length $QR$ is independent of the position of $P$.
[sp]
Let $C$ be the midpoint of $OP$, let $r$ be the radius of the circle, and let $\alpha$ be the angle $AOB$. The circle with centre at $C$ and diameter $OP$ has radius $\frac12r$ and passes through $Q$ and $R$ (because of the right angles). The angle $QCR$ is $2\alpha$. By dropping a perpendicular from $C$ to $QR$, you see that $QR = 2CR\sin\alpha = r\sin\alpha.$ That depends only on $r$ and $\alpha$ and so is independent of the position of $P$.[/sp]
[sp]
Let $C$ be the midpoint of $OP$, let $r$ be the radius of the circle, and let $\alpha$ be the angle $AOB$. The circle with centre at $C$ and diameter $OP$ has radius $\frac12r$ and passes through $Q$ and $R$ (because of the right angles). The angle $QCR$ is $2\alpha$. By dropping a perpendicular from $C$ to $QR$, you see that $QR = 2CR\sin\alpha = r\sin\alpha.$ That depends only on $r$ and $\alpha$ and so is independent of the position of $P$.[/sp]
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- #3
Albert1
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Re: prove the length of QR is fixed
hint:Albert said:
from the diagram it is clear QR=AC (fixed)
can you figure it out ?
if not I will explain it
View attachment 2606
can you figure it out ?
if not I will explain it
View attachment 2606
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