United States — Team Selection Test

Note: We present five different approaches. The first three synthetic solutions are all based on the following simple observation.
We first note that $APBR$ and $AQCR$ are cyclic quadrilaterals. It is easy to see that triangles $APC$ and $ABQ$ are congruent to each other, implying that $\angle APR = \angle APC = \angle ABQ = \angle ABR$. Thus, $APBR$ is a cyclic quadrilateral. Likewise, we can show that $AQCR$ is also cyclic.

Let $\angle PAB = 2x$. Then in isosceles triangle $APB$, $\angle APB = 90^\circ - x$. In cyclic quadrilateral $APBR$, $\angle ARB = 180^\circ - \angle APB = 90^\circ + x$. Likewise, $\angle ARC = 90^\circ + x$. Hence $\angle BRC = 360^\circ - \angle ARB - \angle ARC = 180^\circ - 2x$. It follows that $\angle BOC = 4x$.

First Solution: Reflect $C$ across line $AQ$ to $D$. Then $\angle BAD = 4x + \angle BAC = \angle BAQ$. It is easy to see that triangles $BAD$ and $PAQ$ are congruent, implying that $\angle ADB = \angle AQP = y$.

Note also that $CAD$ and $COB$ are two isosceles triangles with the same vertex angle, and so they are similar to each other. It follows that triangle $CAO$ and $CBD$ are similar by SAS (side-angle-side), implying that $\angle CAO = \angle CDB = z$.

The angle formed by lines $AO$ and $PQ$ is equal to
$$180^\circ - \angle OAQ - \angle AQP = 180^\circ - \angle OAC - \angle CAQ - \angle AQP = 180^\circ - z - 2x - y.$$

Since $AQ$ is perpendicular to the base $CD$ in isosceles triangle $ACD$, we have
$$90^\circ = \angle QAD + \angle CDA = \angle QAD + \angle ADB + \angle BDC = 2x + y + z.$$
Combining the last two equations yields that fact the angle formed by lines $AO$ and $PQ$ is equal to $90^\circ$; that is, $AO \perp PQ$.

Second Solution: We maintain the same notations as in the first solution. Let $M$ be the midpoint of arc $\widehat{BC}$ on the circumcircle of triangle $BOC$. Then $BM = CM$. Since triangles $APC$ and $ABQ$ are congruent, $PC = BQ$. Since $BRMC$ is cyclic, $\angle PCM = \angle RCM = \angle RBM = \angle QBM$. Hence triangles $BMQ$ and $CMP$ are congruent by SAS. It follows triangles $MPQ$ and $MBC$ are similar. Since $\angle BOC = 4x$, $\angle MBC = \angle MCB = x$, and so $\angle MPQ = x$.

Note that both triangles $PAB$ and $MOB$ are isosceles triangles with vertex angle $2x$; that is, they are similar to each other. Hence triangles $BMP$ and $BOA$ are also similar by SAS, implying that $\angle OAB = MPB = s$. We also note that in isosceles triangle $APB$,
$$90^\circ = \angle APB + \angle PAB/2 = \angle APQ + \angle QPM + \angle MPB + \angle PAB/2 = \angle APQ + 2x + s.$$
Putting the above together, we conclude that
$$\angle PAO + \angle APQ = \angle PAB + \angle BAO + \angle APQ = 2x + s + \angle APQ = 90^\circ,$$
that is $AO \perp PQ$.

Third Solution: We consider two rotations:
$\mathbf{R}_1$: a counterclockwise $2x$ (degree) rotation centered at $A$,
$\mathbf{R}_2$: a clockwise $4x$ (degree) rotation centered at $O$.
Let $\mathbf{T}$ denote the composition $\mathbf{R}_1\mathbf{R}_2\mathbf{R}_1$. Then $\mathbf{T}$ is a counterclockwise $2x - 4x + 2x = 0^\circ$ rotation; that is, $\mathbf{T}$ is translation. Note that
$$\mathbf{T}(P) = \mathbf{R}_1(\mathbf{R}_2(\mathbf{R}_1(P))) = \mathbf{R}_1(\mathbf{R}_2(B)) = \mathbf{R}_1(C) = Q,$$
or, $\mathbf{T}$ is the vector translation $\overrightarrow{PQ}$.
Let $A_1 = \mathbf{R}_2(A)$ and $A_2 = \mathbf{R}_1(A_1)$. Then $\mathbf{T}(A) = A_2$; that is, $\overrightarrow{AA_2} = \overrightarrow{PQ}$, or $AA_2 \parallel PQ$.
By the definitions of $\mathbf{R}_2$ and $\mathbf{R}_1$, we know that triangles $OAA_1$ and $A_1AA_2$ are isosceles triangles with respect vertex angles $\angle AOA_1 = 4x$ and $\angle A_1AA_2 = 2x^\circ$. It is routine to compute that $\angle OAA_2 = 90^\circ$; that $AO \perp AA_2$, or $AO \perp PQ$.

