Visual proofs and triangle isosceles

From: 3blue1brown

This article explores several “fake proofs” that demonstrate the limitations of visual proofs and the importance of mathematical rigor. These examples highlight the dangers of hidden assumptions and the subtleties of limiting arguments in mathematics [00:00:04].

Fake Proof: Surface Area of a Sphere = ${\pi }^2{R}^2$

This “proof” attempts to derive the formula for the surface area of a sphere [00:00:11].

The Argument

The method involves subdividing a sphere into vertical slices, similar to an orange [00:00:14]. These “orange wedges” from the northern hemisphere are unraveled to poke upwards, and those from the southern hemisphere are unraveled downwards [00:00:22]. By interlacing these pieces, the goal is to form a shape whose area can be easily calculated [00:00:30].

The base of this combined shape is derived from the circumference of the sphere (an unraveled equator), giving a length of $2\pi R$ [00:00:36]. The “height” of each wedge is considered a quarter of a walk around the sphere, leading to a length of $\pi /2R$ [00:00:45].

The claim is that as the sphere is divided into finer and finer slices, this shape approximates a perfect rectangle [00:00:57]. The area of this “rectangle” would then be $(\pi /2R)\times (2\pi R)$, which simplifies to ${\pi }^2{R}^2$ [00:01:07].

The Flaw

The proof is “completely wrong” [00:01:24]; the true surface area of a sphere is $4\pi R^2$ [00:01:27].

The main issue is that when the orange wedges are flattened, they do not retain their triangular shape if their area is to be preserved; they should bulge outward [00:11:01]. The width across a wedge varies non-linearly (according to a sine curve) as one moves from the top to the bottom [00:12:14]. When interlacing these non-linear edges, there is a persistent overlap that is not eliminated by taking finer subdivisions [00:12:31]. This overlap accounts for the difference between the incorrect ${\pi }^2{R}^2$ and the correct $4\pi R^2$ [00:12:43].

This false proof is similar in spirit to a valid visual proof for the area of a circle where pizza wedges are rearranged into a rectangle with area $\pi R^2$ [00:10:08]. The difference lies in the geometry of curved surfaces versus flat space, governed by concepts like Gaussian curvature [00:14:29]. Flattening a sphere’s parts without losing geometric information is impossible [00:14:36]. Visual proofs involving curved surfaces only work if the limiting pieces get smaller in both directions, appearing locally flat [00:14:48].

Fake Proof: $\pi =4$

This “classic argument” attempts to prove that $\pi$ equals 4 [00:01:42].

The Argument

Start with a circle of radius 1 [00:01:45]. Draw a square whose side lengths are tangent to the circle [00:01:56]. The perimeter of this square is 8 [00:02:00].

A sequence of curves is then generated by iteratively folding in the corners of the previous shape so they “just barely kiss the circle” [00:02:21]. At each fold, the perimeter of the shape remains 8 [00:02:31]. The argument claims that as the number of iterations approaches infinity, this sequence of curves will more and more closely approximate the circle [00:02:10], implying the circumference of the circle (which is $2\pi R=2\pi$) is 8, thus $\pi =4$.

The claim is that the limit of these approximations equals the circle, not just approximates it [00:03:13]. If we parameterize each curve ($c_n(t)$) [00:03:43], the limit of the points ($c_{\infty }(t)={\lim }_{n\to \infty }{c}_n(t)$) is the genuine smooth circular curve [00:04:34]. Also, the limit of the lengths of these curves truly is 8, since each individual curve has a perimeter of 8 [00:04:49].

The Flaw

The fundamental issue is that “there is no reason to expect that the limit of the lengths of the curves is the same as the length of the limits of the curves” [00:05:25]. This example serves as a counter-example [00:05:33].

Even though the sequence of curves visually converges to the circle, and each approximation has a perimeter of 8, the limiting process for the length is not equivalent to the limiting process for the curve itself [00:05:25]. This highlights why careful definition and handling of limits are crucial in mathematical proofs, particularly in calculus when approximating areas or lengths [00:15:09].

Fake Proof: All Triangles are Isosceles

This “Euclid-style proof” aims to demonstrate that any given triangle must be isosceles [00:05:38].

The Argument

Given any triangle ABC, the goal is to prove that side AB is equal to side AC [00:05:50].

1. Draw the perpendicular bisector of side BC. Let its intersection with BC be D [00:06:15].
2. Draw the angle bisector of angle A. Let its intersection with the perpendicular bisector be P [00:06:28].
3. Draw a line from P perpendicular to AC, intersecting at E [00:06:48].
4. Draw a line from P perpendicular to AB, intersecting at F [00:06:57].

The proof then uses congruence relations between triangles:

• Triangle AFP is congruent to triangle AEP (Angle-Side-Angle: AP shared, angles at A are $\alpha$, angles at F and E are 90 degrees) [00:07:06]. This implies AF = AE [00:08:42].
• Draw lines PB and PC [00:07:34]. Triangle PDB is congruent to triangle PDC (Side-Angle-Side: PD shared, BD = DC, angles at D are 90 degrees) [00:07:38]. This implies PB = PC [00:08:04].
• Triangle PFB is congruent to triangle PEC (Side-Side-Angle: PB = PC, PF = PE (from first congruence), angles at F and E are 90 degrees) [00:08:13]. This implies FB = EC [00:08:51].

Finally, by adding these lengths: AF + FB = AE + EC [00:09:01]. Since AF + FB = AB and AE + EC = AC, it follows that AB = AC [00:09:06]. As no assumptions were made about the triangle, this implies any triangle is isosceles, or even equilateral [00:09:21].

The Flaw

The result is “obviously false” [00:05:57]. The flaw is “very subtle” [00:06:10].

The mistake lies in the assumption that the point E (and F) must lie between A and C (or A and B respectively) [00:17:32]. When the angle bisector and perpendicular bisector are constructed accurately, their intersection point P often lies outside the triangle [00:17:08]. If P is outside, then E (and F) can also lie outside the segments AC (or AB) [00:17:43].

For example, if E lies beyond C, then AC would be AE - EC, not AE + EC. The diagram used in the “proof” implicitly assumes the ideal configuration, but this does not hold for all triangles. All the triangle congruences are genuinely true [00:17:22]; the breakdown occurs in the final step of adding segment lengths based on an unstated positional assumption.

Conclusion

These examples underscore that while visual intuition and diagrams can be helpful for understanding, they cannot replace critical thinking and rigorous mathematical proofs [00:17:51]. It is essential to look out for hidden assumptions and edge cases, especially when dealing with limiting arguments or geometric constructions [00:18:06].