GeometryDifficulty 30-8552 tagged problems

Vectors

Vectors is the habit of representing points and directions as vectors to handle lengths, midpoints, and parallelism. 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

$$\text{Midpoint } M \text{ of } AB: \ \vec{M} = \tfrac{1}{2}(\vec{A} + \vec{B}); \qquad \vec{u}\cdot\vec{v} = |\vec{u}|\,|\vec{v}|\cos\theta$$

You write points as position vectors so midpoints, parallelism, and angles become clean vector arithmetic.

Why It Works

1
Represent each point by its position vector from a fixed origin; then $\vec{AB} = \vec{B} - \vec{A}$.
2
A point dividing $AB$ in ratio $t:(1-t)$ is $\vec{P} = (1-t)\vec{A} + t\vec{B}$; setting $t = \tfrac12$ gives the midpoint, and the centroid of a triangle is $\tfrac13(\vec{A}+\vec{B}+\vec{C})$.
3
Parallelism means one vector is a scalar multiple of another; collinearity of $A,B,C$ means $\vec{B}-\vec{A}$ and $\vec{C}-\vec{A}$ are parallel.
4
The dot product encodes length and angle: $\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta$, so $\vec{u}\perp\vec{v} \Leftrightarrow \vec{u}\cdot\vec{v} = 0$ and $|\vec{u}|^2 = \vec{u}\cdot\vec{u}$.

Worked Examples

Example 1

Vectors $\vec{u} = (3, 4)$ and $\vec{v} = (4, -3)$ are given. Show they are perpendicular and find $|\vec{u}|$.

1
Compute the dot product: $\vec{u}\cdot\vec{v} = (3)(4) + (4)(-3) = 12 - 12 = 0$.
2
A zero dot product means $\vec{u} \perp \vec{v}$, so the vectors are perpendicular.
3
Length: $|\vec{u}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Answer:

$\vec{u}\cdot\vec{v} = 0$ so they are perpendicular; $|\vec{u}| = 5$.

Warm-up

The centroid $G$ of $\triangle ABC$ satisfies $\vec{G} = \tfrac13(\vec{A} + \vec{B} + \vec{C})$. If $A = (1,2)$, $B = (4,6)$, $C = (7,1)$, find $G$ and verify $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$.

1
Centroid: $\vec{G} = \tfrac13\big((1,2) + (4,6) + (7,1)\big) = \tfrac13(12, 9) = (4, 3)$.
2
Compute $\vec{GA} = \vec{A} - \vec{G} = (-3,-1)$, $\vec{GB} = (0,3)$, $\vec{GC} = (3,-2)$.
3
Sum: $(-3+0+3,\ -1+3-2) = (0,0) = \vec{0}$, confirming the centroid is the balance point.

Answer:

$G = (4,3)$ and $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$.

Contest level

Prove the parallelogram law $|\vec{u} + \vec{v}|^2 + |\vec{u} - \vec{v}|^2 = 2|\vec{u}|^2 + 2|\vec{v}|^2$, and use it to find the sum of the squares of the diagonals of a parallelogram with sides $3$ and $5$.

1
Expand using $|\vec{w}|^2 = \vec{w}\cdot\vec{w}$: $|\vec{u}+\vec{v}|^2 = |\vec{u}|^2 + 2\vec{u}\cdot\vec{v} + |\vec{v}|^2$ and $|\vec{u}-\vec{v}|^2 = |\vec{u}|^2 - 2\vec{u}\cdot\vec{v} + |\vec{v}|^2$.
2
Add them: the $\pm 2\vec{u}\cdot\vec{v}$ cross terms cancel, leaving $2|\vec{u}|^2 + 2|\vec{v}|^2$.
3
The diagonals of a parallelogram with side vectors $\vec{u}, \vec{v}$ are $\vec{u}+\vec{v}$ and $\vec{u}-\vec{v}$, so the sum of their squares is $2|\vec{u}|^2 + 2|\vec{v}|^2 = 2(3^2) + 2(5^2) = 18 + 50 = 68$.

Answer:

Sum of squared diagonals $= 68$.

Olympiad / Challenge

Let $O$ be the circumcenter and $H$ the orthocenter of $\triangle ABC$. Using vectors from $O$, prove $\vec{OH} = \vec{OA} + \vec{OB} + \vec{OC}$, and deduce that $O$, the centroid $G$, and $H$ are collinear with $OG:GH = 1:2$ (the Euler line).

