Alhazen's problem is amathematical problem inoptics concerningreflection in aspherical mirror. It asks for the point in the mirror where one given point reflects to another.The special case of aconcave spherical mirror is also known asAlhazen's billiard problem, as it can be formulated equivalently as constructing a reflected path from one billiard ball to another on a circularbilliard table. Other equivalent formulations ask for the shortest path from one point to the other that touches the circle, or for anellipse that istangent to the circle and has the given points as its foci.

Although special cases of this problem were studied byPtolemy in the2nd century CE, it is named for the 11th-century Arab mathematicianAlhazen (Hasan Ibn al-Haytham), who formulated it more generally and presented a solution in hisBook of Optics. It has nostraightedge and compass construction; instead, al-Haytham and others includingChristiaan Huygens found solutions involving the intersection ofconic sections. According toRoberto Marcolongo,Leonardo da Vinci invented a mechanical device to solve the problem. Later mathematicians, starting withJack M. Elkin [de] in 1965, solved the problem algebraically as the solution to aquartic equation, and used this equation to prove the impossibility of solving the problem with straightedge and compass.

21st-century researchers have extended this problem and the methods used to solve it to mirrors of other shapes and tonon-Euclidean geometry, and have applied fast computational methods for its solution to modeling light reflection off thelakes of Titan.

The problem comprises drawing lines from two points, meeting at a third point on thecircumference (boundary) of a circle and making equal angles with thenormal at that point (specular reflection). It belongs togeometrical optics (in which light is modeled using rays rather than waves or particles), andcatoptrics, the use of mirrors to control light: it can be used to find the path of aray of light that starts at one point of space, is reflected from a spherical mirror, and passes through a second point. Although this is a three-dimensional problem, it can immediately be reduced to the two-dimensional problem of reflection in a circular mirror in the plane, because its solution lies entirely within the plane formed by the two points and the center of the sphere.^[1]

The same problem can be formulated with the two given points inside the circle instead of outside.^[1] For two points near each other within the circle, ingeneral position, there will be two solutions, but points that are farther apart have four solutions.^[2] Any solution describes the path of abilliards ball reflected within a circular billiards table,^[3][4] asLewis Carroll once suggested for billiards play.^[5] If the two segments of the reflection path are extended tochords of the circle, the two chords make equal angles to the circle and therefore have equal length. Thus, these chords form the two equal sides of anisosceles triangleinscribed within the circle, with the two given points on two sides of this triangle. Another equivalent form of Alhazen's problem asks to construct a triangle with these properties.^[3][6]

Another way of describing the problem, for points inside or outside the circle, is that it seeks anellipse having the two given points as its foci,tangent to the given circle. The point of tangency is the solution point to Alhazen's problem. A ray from one focus of the ellipse to this point of tangency will be reflected by the ellipse to the other focus (focus-to-focus reflection property). Furthermore, because the given circle has the same angle at the point of tangency, it will reflect the same ray in the same way. In an ellipse, all single-reflection paths from one focus to another have equal lengths, so the smallest ellipse tangent to the circle produces the shortest path from one given point to the circle and then to the other point.^[2][7] The idea that light rays follow shortest paths isHero's principle, later reformulated inFermat's principle that light rays follow quickest paths.^[8] More generally, asJames Gregory observed, for any analogous three-dimensional reflection problem, the point of reflection is also a point of tangency of anellipsoid having the source and destination of the reflected ray as its foci.^[9]

Ptolemy included the problem of reflection in a circular mirror in hisOptics (written in the second century CE), but was only able to solve certain special cases;^[10][11] al-Haytham formulated and solved the problem more generally.^[10] Al-Haytham was inspired by Ptolemy's work, and modeled his own book on Ptolemy's, but differed from it in important ways; for instance, Ptolemy used a model of visual perception in which visual rays travel outward from the eye to the objects it sees, while al-Haytham reversed this to the still-used model in which light rays travel inward from objects to the eye.^[12][13]

By the time ofPappus of Alexandria, in the 4th century AD, Greek mathematicians had categorized geometric solutions into three types:straightedge and compass constructions, constructions usingconic sections, andneusis constructions involving a marked ruler, preferring the earlier categories of solution over the later ones.^[14] Ibn al-Haytham's solution is of the second type, usinghyperbola, through which he develops a neusis construction.^[15][16] In his 1881 survey of the problem,Marcus Baker calls al-Haytham's solution "excessively prolix and intricate", and quotesIsaac Barrow as expressing a similar opinion.^[17] Later in the 11th century,Yusuf al-Mu'taman ibn Hud, a king of theTaifa of Zaragoza in Spain, simplified al-Haytham's lemmas somewhat, but did not make a significant advance on the problem.^[18] The work of al-Haytham became known in the rest of Europe through manuscript Latin translations in the 12th or 13th century, and a translation was published inBasel in 1572.^[15] Later geometric solutions byChristiaan Huygens,René-François de Sluse, andGuillaume de l'Hôpital used the same idea of an auxiliaryconic section: a hyperbola for Huygens, aparabola for Sluse,^[17][19][20] and both methods for l'Hôpital. Baker cites Huygens's solution as "the most elegant the problem has ever received".^[17]

