# Largest Area Triangle in the Vesica Piscis

I can place any three points in or on a vesica piscis1. I wish to find the triangle of maximum area. I know the area of the vesica piscis is $$(\frac{2π}{3}-\frac{\sqrt{3}}{2})d^2$$ (where d is the radius of the circles used to form the vesica piscis) and I know the area of a triangle containing the two points of intersection and a cusp point or a triangle containing two cusp points and a point of intersection has area $$\frac{\sqrt{3}}{4}d^2$$. Getting a percentage for area coverage of these triangles gives ~35.3%. I believe that there is definitely bigger triangles.

What approaches can I take to find the triangle with the largest area (or if there are multiple, list them all)? I see this article here: Largest area of the triangle that can be inscribed in ellipse, but I am wondering if there is also a more elementary way to do it than to go the Lagrange Multiplier's route. I also see in the answer here (Do Any "Geometrical Inscription Theorems" Exist?) that the three points should lie on the boundary, but I don't know a general method to go about this

EDIT.

I'm inserting here a purely geometrical solution, the original reasoning can be seen at the end.

I'll repeatedly make use of the following result: if we have a line $$r$$ and an arc of circle $$\gamma$$, and the point $$Q\in\gamma$$ which is the farthest from $$r$$ is not an endpoint of $$\gamma$$, then the tangent at $$Q$$ is parallel to $$r$$.

The vesica piscis is composed of two arcs, with centers $$O$$ and $$O'$$. Let's consider the case when the vertices $$ABC$$ of an inscribed triangle are all internal points of the arcs: $$A$$ on arc $$O$$ and $$BC$$ on arc $$O'$$. If $$ABC$$ is the triangle of maximum area, then $$B$$ must be the point on the arc which is the farthest from line $$AC$$ and by the above result the tangent at $$B$$ must be parallel to $$AC$$, i.e. radius $$O'B$$ must be perpendicular to $$AC$$ and thus lies on an altitude of triangle $$ABC$$. By the same argument we also get $$O'C\perp AB$$, hence $$O'C$$ also lies on an altitude and $$O'$$ is the orthocenter of the triangle.

But we also have $$OA\perp BC$$, implying that altitude $$OA$$ must pass through $$O'$$, and that can happen only if $$A=O'$$. Triangle $$ABC$$ is then isosceles and right angled at $$A$$, as shown in figure below. Its area is: $$area_1={1\over2}d^2.$$

Let's now consider the case when one of the vertices (e.g. $$A$$) lies at an endpoint of the arcs, while $$B$$ and $$C$$ are internal, each on a different arc as in figure below (it's easy to show that we don't get maximum area if $$B$$ and $$C$$ lie on the same arc). Moreover, let $$A'$$ be the other endpoint and $$M$$ the midpoint of $$OO'$$ (see figure below).

For the area to be maximum we need as before $$O'B\perp AC$$ and $$OC\perp AB$$. If we set $$\alpha=\angle O'OC$$ we can notice that triangles $$OFD$$ and $$AFM$$ have both a right angle and a pair of equal vertical angles, hence $$\angle MAF=\angle DOF=\alpha$$. Moreover, $$\angle O'AC={1\over2}\angle O'OC={1\over2}\alpha$$, because they are an inscribed and central angle subtending the same arc in circle $$O'$$.

We can analogously set $$\alpha'=\angle OO'B$$ to find: $$\angle GAM=\alpha'$$ and $$\angle OAB={1\over2}\alpha'$$. But $$\angle OAM=\angle O'AM=30°$$, hence we have: $$\alpha+{1\over2}\alpha'=30° \quad\text{and}\quad \alpha'+{1\over2}\alpha=30°.$$ This system of equation can be easily solved to get: $$\alpha=\alpha'=20°.$$

Hence the largest triangle in this case is isosceles and symmetric about line $$AA'$$. To compute its sides, note that from $$\angle ABA'=120°$$ we get $$\angle BA'A=60°-\alpha$$ and $$AB=AC=2d\sin(60°-\alpha)$$. Its area is then $$area_2={1\over2}AB\cdot AC\cdot\sin(\alpha+\alpha')= {1\over2}\big(2d\sin(60°-\alpha)\big)^2\sin2\alpha= 2d^2\sin^340°\approx 0.53 d^2.$$

Finally, we can consider the case when two points (e.g. $$A$$ and $$B$$) lie on the endpoints of the arcs. In that case point $$C$$ must be either $$O$$ or $$O'$$, for the tangent at $$C$$ to be parallel to $$AB$$. The area of triangle $$ABC$$ is: $$area_1={1\over2}d^2\sin120°={\sqrt3\over4}d^2.$$

In summary, the largest area occurs in the second case.

