Library / Bulgarian Mathematical Competitions /17 of 92

Bulgarian Mathematical Competitions

Geometry Difficulty 5.3 AIME, harder Prove it Bulgaria

Problem:
Let $ABCD$ be a parallelogram and let $\angle BAD$ be acute. Denote by $E$ and $F$ the feet of the perpendiculars from the vertex $C$ to the lines $AB$ and $AD$ , respectively. A circle through $D$ and $F$ is tangent to the diagonal $AC$ at a point $Q$ and a circle through $B$ and $E$ is tangent to the segment $QC$ at its midpoint $P$ . Find the length of diagonal $AC$ if $AQ=1$ .

Solution

Solution:
Let $DH \perp AC$ ( $H \in AC$ ). Then $\triangle AHD \sim \triangle AFC$ and $\triangle CHD \sim \triangle AEC$ . Hence
$\begin{aligned} AC^{2} &= AH \cdot AC + HC \cdot AC \\ &= AF \cdot AD + AE \cdot CD \\ &= AQ^{2} + AE \cdot AB = AQ^{2} + AP^{2} \end{aligned}$
Setting $QP = PC = x$ , we get the equation

$(1+2x)^{2} = 1 + (1+x)^{2} \Longleftrightarrow 3x^{2} + 2x - 1 = 0$ , which has a unique positive root $x = \frac{1}{3}$ . Therefore $AC = 1 + 2x = \frac{5}{3}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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