In al-Haytham's solution, the hyperbola is used within a construction of the angle of reflection, after which the point of reflection is easy to find. Al-Haytham further subdivides the problem into cases, according to the number of reflected images (one for a convex mirror but up to four for a concave mirror), and solves each case separately.^[15][16] Instead, Huygens finds a hyperbola that directly solves the problem in all cases: the reflection points are points of intersection between this hyperbola and the given circle. This hyperbola can be characterized in many ways; one way involvesinversive geometry.^[21] Thelocus of points${\displaystyle L}$ at which the two lines to the given points cross, at equal angles, a circle concentric to the given one, is acubic curve containing both given points. The inversion of${\displaystyle L}$ through the given circle^[a] is arectangular hyperbola passing through the two points inverse to the given points and centered at the midpoint of the two inverse points. Its asymptotic lines are parallel to and perpendicular to theangle bisector of the angle subtended by the given points (or their inverses) at the center of the circle. The intersections of this hyperbola with the given circle include the desired solution point or points.^[21][22]

The number of solutions, for points inside the circle, can also be determined geometrically. In general terms, pairs of given points that are near each other within the circle, and near to the center of the circle, have two reflection points; pairs of points that are far apart and far from the center have four reflections. If one given point is fixed, the positions of the other point that produce two reflections are separated from the positions that produce four reflections by thecaustic generated by the reflections of a light source at the fixed point. On the caustic itself, away from its cusps, there are three reflections. At the cusps there are only two reflections.^[2]

The impossibility of a straightedge and compass solution was finally proven in 1965, using algebraic methods, by Jack M. Elkin (anactuary).^[23][10] A similar impossibility proof was rediscovered in 1997 byOxford mathematicianPeter M. Neumann.^[1][24] The neusis construction can also be carried out usingorigami folds following theHuzita–Hatori axioms,^[25][26] andRoger C. Alperin has argued through algebraic methods that the problem can be solved by straightedge, compass, andangle trisector, but without providing an explicit construction.^[27]

According toRoberto Marcolongo, a mechanical solution was presented byLeonardo da Vinci, after he failed to find a mathematical solution. The solution, as reconstructed by Marcolongo, takes the form of amechanical linkage that, when placed with its tip pinned to the circle center and the two given points allowed to slide along its arms, always maintains equal angles to these points at the hinge point of the two arms. Therefore, if the mechanism is moved in order to place this hinge point on the given circle, the solution will be obtained at this point.^{[28][29][30][31]}

Later mathematicians such asJames Gregory and many others attempted to find an algebraic solution to the problem, using various methods, includinganalytic methods of geometry and derivation bycomplex numbers.^[10][32][17] An algebraic solution to the problem was finally found in 1965 by Elkin, by means of aquartic polynomial.^[23]Other solutions were rediscovered later: in 1989, by Harald Riede;^[33] in 1990 (submitted in 1988), by Miller and Vegh;^[34]and in 1992, by John D. Smith^[10]and also by Jörg Waldvogel.^[22]

Waldvogel simplifies the algebra by formulating the problem for theunit circle and two given points${\displaystyle a}$ and${\displaystyle b}$ in thecomplex plane. With the aid of Huygens's hyperbola Waldvogel derives a quartic equation for the reflection point${\displaystyle \zeta }$,$${\displaystyle \overline{ab}{\zeta }^4-(\overline{a}+\overline{b}){\zeta }^3+(a+b)\zeta -ab=0,}$$ where${\displaystyle \overline{a}}$ and${\displaystyle \overline{b}}$ are thecomplex conjugates of${\displaystyle a}$ and${\displaystyle b}$.^[22] Someroots of this equation might not lie on the unit circle, or fail to give a valid reflection path, but the valid solutions can all be found among the roots. The root on the unit circle minimizing the total distance to the given points,${\displaystyle |a-\zeta |+|\zeta -b|}$, is always a valid solution.^[32] With some further manipulation this can be reformulated as an equation involving real numbers instead of complex numbers.^[22] Elkin instead formulates the problem in terms of thesquared Euclidean distances among the given points and the center of the given unit circle, and finds an "inelegant, asymmetrical" quartic equation for the squared distance from one given point to the reflection point, with combinations of the other squared distances as its coefficients.^[23]

