Olympiad MethodsDifficulty 70-10018 tagged problems

Projective Geometry

Projective Geometry is the habit of using projection-friendly theorems and transformations to simplify incidence geometry. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$(A,B;C,D) = \frac{AC/CB}{AD/DB} \;\text{ is invariant under projection; a harmonic range has } (A,B;C,D) = -1.$$

Projecting points from one line to another (through a point or onto another line) preserves the cross-ratio of any four collinear points, so cross-ratio is the fundamental invariant for incidence problems.

Why It Works

1
Assign each of four collinear points $A,B,C,D$ signed positions; the cross-ratio $(A,B;C,D) = \dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}}$ (equivalently $\frac{AC/CB}{AD/DB}$) is a ratio of ratios.
2
A central projection from a point $O$ sends the four points to four points on another line; using the sine area formula for the triangles $OAC,\,OBD,\dots$, the $O$-dependent factors cancel, leaving the cross-ratio unchanged.
3
Because cross-ratio survives projection, any configuration property expressible through it (collinearity, concurrency, tangency) transfers between a hard figure and a simpler projected one.
4
The special value $-1$ defines a harmonic range/pencil; harmonic conjugates arise from complete quadrilaterals and polar lines, which is why $-1$ is the workhorse value in olympiad projective arguments.

Worked Examples

Example 1

On a line, points $A, B, C, D$ are placed so that $(A,B;C,D) = -1$ (a harmonic range). If $A=0$, $B=6$, and $C=2$ on a number line, find $D$.

1
Use the signed cross-ratio $(A,B;C,D) = \dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}} = -1$ with coordinates measured as differences.
2
Compute the known signed lengths from $A=0$, $B=6$, $C=2$: $\overline{AC} = 2-0 = 2$, $\overline{BC} = 2-6 = -4$.
3
Let $D = d$; then $\overline{BD} = d - 6$ and $\overline{AD} = d - 0 = d$. The equation is $\dfrac{2(d-6)}{d\,(-4)} = -1$.
4
Solve: $2(d-6) = (-1)\cdot d \cdot(-4) = 4d$, so $2d - 12 = 4d$, giving $2d = -12$ and $d = -6$. Verify: $(0,6;2,-6) = \dfrac{(2)(-6-6)}{(-6)(2-6)} = \dfrac{2\cdot(-12)}{(-6)(-4)} = \dfrac{-24}{24} = -1.$ ✓

Answer:

$D = -6$.

Contest level

On a number line, $A=1$, $B=2$, $C=4$, $D=8$. Compute the cross-ratio $(A,B;C,D)$, then verify it is unchanged by the projective map $x \mapsto 1/x$.

1
Signed lengths: $\overline{AC} = 4-1 = 3$, $\overline{BD} = 8-2 = 6$, $\overline{AD} = 8-1 = 7$, $\overline{BC} = 4-2 = 2$.
2
Cross-ratio $(A,B;C,D) = \dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}} = \dfrac{3\cdot 6}{7\cdot 2} = \dfrac{18}{14} = \dfrac{9}{7}$.
3
Apply $x\mapsto 1/x$ (a Möbius/projective transformation of the line): $A'=1$, $B'=\tfrac12$, $C'=\tfrac14$, $D'=\tfrac18$. Then $\overline{A'C'} = -\tfrac34$, $\overline{B'D'} = -\tfrac38$, $\overline{A'D'} = -\tfrac78$, $\overline{B'C'} = -\tfrac14$.
4
New cross-ratio $= \dfrac{(-3/4)(-3/8)}{(-7/8)(-1/4)} = \dfrac{9/32}{7/32} = \dfrac{9}{7}$ — unchanged, as projective invariance guarantees. ✓

Answer:

$(A,B;C,D) = \dfrac{9}{7}$, preserved under $x\mapsto 1/x$.

Olympiad / Challenge

Let $(A,B;C,D) = -1$ be a harmonic range and let $M$ be the midpoint of $AB$. Prove $MC \cdot MD = MA^2$ (signed), the harmonic 'midpoint' relation.