ORIGINAL REASONING.

Assume WLOG that points $$A$$ and $$B$$ are chosen on the right side of the vesica, while $$C$$ is on the left side (obviously we cannot have maximum area if all three points lie on the same side). For given positions of $$A$$ and $$B$$ we get the maximum area if the tangent at $$C$$ is parallel to $$AB$$, i.e. if $$OC\perp AB$$, where $$O$$ is the center of the right circle (see figure below).

Hence if we choose $$C$$ at will and some point $$D$$ on $$OC$$, points $$A$$, $$B$$ of the largest triangle will be the intersections between the left circle (of center $$O'$$) and the perpendicular to $$OC$$ at $$D$$. Note that point $$D$$ must be close enough to $$O$$ so that point $$A$$ won't lie past the intersection point of the circle.

Let $$H$$ be the projection of $$O'$$ on $$AB$$ and $$d$$ the radius of the circles. If we set $$\alpha=\angle O'OC$$ and $$x=OD$$, then by triangle similitude we get: $$O'H:x=\left({d-{x\over\cos\alpha}}\right):{x\over\cos\alpha}$$ that is: $$O'H=d\cos\alpha-x.$$ We can then find base $$AB$$ of the triangle: $$AB=2\sqrt{d^2-O'H^2}=2\sqrt{d^2-(d\cos\alpha-x)^2}.$$ The area of triangle $$ABC$$ is then: $$area={1\over2}AB\cdot CD=(d-x)\sqrt{d^2-(d\cos\alpha-x)^2}.$$ If $$x$$ is kept constant we get the largest area when $$\cos\alpha$$ has the lowest possible value, i.e. when $$\alpha$$ has the greatest possible value (note that $$0\le\alpha\le60°$$ and $$\cos\alpha$$ is then positive and decreasing). But the largest possible value is attained by $$\alpha$$ when point $$A$$ lies at the intersection of the circles: in the following we can then study only that case.

Let's suppose then that $$A$$ lies at an intersection of the circles and name $$A'$$ the other intersection point (see figure below). Note that $$\angle A'AB=\alpha$$ and $$\angle ABA'=120°$$, so that $$\angle AA'B=60°-\alpha$$. We can then compute: $$AB=2d\sin(60°-\alpha).$$ From the equality $$AB=2\sqrt{d^2-(d\cos\alpha-x)^2}$$ we then get: $$x=d\cos\alpha-d\cos(60°-\alpha)$$ and we we can finally write an expression for the area as a function of $$\alpha$$: $$area={1\over2}AB\cdot (d-x)=d^2\sin(60°-\alpha) (1-\cos\alpha+\cos(60°-\alpha)).$$ Expanding, this can be rewritten as: $${area\over d^2}={\sqrt3\over2}\cos\alpha-{1\over2}\sin\alpha +\cos\alpha\sin\alpha-{\sqrt3\over4}.$$

Now we only need to show that the maximum of this expression is attained for $$\alpha=20°$$. Note that it vanishes for $$\alpha=60°$$ and $$\alpha=-120°$$, hence there must be an absolute maximum between those values, corresponding to a stationary point.

Differentiating the above expression with respect to $$\alpha$$ and equating the result to zero, we get the equation: $$2\cos^2\alpha-{1\over2}\cos\alpha-1={\sqrt3\over2}\sin\alpha.$$ Squaring both sides we obtain $$(2\cos^2\alpha-{1\over2}\cos\alpha-1)^2={3\over4}(1-\cos^2\alpha),$$ that is: $$16c^4-8c^3-12c^2+4c+1=0,$$ where I set $$c=\cos\alpha$$. This equation has $$c={1\over2}$$ as solution, which doesn't correspond to an acceptable value for $$\alpha$$. Hence we can divide the l.h.s. by $$2c-1$$ to get an equivalent cubic equation: $$8c^3-6c-1=0.$$ But from the cosine triplication formula: $$\cos3x=4\cos^3x-3\cos x$$ we get: $$4\cos^320°-3\cos20°=\cos60°={1\over2}$$ implying that $$c=\cos20°$$ is a solution of the above equation. The other two solutions can be shown to be negative, hence the only acceptable stationary point is $$\alpha=20°$$, as it was to be proved.