The algebraic solution of this problem allows the use ofGalois theory to prove that, for certain easily constructed inputs, the solution point has coordinates that are notconstructible numbers, and therefore that the problem has no straightedge and compass solution.^[7][23][35] Elkin considers two given points with squared distances${\displaystyle \frac{1}{2}}$ and${\displaystyle \frac{1}{3}}$ from the center of a unit circle and${\displaystyle \frac{1}{2}}$ from each other, and uses theresolvent cubic of his algebraic solution to show that it is not constructible.^[23] Alternatively,Carréga & Haddad (2016) show that the two given points${\displaystyle (-\frac{1}{10},-\frac{1}{10})}$ and${\displaystyle (\frac{1}{6},-\frac{1}{6})}$ have reflection points on the unit circle whose coordinates come from roots of the polynomial${\displaystyle x^4-x-1}$, which has thesymmetric group on four elements as itsGalois group. It follows that the reflection point cannot be constructed with straightedge and compass. Carréga and Haddad generalize this example to pairs of rational points whose reflection points give the roots of anyStewart polynomial${\displaystyle x^4-rx-1}$ and show that this leads to unconstructible points whenever${\displaystyle r}$ is aprime number.^[35]

Isaac Barrow, in a set of lectures in 1669, used a trigonometric equation inpolar coordinates to describe thecubic curve${\displaystyle L}$ inverse to Huygens's hyperbola. It has the same property as the hyperbola: it intersects the given circle at the reflection point or points that solve the problem.^[10]

Gander & Gruntz (1992) derive a messy equation involving square roots of polynomials oftrigonometric functions for theangle around the circle of the reflection point. They suggest the use ofNewton's method to solve this equation numerically, but this involves thederivative of one side of the equation, and they write that finding this derivative explicitly is "certainly not the way to go". Instead, for this part they useautomatic differentiation.^[36]

For inputs at which the segment from one given point to the reflection point is perpendicular to the segment between the other given point and the circle center, the Huygens hyperboladegenerates to two lines. This degeneracy can lead certain numerical solutions of Alhazen's problem to become unstable near these inputs.^[37] Additional forms of instability can arise when the circle radius is much smaller than its distance to the two given points. To avoid these issues,Miller, Barnes & MacKenzie (2021) reformulate the problem as a trigonometric equation with, in general, 16 solutions, but for which it is possible to predetermine which of these solutions is the desired one. This leads to an iterative numerical solution that is fast and robust, which they apply inplanetary science to modeling light reflection off thelakes of Titan.^[38]

An iterative numerical solution is also possible based on either thelaw of reflection orFermat's principle.^[39]

Certain special cases admit simpler solutions. For two given points equidistant from the center of the circle, the reflection point or points occur where the circle is crossed by theperpendicular bisector of the two points. And for two given points that lie on a single diameter of the circle, there are one or two reflection points where this diameter crosses the circle.

As well, when the two points on the diameter are interior to the circle, there may be two more reflection points, where the given circle is crossed by anApollonian circle through the center of the circle. This circle is the locus of points whose ratio of distances to the two given points is constant. If the Apollonian circle crosses the given circle, reflection points occur at the crossings; however, it may remain entirely within the given circle, in which case the only reflection points are the endpoints of the diameter.^[23]

When the distance to the circle is much greater for one of the given points that it can be assumed practically infinite, a one-finite solution exists whose correct root branch can be determineda priori.^[38]

As well as spherical mirrors, al-Haytham also studied conical and cylindrical mirrors,^[15] which can be reduced in the same way as a spherical mirror to reflection in a circular mirror in the plane.^[37] Researchers have extended Alhazen's problem to general rotationally symmetricquadric mirrors, including hyperbolic, parabolic and elliptical mirrors.^[40] They showed that the mirror reflection point can be computed by solving an eighth-degree equation in the most general case. If the camera (eye) is placed on the axis of the mirror, the degree of the equation reduces tosix.^[41]

Alhazen's problem can also be extended to multiplerefractions from a spherical ball. Given a light source and a spherical ball of certainrefractive index, the closest point on the spherical ball where the light is refracted to the eye of the observer can be obtained by solving a tenth-degree equation.^[41]

Another direction for generalization is tonon-Euclidean geometry. In thehyperbolic plane, as in theEuclidean plane, it is not possible to solve the problem using only a straightedge and compass.^[6]

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