1
Place $M$ at the origin so $A = -m$, $B = m$, and let $C = c$, $D = d$ be the coordinates of $C, D$.
2
Harmonic condition: $\dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}} = \dfrac{(c+m)(d-m)}{(d+m)(c-m)} = -1$.
3
Cross-multiply: $(c+m)(d-m) = -(d+m)(c-m)$. Expanding, $cd - cm + dm - m^2 = -cd + dm - cm + m^2$; the $-cm + dm$ terms cancel, leaving $cd - m^2 = -cd + m^2$, so $2cd = 2m^2$, i.e. $cd = m^2$.
4
Therefore $MC\cdot MD = c\cdot d = m^2 = MA^2$. (Check on $A=0,B=6,C=2,D=-6$: $M=3$, $MC\cdot MD = (-1)(-9) = 9 = (-3)^2 = MA^2$.) ✓

Answer:

$MC\cdot MD = MA^2$ for any harmonic range with $M$ the midpoint of $AB$.

Going Deeper

Cross-ratio extends to four concurrent lines (a pencil) and to four points on a conic, where it is again projection-invariant; this is the basis of pole–polar duality and the projective treatment of conics.
Harmonic ranges $(=-1)$ are everywhere in olympiad geometry: complete quadrilaterals, the polar of a point w.r.t. a circle, and the foot of an angle bisector with its external partner all produce harmonic conjugates — letting you 'project to infinity' to trivialize a configuration.
Pitfall: cross-ratio uses SIGNED lengths and a fixed order of the four points; swapping the roles changes its value (the six values $\lambda, \tfrac1\lambda, 1-\lambda,\dots$ form an orbit). Dropping signs or reordering carelessly turns $-1$ into $2$ or $\tfrac12$ and breaks the harmonic argument.

Spot the Signal

Look for problems where the key step is using projection-friendly theorems and transformations to simplify incidence geometry.
You can describe the hard part as using projection-friendly theorems and transformations to simplify incidence geometry, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for projective geometry, then rewrite the givens around it.
Name the relation that makes Projective Geometry legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Projective Geometry just because the surface looks familiar; verify the required condition first.
Applying Projective Geometry because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let projective geometry reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using projection-friendly theorems and transformations to simplify incidence geometry.

Back to techniques

Olympiad MethodsDifficulty 70-10018 tagged problems

Projective Geometry

Practice this technique Match my level

The Key Result

$$(A,B;C,D) = \frac{AC/CB}{AD/DB} \;\text{ is invariant under projection; a harmonic range has } (A,B;C,D) = -1.$$

Why It Works

1
Assign each of four collinear points $A,B,C,D$ signed positions; the cross-ratio $(A,B;C,D) = \dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}}$ (equivalently $\frac{AC/CB}{AD/DB}$) is a ratio of ratios.
2
A central projection from a point $O$ sends the four points to four points on another line; using the sine area formula for the triangles $OAC,\,OBD,\dots$, the $O$-dependent factors cancel, leaving the cross-ratio unchanged.
3
Because cross-ratio survives projection, any configuration property expressible through it (collinearity, concurrency, tangency) transfers between a hard figure and a simpler projected one.
4
The special value $-1$ defines a harmonic range/pencil; harmonic conjugates arise from complete quadrilaterals and polar lines, which is why $-1$ is the workhorse value in olympiad projective arguments.

Worked Examples

Example 1

On a line, points $A, B, C, D$ are placed so that $(A,B;C,D) = -1$ (a harmonic range). If $A=0$, $B=6$, and $C=2$ on a number line, find $D$.

1
Use the signed cross-ratio $(A,B;C,D) = \dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}} = -1$ with coordinates measured as differences.
2
Compute the known signed lengths from $A=0$, $B=6$, $C=2$: $\overline{AC} = 2-0 = 2$, $\overline{BC} = 2-6 = -4$.
3
Let $D = d$; then $\overline{BD} = d - 6$ and $\overline{AD} = d - 0 = d$. The equation is $\dfrac{2(d-6)}{d\,(-4)} = -1$.
4
Solve: $2(d-6) = (-1)\cdot d \cdot(-4) = 4d$, so $2d - 12 = 4d$, giving $2d = -12$ and $d = -6$. Verify: $(0,6;2,-6) = \dfrac{(2)(-6-6)}{(-6)(2-6)} = \dfrac{2\cdot(-12)}{(-6)(-4)} = \dfrac{-24}{24} = -1.$ ✓

Answer:

$D = -6$.