1
Take $O$ as origin, so $|\vec{A}| = |\vec{B}| = |\vec{C}| = R$. Define $\vec{H} = \vec{A} + \vec{B} + \vec{C}$ and show this point is the orthocenter.
2
Check $AH \perp BC$: $\vec{AH} = \vec{H} - \vec{A} = \vec{B} + \vec{C}$ and $\vec{BC} = \vec{C} - \vec{B}$. Their dot product is $(\vec{B}+\vec{C})\cdot(\vec{C}-\vec{B}) = |\vec{C}|^2 - |\vec{B}|^2 = R^2 - R^2 = 0$.
3
By symmetry the analogous perpendicularities hold, so $\vec{H} = \vec{A}+\vec{B}+\vec{C}$ is indeed the orthocenter: $\vec{OH} = \vec{A}+\vec{B}+\vec{C}$.
4
Since $\vec{OG} = \tfrac13(\vec{A}+\vec{B}+\vec{C}) = \tfrac13\vec{OH}$, the points $O, G, H$ are collinear and $OG:GH = \tfrac13 : \tfrac23 = 1:2$.

Answer:

$\vec{OH} = \vec{OA}+\vec{OB}+\vec{OC}$, and $O, G, H$ are collinear with $OG:GH = 1:2$.

Going Deeper

Generalization: a point dividing a segment, the centroid, and any weighted barycenter are all affine combinations $\sum w_i \vec{P_i}$ with $\sum w_i = 1$ — vectors unify ratio geometry and mass points under one operation. Choosing the origin AT a key center (circumcenter $O$, as in the Euler-line proof) collapses lengths to the constant $R$ and kills most of the algebra.
Where it appears: vectors shine on centroid/orthocenter/Euler-line results, midpoint and parallelism proofs, and any 'prove these three points are collinear' problem (show one displacement is a scalar multiple of another). The dot product handles every length-and-angle condition at once.
Pitfall: position vectors depend on the origin, but DISPLACEMENT vectors $\vec{B}-\vec{A}$ do not. Mixing the two — e.g. asserting $\vec{A}\cdot\vec{B} = 0$ means $OA \perp OB$ only because $O$ is the origin — leads to false conclusions if the origin moves. State your origin and stick to it.

Spot the Signal

Look for problems where the key step is representing points and directions as vectors to handle lengths, midpoints, and parallelism.
You can describe the hard part as representing points and directions as vectors to handle lengths, midpoints, and parallelism, 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 diagram relation that calls for vectors, then rewrite the givens around it.
Name the relation that makes Vectors 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 Vectors just because the surface looks familiar; verify the required condition first.
Applying Vectors 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 vectors reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is representing points and directions as vectors to handle lengths, midpoints, and parallelism.

Back to techniques

GeometryDifficulty 30-8552 tagged problems

Vectors

Practice this technique Match my level

The Key Result

$$\text{Midpoint } M \text{ of } AB: \ \vec{M} = \tfrac{1}{2}(\vec{A} + \vec{B}); \qquad \vec{u}\cdot\vec{v} = |\vec{u}|\,|\vec{v}|\cos\theta$$

You write points as position vectors so midpoints, parallelism, and angles become clean vector arithmetic.

Why It Works

1
Represent each point by its position vector from a fixed origin; then $\vec{AB} = \vec{B} - \vec{A}$.
2
A point dividing $AB$ in ratio $t:(1-t)$ is $\vec{P} = (1-t)\vec{A} + t\vec{B}$; setting $t = \tfrac12$ gives the midpoint, and the centroid of a triangle is $\tfrac13(\vec{A}+\vec{B}+\vec{C})$.
3
Parallelism means one vector is a scalar multiple of another; collinearity of $A,B,C$ means $\vec{B}-\vec{A}$ and $\vec{C}-\vec{A}$ are parallel.
4
The dot product encodes length and angle: $\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta$, so $\vec{u}\perp\vec{v} \Leftrightarrow \vec{u}\cdot\vec{v} = 0$ and $|\vec{u}|^2 = \vec{u}\cdot\vec{u}$.

Worked Examples

Example 1

Vectors $\vec{u} = (3, 4)$ and $\vec{v} = (4, -3)$ are given. Show they are perpendicular and find $|\vec{u}|$.

1
Compute the dot product: $\vec{u}\cdot\vec{v} = (3)(4) + (4)(-3) = 12 - 12 = 0$.
2
A zero dot product means $\vec{u} \perp \vec{v}$, so the vectors are perpendicular.
3
Length: $|\vec{u}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

Answer:

$\vec{u}\cdot\vec{v} = 0$ so they are perpendicular; $|\vec{u}| = 5$.

