Math Core

Lines from the vertices of a triangle to the opposite sides (called cevians) show up all over the AIME. Two classical theorems tell you exactly how the pieces of the sides relate, and mass points turn those theorems into quick arithmetic. Once you have these tools, a whole family of "find the ratio" and "find the area" problems becomes routine.

Ceva's theorem

Let $D$ , $E$ , $F$ be points on sides $BC$ , $CA$ , $AB$ of triangle $ABC$ . The cevians $AD$ , $BE$ , $CF$ are concurrent exactly when a product of ratios equals $1$ .

Ceva's theorem

Cevians $AD$ , $BE$ , $CF$ of triangle $ABC$ meet at one point if and only if

$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.$

Read it by walking around the triangle: $A \to F \to B \to D \to C \to E \to A$ .

Proof (areas). Suppose the cevians meet at $P$ . Triangles with the same height have areas in the ratio of their bases, so

$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[PBD]}{[PDC]}.$

If $\dfrac{x}{y} = \dfrac{u}{v}$ , then also $\dfrac{x}{y} = \dfrac{x - u}{y - v}$ . Subtracting,

$\frac{BD}{DC} = \frac{[ABD] - [PBD]}{[ADC] - [PDC]} = \frac{[ABP]}{[CAP]}.$

In the same way, $\dfrac{CE}{EA} = \dfrac{[BCP]}{[ABP]}$ and $\dfrac{AF}{FB} = \dfrac{[CAP]}{[BCP]}$ . Multiply the three: everything cancels and the product is $1$ .

For the converse, suppose the product is $1$ . Let $AD$ and $BE$ meet at $P$ , and let line $CP$ hit $AB$ at $F'$ . By the part already proved, $\dfrac{AF'}{F'B} \cdot \dfrac{BD}{DC} \cdot \dfrac{CE}{EA} = 1$ , so $\dfrac{AF'}{F'B} = \dfrac{AF}{FB}$ . Only one point on segment $AB$ divides it in a given ratio, so $F' = F$ . $\blacksquare$

Menelaus's theorem

Now take a straight line (a transversal) that crosses lines $AB$ , $BC$ , $CA$ at $F$ , $D$ , $E$ . A line can't cross all three sides of a triangle internally, so either two points are on the sides and one is on an extension, or all three are on extensions.

Menelaus's theorem

Points $F$ , $D$ , $E$ on lines $AB$ , $BC$ , $CA$ are collinear if and only if

$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1$

(using unsigned lengths, with an odd number of the points on extensions).

Proof. Suppose $F$ , $D$ , $E$ lie on a line $\ell$ . Drop perpendiculars from $A$ , $B$ , $C$ to $\ell$ , with lengths $h_A$ , $h_B$ , $h_C$ . The right triangles formed at $F$ are similar, so $\dfrac{AF}{FB} = \dfrac{h_A}{h_B}$ . Likewise $\dfrac{BD}{DC} = \dfrac{h_B}{h_C}$ and $\dfrac{CE}{EA} = \dfrac{h_C}{h_A}$ . The product telescopes to $1$ . The converse works just like Ceva's converse. $\blacksquare$

The formula is the same as Ceva's. The difference is the picture: Ceva is about three lines through one point; Menelaus is about three points on one line.

Mass points

Mass points package Ceva and Menelaus into a mechanical procedure. Put a weight at each vertex so that each cevian's foot is the balance point (center of mass) of the two vertices it sits between.

Definition

Mass points

If $D$ is on segment $BC$ and masses $m_B$ , $m_C$ sit at $B$ and $C$ , then $D$ is the balance point when

$m_B \cdot BD = m_C \cdot DC,$

and then $D$ gets mass $m_B + m_C$ . The heavier end is closer to the balance point.

Why it works. Treat each mass point as the vector $m\,X$ . The center of mass of the whole system $m_A A + m_B B + m_C C$ (divided by the total mass) can be computed in stages: first combine $B$ and $C$ into mass $m_B + m_C$ at $D$ , then combine that with $A$ . So the center of mass lies on segment $AD$ , dividing it so that

$\frac{AP}{PD} = \frac{m_D}{m_A} = \frac{m_B + m_C}{m_A}.$

Combining in a different order shows it lies on $BE$ and $CF$ too. That is a physical proof of Ceva, and it gives every ratio along every cevian at once.