• Excellent work, quick question before I accept the answer. In the first paragraph, you say that the maximum is achieved when the tangent at C is parallel to AB. Why can we assume that? Commented Jul 9 at 4:05
• @WakkaTrout For in that case the distance from $C$ to $AB$ is maximum, as all the other points on the arc are inside the stripe formed by $AB$ with the parallel. But I've found a much simpler proof, I'll update my answer as soon as I can. Commented Jul 9 at 4:25
• @WakkaTrout I completely revised my answer, to give a purely geometric construction of the largest triangle. Let me know if something is not clear. Commented Jul 9 at 15:00
• I like how you now imply the location of B with alpha = alpha' which was not implied before (which actually shows Daniel was correct below) and how it a lot simpler. Nice job. Commented Jul 9 at 18:10
• Please have a look at the solution I just posted. Commented Jul 15 at 15:36

For triangles with an edge parallel to the line connecting the centers of the circles, the largest is shown in the image below. I expect this to be the largest in general.

• That is kind of what I was thinking too. Do you have the source for this still, I kind of want to play around with it to see if I can find other configurations with similar coverage Commented Jul 8 at 3:14
• @WakkaTrout It is simple enough to reconstruct: here you go. Three points are constrained to the arcs, one on the left and two on the right. Commented Jul 8 at 4:02
• Based on some offline use of Mathematica, this does indeed appear to be the largest area possible. Commented Jul 8 at 4:46
• Playing around with the geogebra file that @DanielMathias provided: with one of the points at the cusp, I could get the triangle to be at most 50% with some change. Only when the points were nearing the intersection points did it get larger. I am inclined to agree with this, wish I could give an upvote, but not enough reputation to do it. I do not want to accept it as the answer though as it still doesn't seem solved, but thanks! Commented Jul 8 at 4:57

Here is a solution in the "17th century spirit" where extremal solutions were found based on the computation of infinitesimal quantities.

I assume that we look for an optimal solution under the form of an isosceles triangle with its basis parallel to the line of centers.

Consider the following figure in the vicinity of an optimal solution $$\alpha$$ ; moving $$M$$ into $$M'$$ (i.e., changing $$\alpha$$ into $$\alpha + d\alpha$$) should give a "statu quo" : infinitesimal areas $$area(MAM')$$ and $$area(HMM'H')$$ should be the same (the loss on one side is balanced with the gain on the other), neglecting a second order area, of course.

Fig. 1 : The red and the blue infinitesimal areas must be equal when triangle $$ANM$$ has a maximal area, i.e., for the optimal $$\alpha$$.

First of all, the inscribed angle theorem gives

$$\text{angle}(MOM')=2 d\alpha\tag{1}$$

Moreover, with the notations of the figure :

$$\frac{x}{y}=\tan \alpha \tag{2}$$

giving

$$AM^2=x^2+y^2=y^2(1+(\tan \alpha)^2)=\frac{y^2}{(\cos \alpha)^2}\tag{3}$$

As a consequence :

$$area(MAM')=\tfrac12 AM^2 d\alpha = \frac{y^2}{2 (\cos \alpha)^2}d\alpha \tag{4}$$

Besides, considering $$HMM'H'$$ as a rectangle and using (2) :

$$area(HMM'H')=HM \times MM' \ \cos \alpha = x (2 d\alpha) \cos \alpha = 2y \sin \alpha d \alpha \tag{5}$$

Equating (4) and (5), we get :

$$y=4 \sin \alpha (\cos \alpha)^2 \tag{6}$$

Using (2) again, (6) yields :

$$x=4 \cos \alpha (\sin \alpha)^2 \tag{7}$$

But point $$M$$ with coordinates $$(x+\tfrac12,y-\tfrac{\sqrt{3}}{2})$$ belongs to the unit circle. Therefore :

$$(x+\tfrac12)^2+(y-\tfrac{\sqrt{3}}{2})^2=1$$

Otherwise said, using (6) and (7), we must have :

$$(4 \cos \alpha (\sin \alpha)^2 +\tfrac12)^2+(4 \sin \alpha (\cos \alpha)^2-\tfrac{\sqrt{3}}{2})^2=1\tag{8}$$

whose unique solution in the range $$[0, \tfrac{\pi}{4}]$$ can be shown to be :

$$\alpha = \tfrac{\pi}{9} = 20°\tag{9}$$

Proof of (9) : The expansion of (8) gives, once $$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$$ has been factored out :

$$\sin(2\alpha)=\frac{\sqrt{3}}{2}\cos(\alpha)-\frac12\sin(\alpha)$$

$$\sin(2\alpha)=\sin(\tfrac{\pi}{3})\cos(\alpha)-\cos(\tfrac{\pi}{3})\sin(\alpha)$$

$$\sin(2\alpha)=\sin(\tfrac{\pi}{3}-\alpha)$$

with a unique solution in $$(0,\tfrac{\pi}{4})$$ given by $$2\alpha=\tfrac{\pi}{3}-\alpha$$, i.e., $$\alpha=\tfrac{\pi}{9}$$.

