Library / Japan Mathematical Olympiad /48 of 84

Japan Mathematical Olympiad

Geometry Difficulty 6.8 National Olympiad Prove it Japan

Let $ABC$ be a right triangle with $\angle ABC = 90^\circ$ . Points $P, Q, R$ lie on the sides $BC, CA, AB$ , respectively, in such a way that conditions
$AQ : QC = 2 : 1, \quad AR = AQ, \quad QP = QR, \quad \angle PQR = 90^\circ$
are satisfied. Find the value of $AR$ if $CP = 1$ . Here for a line segment $XY$ its length is also represented by $XY$ .

Solution

Take a point $C'$ on the line $AR$ in such a way that the points $A, R, C'$ lie on the line in this order and $RC' = 1$ is satisfied. From the hypothesis of the problem, it then follows that $RC' = PC, RQ = PQ$ are satisfied, and furthermore, we obtain the fact that the triangles $C'RQ$ and $CPQ$ are congruent since
$\begin{aligned} \angle C'RQ &= 360^\circ - (\angle RQP + \angle QPB + \angle PBR) = 360^\circ - (90^\circ + \angle QPB + 90^\circ) \\ &= 180^\circ - \angle QPB = \angle CPQ. \end{aligned}$
Consequently, we have $\angle AQC' = \angle AQR + \angle RQC' = \angle AQR + \angle PQC = 90^\circ$ . Let $AR = AQ = 2x$ . Then, we have $QC = QC' = x$ , and by applying the Pythagorean Theorem to the triangle $AQC'$ , we get $AC' = \sqrt{AQ^2 + (QC')^2} = \sqrt{5}x$ . From $1 = RC' - AR = (\sqrt{5} - 2)x$ we have $x = \frac{1}{\sqrt{5}-2}$ , and therefore,
$AR = 2x = \frac{2}{\sqrt{5}-2} = 2\sqrt{5} + 4.$

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.