Triangle ABC with A at the top, B at the lower left, C at the lower right. Cevian AD goes to side BC and cevian BE goes to side CA; they cross at P.

Worked example: Ratios along two cevians

In triangle $ABC$ , $D$ is on $BC$ with $BD : DC = 2 : 1$ and $E$ is the midpoint of $CA$ . Cevians $AD$ and $BE$ meet at $P$ . Find $AP : PD$ and $BP : PE$ .

Balance $D$ : $m_B \cdot 2 = m_C \cdot 1$ , so take $m_B = 1$ , $m_C = 2$ . Balance $E$ (a midpoint): $m_A = m_C = 2$ .

Now $m_D = 1 + 2 = 3$ and $m_E = 2 + 2 = 4$ . So

$\frac{AP}{PD} = \frac{m_D}{m_A} = \frac{3}{2}, \qquad \frac{BP}{PE} = \frac{m_E}{m_B} = \frac{4}{1}.$

Worked example: Ceva finds the third ratio

Cevians $AD$ , $BE$ , $CF$ are concurrent, with $AF : FB = 1 : 2$ and $BD : DC = 3 : 1$ . Find $CE : EA$ .

Ceva gives $\dfrac{1}{2} \cdot \dfrac{3}{1} \cdot \dfrac{CE}{EA} = 1$ , so $\dfrac{CE}{EA} = \dfrac{2}{3}$ .

Worked example: A transversal with Menelaus

In triangle $ABC$ , $F$ is on $AB$ with $AF : FB = 1 : 2$ and $E$ is the midpoint of $CA$ . Line $FE$ meets line $BC$ at $D$ . Where is $D$ ?

Menelaus: $\dfrac{AF}{FB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CE}{EA} = \dfrac{1}{2} \cdot \dfrac{BD}{DC} \cdot 1 = 1$ , so $BD = 2\,DC$ . Since the line crosses two sides internally, $D$ is on the extension of $BC$ past $C$ , with $BD = 2\,DC$ . So $CD = BC$ : the point $C$ is the midpoint of $BD$ .

Splitting masses

What if a line crosses a cevian but doesn't pass through a vertex? Give the vertex it "splits" two separate masses, one for each side it touches.

Worked example: A line crossing a cevian

In triangle $ABC$ , $D$ is the midpoint of $BC$ . Point $E$ is on $AB$ with $AE : EB = 1 : 2$ , and $F$ is on $AC$ with $AF : FC = 1 : 1$ . Segment $EF$ meets $AD$ at $P$ . Find $AP : PD$ .

Put $m_B = m_C = 1$ , so $m_D = 2$ . Split $A$ 's mass into a part $a_B$ used on side $AB$ and a part $a_C$ used on side $AC$ :

On $AB$ : $a_B \cdot AE = m_B \cdot EB$ , so $a_B = 2$ .
On $AC$ : $a_C \cdot AF = m_C \cdot FC$ , so $a_C = 1$ .

The total mass at $A$ is $a_B + a_C = 3$ , and $P$ is the balance point of the whole system on $AD$ :

$\frac{AP}{PD} = \frac{m_D}{m_A} = \frac{2}{3}.$

This works because $E$ (mass $a_B + m_B$ ) and $F$ (mass $a_C + m_C$ ) together have the same center of mass as $A$ (mass $3$ ) and $D$ (mass $2$ ): both are the center of mass of the same four weights.

Common mistake

Mass points give ratios along a segment between two massed points. The mass at a point is the sum of the masses it balances, and the ratio is always "far mass over near mass": $\dfrac{AP}{PD} = \dfrac{m_D}{m_A}$ , not $\dfrac{m_A}{m_D}$ . Check with common sense: the heavier endpoint is closer to $P$ .

Tip

For areas, turn ratios into area fractions: if $X$ is on cevian $AD$ , then $[ABX] = \dfrac{AX}{AD} \cdot \dfrac{BD}{BC} \cdot [ABC]$ . And when all else fails, coordinates with $B = (0, 0)$ and $C$ on the $x$ -axis will check any mass-point answer.

Practice

Practice 1

Cevians $AD$ , $BE$ , $CF$ of triangle $ABC$ are concurrent, with $AF : FB = 2 : 3$ and $BD : DC = 7 : 4$ . If $AC = 26$ , find $CE \cdot EA$ .