Exploration
How big can a cevian triangle be?
Three lines meet. Three side points make a triangle. Move the meeting point—and discover why a quarter is the most you can get.
Pick a point \(O\) inside a triangle \(ABC\). Draw a line from each vertex through \(O\), continuing to the opposite side. Call the three side points \(X,Y,Z\), and join them.
The green triangle \(XYZ\) moves as \(O\) moves. Sometimes it becomes surprisingly large. Sometimes it almost disappears.
How much of the original triangle can it occupy?
A segment from a vertex to the opposite side is a cevian. The three cevians here are concurrent: they meet at one point. Their side points form the cevian triangle. We require \(O\) to lie strictly inside \(ABC\).
Choosing arbitrary points on the three sides is a different problem. Here they must come from the same meeting point.
The answer: one quarter, and only at the midpoints
For every such configuration,
\[\frac{[XYZ]}{[ABC]}\leq\frac14.\]The brackets mean area. Equality holds exactly when \(X,Y,Z\) are the side midpoints, so \(O\) is the centroid. The midpoint triangle and the three corner triangles then all have the same area.
Try Show the maximum, then change the shape of the outer triangle. The answer survives stretching and slanting.
Here is the complete proof. The geometric part finds the area; one small inequality finishes the job.
1. Concurrency creates a balance
Measure the fractions of the three sides:
\[\begin{aligned} \alpha&=\frac{BX}{BC},\\ \beta&=\frac{CY}{CA},\\ \gamma&=\frac{AZ}{AB}. \end{aligned}\]All three lie between \(0\) and \(1\). Ceva’s condition is
\[\frac{\alpha}{1-\alpha}\, \frac{\beta}{1-\beta}\, \frac{\gamma}{1-\gamma}=1,\]or, equivalently,
\[\alpha\beta\gamma =(1-\alpha)(1-\beta)(1-\gamma).\]Why does concurrency give this condition? We can see it using just areas.
Three areas make the ratios balance
1.667 × 1.600 × 0.375 = 1
Put \(p=[OBC]\), \(q=[OCA]\), and \(r=[OAB]\). The triangles \(ABO\) and \(ACO\) share the base \(AO\). Replacing that base by \(AX\) multiplies both areas by the same factor. The triangles \(ABX\) and \(ACX\), in turn, have the same altitude to \(BC\). Therefore
\[\frac rq =\frac{[ABO]}{[ACO]} =\frac{[ABX]}{[ACX]} =\frac{BX}{XC}.\]Repeating the argument at the other two vertices gives
\[\frac{CY}{YA}=\frac pr,\qquad \frac{AZ}{ZB}=\frac qp.\]Multiplying these three ratios cancels \(p,q,r\) and gives \(1\). This proves the part of Ceva’s theorem we need.
2. Find the area by removing the corners
The inner triangle is what remains after subtracting the three corner triangles.
Subtract the three corners
20.98% of ABCThe three corner areas and the green area add to the whole triangle. Move O above to change this partition.
For example, \(AYZ\) and \(ABC\) share the angle at \(A\). Since a triangle’s area is half the product of two sides and the sine of their included angle,
\[\frac{[AYZ]}{[ABC]} =\frac{AY}{AC}\frac{AZ}{AB} =(1-\beta)\gamma.\]The same argument at \(B\) and \(C\) gives
\[\begin{aligned} \frac{[BZX]}{[ABC]}&=(1-\gamma)\alpha,\\ \frac{[CXY]}{[ABC]}&=(1-\alpha)\beta. \end{aligned}\]Let \(R=[XYZ]/[ABC]\). The four areas partition \(ABC\), so
\[\begin{aligned} R&=1-\gamma(1-\beta)\\ &\quad-\alpha(1-\gamma)-\beta(1-\alpha)\\ &=1-\alpha-\beta-\gamma\\ &\quad+\alpha\beta+\beta\gamma+\gamma\alpha. \end{aligned}\]Now expand the product from Ceva’s condition:
\[\begin{aligned} (1-\alpha)(1-\beta)(1-\gamma) &=R-\alpha\beta\gamma. \end{aligned}\]That product also equals \(\alpha\beta\gamma\). Hence the exact area formula is
\[\boxed{R=2\alpha\beta\gamma.}\]The factor 2 matters: at the midpoints, this gives \(2(1/2)^3=1/4\).
3. The small trick that proves the maximum
Use the ratios of the two pieces of each side:
\[\begin{aligned} a&=\frac{\alpha}{1-\alpha},\\ b&=\frac{\beta}{1-\beta},\\ c&=\frac{\gamma}{1-\gamma}. \end{aligned}\]They are positive, and concurrency says \(abc=1\). Solving for the side fractions gives
\[\begin{aligned} \alpha&=\frac{a}{1+a},\\ \beta&=\frac{b}{1+b},\\ \gamma&=\frac{c}{1+c}. \end{aligned}\]Substitute them into the area formula:
\[\begin{aligned} R&=\frac{2abc}{(1+a)(1+b)(1+c)}\\ &=\frac{2}{(1+a)(1+b)(1+c)}. \end{aligned}\]Now pair each ratio with \(1\). For any \(t>0\),
\[1+t-2\sqrt t=(\sqrt t-1)^2\geq0.\]Thus \(1+t\geq2\sqrt t\), with equality only when \(t=1\). This is the two-number arithmetic–geometric mean inequality, with its entire proof written in one line.
Multiply the inequalities for \(a,b,c\):
\[\begin{aligned} (1+a)(1+b)(1+c) &\geq8\sqrt{abc}\\ &=8. \end{aligned}\]Therefore
\[\boxed{R\leq\frac28=\frac14.}\]Equality requires all three squared gaps to vanish: \(a=b=c=1\). This means \(\alpha=\beta=\gamma=1/2\), so all three side points are midpoints. Conversely, the midpoints give equality. The proof is complete.
