Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV / 1 / a

Write $\mathbf{u}=\mathbf{b}-\mathbf{a}$. Since the triangle is nondegenerate, $\mathbf{u}\ne 0$. Every point on the line $AB$ has a unique position vector $\mathbf{p}=\mathbf{a}+t\mathbf{u}$. The condition that $CP$ is orthogonal to $AB$ is the dot product equationIts coefficient of $t$ is $-\Vert \mathbf{u}{\Vert }^2\ne 0$, so it has exactly one solution. The foot of a perpendicular is thereforeThis is the orthogonal projection of $C$ onto the line $AB$, expressed using the Euclidean norm.

With $t$ as in i, the foot of a perpendicular lies strictly between the endpoints precisely when $0<t<1$. Since the denominator is positive, this givesEquivalently, both dot productsare positive: the angles of the triangle at $A$ and $B$ are both less than $\pi /2$. If “between” includes the endpoints, replace both strict inequalities by non-strict ones; equality gives $P=A$ or $P=B$.

The altitude of a triangle through $A$ has equation $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$, and the altitude of a triangle through $B$ has equation $(\mathbf{b}-\mathbf{d})\cdot (\mathbf{a}-\mathbf{c})=0$. Their normal vectors $\mathbf{b}-\mathbf{c}$ and $\mathbf{a}-\mathbf{c}$ are linearly independent because the triangle is nondegenerate. Consequently these two lines meet in exactly one point $D$.

Substitute its position vector $\mathbf{d}$ in the supplied identity. Its first term vanishes by the first altitude of a triangle equation, and its second term vanishes by the second altitude of a triangle equation. HenceThis is exactly the condition that $D$ lies on the altitude of a triangle through $C$. The three lines therefore meet in a unique point, the orthocenter. This argument includes right triangles and triangles whose orthocenter lies outside them.