• Very nice! (+1) Commented Jul 15 at 16:48
• That is very nice, will have to think about it. Great! Commented Jul 15 at 21:38
• Can we go backwards and show that if the equilateral triangle has half angle $20^{\circ}$, then the heights from the bases pass though the centers of the circles ? I like that the angle is a piece of $\pi$. Commented Jul 15 at 23:18
• @orangeskid Do you mean that the altitude issued for example from $M$ (in the case $M$ is in the optimal placement) passes through $O$ ? Commented Jul 16 at 20:07
• @JeanMarie: Yes. Now I realize that all can be done with angle chasing, even finding the angles. There is one type of critical triangle, a right one, vertex a center, and another one with vertex at a singular point ( other ones are not possible, since an orthocenter has to be one of the centers). Also now it seems we could solve the problem for the more general Vesica Pisci ( equal circles, but radius distance different), also with angle chasing. Will add a picture in my answer to clarify. Commented Jul 16 at 20:18

Consider a bounded and closed (i.e. compact) region in the plane, not contained in a line. There exists a triangle with largest area with vertices in the figure, $$\Delta ABC$$. Now if we keep $$B$$, $$C$$ fixed and vary $$A=X$$ we cannot increase the area of $$\Delta XBC$$. It follows that the parallel through $$A$$ to BC leaves the region to the side of $$BC$$. Same can be said for $$B$$, $$C$$. This means there exist a bigger triangle ( double the size) $$\Delta A'B'C'$$ such that the region is inscribed in it, touching the sides at the midpoints $$A$$, $$B$$, $$C$$. Note that critical triangle is not necessarily a triangle of largest area. Indeed, we could take the region a triangle $$\Delta P'Q'R'$$, and $$\Delta ABC$$ to be the midpoint triangle.

In the concrete case of the vesica piscis, assume that it is bounded by the circles centered at points $$(\pm 1, 0)$$ of radius $$2$$. We'll describe a critical triangle that might be of largest area. Consider $$\Delta A'B'C'$$ with vertices at points $$(\pm a, \sqrt{3})$$ and $$(0,- b)$$, where $$a$$ is the positive root of $$a^3+6 a^2-24$$ and $$b$$ is the smallest positive root of $$b^4-2 \sqrt{3} b^3-12 b^2+14 \sqrt{3} b+ 51$$. The midpoint triangle $$\Delta ABC$$ is critical and has area about $$2.125$$

$$\bf{Added:}$$

Jean Marie notified me of his beautiful solution. He shows that the isosceles triangle with angles $$40^{\circ}$$, $$70^{\circ}$$, $$70^{\circ}$$ with vertex at one angle point is critical (also shown by Intelligenti Pauca and Daniel Mathias). This is equivalent to: the heights from the base points pass through the centers of the circles, which can be showed using this older question of mine.

$$\bf{Added:}$$

There are two types of critical triangles. One is a right angles one, with vertex at one of the centers, the other is isosceles and has the vertex at one of the two singular points.

Let's see how we can prove that. First of all, one will see that if no vertex is singular, then the orthocenter must be one of the centers, hence it is of the first type. Now we analyse the case " one vertex is singular" . With Daniel Mathias, we denote the unknown angles by $$\alpha$$, $$\beta$$. Now the angles with red vertex are right, so we get the system

$$\alpha + \frac{\pi}{3} + \frac{\beta}{2} = \frac{\pi}{2} \\ \beta + \frac{\pi}{3} + \frac{\alpha}{2} = \frac{\pi}{2}$$

with unique solution : $$\alpha = \beta = \frac{\pi}{9}$$. Incidentally, this easy angle chasing also solves a problem with a quadrilateral with diagonal perpendicular mentioned about. Also, we can also find the two types of critical triangle for the generalized vesica pisci ( equal radiuses $$\ne$$ center distance).

• Please have a look at the solution I just posted. Commented Jul 15 at 15:35