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Problem 1161

AMC 12 late, AIME early

Geometry Difficulty 4.9 Prove it South African Mathematical Olympiad Third Round · South Africa

Consider a triangle $ABC$ with $BC = 3$ . Choose a point $D$ on $BC$ such that $BD = 2$ . Find the value of
$AB^2 + 2AC^2 - 3AD^2.$

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Drop the altitude from $A$ to $BC$ , and let $F$ be its foot. Furthermore, suppose that $BF = x$ (if $F$ lies on the extension of $BC$ beyond $B$ , assign a negative sign to $x$ ) and $AF = y$ . Then, by the Pythagorean theorem,
$AB^2 = BF^2 + AF^2 = x^2 + y^2,$
$AC^2 = CF^2 + AF^2 = (3-x)^2 + y^2,$
$AD^2 = DF^2 + AF^2 = (2-x)^2 + y^2.$
It follows that
$\begin{aligned} AB^2 + 2AC^2 - 3AD^2 &= x^2 + y^2 + 2(3-x)^2 + 2y^2 - 3(2-x)^2 - 3y^2 \\ &= x^2 + y^2 + 18 - 12x + 2x^2 + 2y^2 - 12 + 12x - 3x^2 - 3y^2 \\ &= 6, \end{aligned}$
regardless of the values of $x$ and $y$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.