Every possible meeting point
There is nothing special about an equilateral triangle here. The formulas involve side fractions, so the outer triangle’s angles and side lengths have disappeared from the answer.
One dimension higher: a tetrahedron
A triangle has three vertices and three opposite sides. A tetrahedron has four vertices and four opposite triangular faces.
Choose an interior point \(O\). From each vertex \(A_i\), continue the line through \(O\) to the opposite face, reaching \(X_i\). The four contact points form an inner tetrahedron.
The largest possible volume is \(1/27\) of the outer volume, attained when the contact points are the centers of the four faces.
At this configuration, \(O\) is the centroid \(G\). Each face center is obtained from its opposite vertex by reflecting through \(G\) and shrinking lengths by a factor of \(1/3\). Volume therefore shrinks by \((1/3)^3=1/27\).
That explains the value at the candidate maximum. To prove that every other configuration is smaller, we need one more argument.
The complete proof in any dimension
An \(n\)-simplex is the higher-dimensional version of a triangle: it has \(n+1\) affinely independent vertices \(A_1,\ldots,A_{n+1}\). A triangle has \(n=2\); a tetrahedron has \(n=3\).
Choose an interior point \(O\), and let \(X_i\) be the intersection of \(A_iO\) with the opposite facet. For \(n\geq2\), the general result is
\[\boxed{ \frac{\operatorname{Vol}_n(X_1\cdots X_{n+1})} {\operatorname{Vol}_n(A_1\cdots A_{n+1})} \leq\frac{1}{n^n}. }\]Equality holds exactly when every contact point is the centroid of its opposite facet. In two dimensions, these facet centers are the side midpoints.
Describe the common point by weights
Every interior point has unique positive barycentric weights:
\[\begin{gathered} O=\sum_{i=1}^{n+1}w_iA_i,\\ w_i>0,\qquad \sum_iw_i=1. \end{gathered}\]This says that \(O\) is a weighted average of the vertices. Isolating the contribution from \(A_i\) gives
\[O=w_iA_i+(1-w_i)X_i,\]where
\[X_i=\sum_{j\ne i}\frac{w_j}{1-w_i}A_j.\]The coefficients in this last sum are positive and add to \(1\), so \(X_i\) is inside the opposite facet. The previous equation puts \(A_i,O,X_i\) on one line, with \(O\) between the endpoints. Thus these formulas describe every configuration in the theorem.
Calculate the volume ratio
Let \(P_A\) be the square matrix whose \(i\)th row contains the \(n\) coordinates of \(A_i\) followed by \(1\). Define \(P_X\) similarly. The coordinate formula above says
\[\begin{gathered} P_X=MP_A,\\[.5em] M_{ij}= \begin{cases} 0,&i=j,\\ \dfrac{w_j}{1-w_i},&i\ne j. \end{cases} \end{gathered}\]The volume of a simplex is the absolute value of this coordinate determinant divided by \(n!\). Taking determinants cancels both \(n!\) and the outer determinant, so the volume ratio is \(\lvert\det M\rvert\).
We can evaluate it without a long expansion. Let \(J\) be the \((n+1)\times(n+1)\) all-ones matrix. Multiplying row \(i\) of \(M\) by \(1-w_i\), then dividing column \(j\) by \(w_j\), leaves exactly \(J-I\). Therefore
\[|\det M| =|\det(J-I)|\, \frac{\prod_iw_i}{\prod_i(1-w_i)}.\]The matrix \(J-I\) multiplies the all-ones vector by \(n\). On the \(n\)-dimensional subspace of vectors whose components sum to zero, it multiplies by \(-1\). Its determinant is consequently \(n(-1)^n\). We obtain the exact formula
\[\boxed{ R_n=n\prod_{i=1}^{n+1}\frac{w_i}{1-w_i}. }\]Bound the denominator
For each \(i\), apply AM–GM to the other \(n\) weights:
\[\begin{aligned} 1-w_i&=\sum_{j\ne i}w_j\\ &\geq n\left(\prod_{j\ne i}w_j\right)^{1/n}. \end{aligned}\]Multiply all \(n+1\) inequalities. Each weight appears in exactly \(n\) of the products, each time with exponent \(1/n\). Its total exponent is therefore \(1\):
\[\prod_i(1-w_i)\geq n^{n+1}\prod_iw_i.\]Substituting into the exact volume formula finishes the bound:
\[R_n \leq\frac{n}{n^{n+1}} =\boxed{\frac{1}{n^n}}.\]For equality, every AM–GM comparison must be an equality. Since \(n\geq2\), this forces all the weights to be equal: \(w_i=1/(n+1)\). Conversely, equal weights make every inequality an equality, and \(X_i\) is then the average of the other vertices—the opposite facet’s centroid.
Indeed, writing \(G\) for the outer centroid,
\[X_i=G-\frac1n(A_i-G).\]So the maximizing simplex is reflected through its centroid and scaled by \(1/n\) in every direction. Its volume is exactly \(1/n^n\) of the original. This proves both the general bound and its equality case.
The one-dimensional case just swaps the two endpoints of a segment; its length ratio is always \(1\), so it has no unique maximizing meeting point.
The same balance in every dimension
The geometry lets us move a point freely. Concurrency ties the resulting side or face points together. Once that constraint is written down, balancing the weights makes the inner figure largest.
For triangles, the ceiling is a quarter. For tetrahedra, it is a twenty-seventh. In every dimension, the maximizing picture comes from the same choice: equal weights, meeting at the centroid.
For related background, see Steven Landy’s A Generalization of Ceva’s Theorem to Higher Dimensions, The American Mathematical Monthly 95 (1988), 936–939.