Warm-up

1
Centroid: $\vec{G} = \tfrac13\big((1,2) + (4,6) + (7,1)\big) = \tfrac13(12, 9) = (4, 3)$.
2
Compute $\vec{GA} = \vec{A} - \vec{G} = (-3,-1)$, $\vec{GB} = (0,3)$, $\vec{GC} = (3,-2)$.
3
Sum: $(-3+0+3,\ -1+3-2) = (0,0) = \vec{0}$, confirming the centroid is the balance point.

Answer:

$G = (4,3)$ and $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$.

Contest level

1
Expand using $|\vec{w}|^2 = \vec{w}\cdot\vec{w}$: $|\vec{u}+\vec{v}|^2 = |\vec{u}|^2 + 2\vec{u}\cdot\vec{v} + |\vec{v}|^2$ and $|\vec{u}-\vec{v}|^2 = |\vec{u}|^2 - 2\vec{u}\cdot\vec{v} + |\vec{v}|^2$.
2
Add them: the $\pm 2\vec{u}\cdot\vec{v}$ cross terms cancel, leaving $2|\vec{u}|^2 + 2|\vec{v}|^2$.
3
The diagonals of a parallelogram with side vectors $\vec{u}, \vec{v}$ are $\vec{u}+\vec{v}$ and $\vec{u}-\vec{v}$, so the sum of their squares is $2|\vec{u}|^2 + 2|\vec{v}|^2 = 2(3^2) + 2(5^2) = 18 + 50 = 68$.

Answer:

Sum of squared diagonals $= 68$.

Olympiad / Challenge

1
Take $O$ as origin, so $|\vec{A}| = |\vec{B}| = |\vec{C}| = R$. Define $\vec{H} = \vec{A} + \vec{B} + \vec{C}$ and show this point is the orthocenter.
2
Check $AH \perp BC$: $\vec{AH} = \vec{H} - \vec{A} = \vec{B} + \vec{C}$ and $\vec{BC} = \vec{C} - \vec{B}$. Their dot product is $(\vec{B}+\vec{C})\cdot(\vec{C}-\vec{B}) = |\vec{C}|^2 - |\vec{B}|^2 = R^2 - R^2 = 0$.
3
By symmetry the analogous perpendicularities hold, so $\vec{H} = \vec{A}+\vec{B}+\vec{C}$ is indeed the orthocenter: $\vec{OH} = \vec{A}+\vec{B}+\vec{C}$.
4
Since $\vec{OG} = \tfrac13(\vec{A}+\vec{B}+\vec{C}) = \tfrac13\vec{OH}$, the points $O, G, H$ are collinear and $OG:GH = \tfrac13 : \tfrac23 = 1:2$.

Answer:

$\vec{OH} = \vec{OA}+\vec{OB}+\vec{OC}$, and $O, G, H$ are collinear with $OG:GH = 1:2$.

Going Deeper

Generalization: a point dividing a segment, the centroid, and any weighted barycenter are all affine combinations $\sum w_i \vec{P_i}$ with $\sum w_i = 1$ — vectors unify ratio geometry and mass points under one operation. Choosing the origin AT a key center (circumcenter $O$, as in the Euler-line proof) collapses lengths to the constant $R$ and kills most of the algebra.
Where it appears: vectors shine on centroid/orthocenter/Euler-line results, midpoint and parallelism proofs, and any 'prove these three points are collinear' problem (show one displacement is a scalar multiple of another). The dot product handles every length-and-angle condition at once.
Pitfall: position vectors depend on the origin, but DISPLACEMENT vectors $\vec{B}-\vec{A}$ do not. Mixing the two — e.g. asserting $\vec{A}\cdot\vec{B} = 0$ means $OA \perp OB$ only because $O$ is the origin — leads to false conclusions if the origin moves. State your origin and stick to it.

Spot the Signal

Look for problems where the key step is representing points and directions as vectors to handle lengths, midpoints, and parallelism.
You can describe the hard part as representing points and directions as vectors to handle lengths, midpoints, and parallelism, 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 diagram relation that calls for vectors, then rewrite the givens around it.
Name the relation that makes Vectors 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 Vectors just because the surface looks familiar; verify the required condition first.
Applying Vectors 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 vectors reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is representing points and directions as vectors to handle lengths, midpoints, and parallelism.