Contest level

On a number line, $A=1$, $B=2$, $C=4$, $D=8$. Compute the cross-ratio $(A,B;C,D)$, then verify it is unchanged by the projective map $x \mapsto 1/x$.

1
Signed lengths: $\overline{AC} = 4-1 = 3$, $\overline{BD} = 8-2 = 6$, $\overline{AD} = 8-1 = 7$, $\overline{BC} = 4-2 = 2$.
2
Cross-ratio $(A,B;C,D) = \dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}} = \dfrac{3\cdot 6}{7\cdot 2} = \dfrac{18}{14} = \dfrac{9}{7}$.
3
Apply $x\mapsto 1/x$ (a Möbius/projective transformation of the line): $A'=1$, $B'=\tfrac12$, $C'=\tfrac14$, $D'=\tfrac18$. Then $\overline{A'C'} = -\tfrac34$, $\overline{B'D'} = -\tfrac38$, $\overline{A'D'} = -\tfrac78$, $\overline{B'C'} = -\tfrac14$.
4
New cross-ratio $= \dfrac{(-3/4)(-3/8)}{(-7/8)(-1/4)} = \dfrac{9/32}{7/32} = \dfrac{9}{7}$ — unchanged, as projective invariance guarantees. ✓

Answer:

$(A,B;C,D) = \dfrac{9}{7}$, preserved under $x\mapsto 1/x$.

Olympiad / Challenge

Let $(A,B;C,D) = -1$ be a harmonic range and let $M$ be the midpoint of $AB$. Prove $MC \cdot MD = MA^2$ (signed), the harmonic 'midpoint' relation.

1
Place $M$ at the origin so $A = -m$, $B = m$, and let $C = c$, $D = d$ be the coordinates of $C, D$.
2
Harmonic condition: $\dfrac{\overline{AC}\cdot\overline{BD}}{\overline{AD}\cdot\overline{BC}} = \dfrac{(c+m)(d-m)}{(d+m)(c-m)} = -1$.
3
Cross-multiply: $(c+m)(d-m) = -(d+m)(c-m)$. Expanding, $cd - cm + dm - m^2 = -cd + dm - cm + m^2$; the $-cm + dm$ terms cancel, leaving $cd - m^2 = -cd + m^2$, so $2cd = 2m^2$, i.e. $cd = m^2$.
4
Therefore $MC\cdot MD = c\cdot d = m^2 = MA^2$. (Check on $A=0,B=6,C=2,D=-6$: $M=3$, $MC\cdot MD = (-1)(-9) = 9 = (-3)^2 = MA^2$.) ✓

Answer:

$MC\cdot MD = MA^2$ for any harmonic range with $M$ the midpoint of $AB$.

Going Deeper

Cross-ratio extends to four concurrent lines (a pencil) and to four points on a conic, where it is again projection-invariant; this is the basis of pole–polar duality and the projective treatment of conics.
Harmonic ranges $(=-1)$ are everywhere in olympiad geometry: complete quadrilaterals, the polar of a point w.r.t. a circle, and the foot of an angle bisector with its external partner all produce harmonic conjugates — letting you 'project to infinity' to trivialize a configuration.
Pitfall: cross-ratio uses SIGNED lengths and a fixed order of the four points; swapping the roles changes its value (the six values $\lambda, \tfrac1\lambda, 1-\lambda,\dots$ form an orbit). Dropping signs or reordering carelessly turns $-1$ into $2$ or $\tfrac12$ and breaks the harmonic argument.

Spot the Signal

Look for problems where the key step is using projection-friendly theorems and transformations to simplify incidence geometry.
You can describe the hard part as using projection-friendly theorems and transformations to simplify incidence geometry, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for projective geometry, then rewrite the givens around it.
Name the relation that makes Projective Geometry legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Projective Geometry just because the surface looks familiar; verify the required condition first.
Applying Projective Geometry because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let projective geometry reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using projection-friendly theorems and transformations to simplify incidence geometry.