Fourth Solution: (By Ian Le) In this solutions, let each lowercase letter denote the number assigned to the point labeled with the corresponding uppercase letter. We further assume that $A$ is origin; that is, let $a = 0$. Let $\omega = e^{2xi}$ (or $\omega = \cos(2x) + i\sin(2x)$, and $\omega^{-1} = \cos(2x) - i\sin(2x)$). Then because $O$ lies on the perpendicular bisector of $BC$ and $\angle BOC = 4x$,
$$o = c + \frac{(b-c)i}{2\omega \sin(2x)} = c + \frac{bi}{2\omega \sin(2x)} - \frac{ci}{2\omega \sin(2x)}.$$
Note that
$$c - \frac{ci}{2\omega \sin(2x)} = c + \frac{c\omega^{-1}}{2i \sin(2x)} = \frac{c(\omega^{-1} + 2i \sin(2x))}{2i \sin(2x)} = \frac{c\omega}{2i \sin(2x)},$$
Combining the last two equations gives
$$o = \frac{bi}{2\omega \sin(2x)} + \frac{c\omega}{2i \sin(2x)} = -\frac{b}{2i\omega \sin(2x)} + \frac{c\omega}{2i \sin(2x)} = \frac{1}{2i \sin(2x)} \left( c\omega - \frac{b}{\omega} \right).$$
Now we note that $p = \frac{b}{\omega}$ and $q = c\omega$. Consequently, we obtain
$$\frac{q-p}{o-a} = 2i \sin(2x),$$
which is clearly a pure imaginary number; that is, $OA \perp PQ$.

Fifth Solution: (By Lan Le) In this solutions, we set $BC = a$, $AB = c$, $CA = b$, $A = \angle BAC$, $B = \angle ABC$, and $C = \angle BCA$. We use the fact that
$$OA \perp PQ \text{ if and only if } AP^2 - AQ^2 = OP^2 - OQ^2.$$
Clearly $AP^2 - AQ^2 = c^2 - b^2$. It remains to show that
$$OP^2 - OQ^2 = c^2 - b^2. \qquad (*)$$
In isosceles triangles APB and BOC, $BP = 2c \sin x$ and $BO = \frac{a}{2 \sin(2x)}$. Note that $\angle PBA + \angle ABC + \angle CBO = 90^\circ - x + B + 90^\circ - 2x = 180^\circ + B - 3x$. Applying the law of cosines to triangle PBO yields
$$OP^2 = 4c^2 \sin^2 x + \frac{a^2}{4 \sin^2(2x)} + \frac{ac \cos(B - 3x)}{\cos x}.$$
In exactly the same way, we can show that
$$OQ^2 = 4b^2 \sin^2 x + \frac{a^2}{4 \sin^2(2x)} + \frac{ab \cos(C - 3x)}{\cos x}.$$
Hence
$$OP^2 - OQ^2 = 4(c^2 - b^2) \sin^2 x + \frac{a}{\cos x} (c \cos(B - 3x) - b \cos(C - 3x)). \quad (\dagger)$$
Using Addition and Subtraction formulas and the law of sines (more precisely, $c \sin B = b \sin C$), we have
$$\begin{aligned} & c \cos(B - 3x) - b \cos(C - 3x) \\ = & c \cos(3x) \cos B + c \sin(3x) \sin B - b \cos(3x) \cos C - b \sin(3x) \sin C \\ = & \cos(3x)(c \cos B - b \cos C). \end{aligned}$$
Substituting the last equation into $(\dagger)$ gives
$$OP^2 - OQ^2 = 4(c^2 - b^2) \sin^2 x + \frac{\cos 3x}{\cos x} (ac \cos B - ab \cos C).$$
Note that
$$ac \cos B - ab \cos C = c(a \cos B + b \cos A) - b(a \cos C + c \cos A) = c^2 - b^2.$$
Combining the last equations gives
$$OP^2 - OQ^2 = (c^2 - b^2) \left( 4 \sin^2 x + \frac{\cos 3x}{\cos x} \right).$$
By the Triple-angle formulas, we have $\cos 3x = 4 \cos^3 x - 3 \cos x$, and so
$$OP^2 - OQ^2 = (c^2 - b^2)(4 \sin^2 x + 4 \cos^2 x - 3) = c^2 - b^2,$$
which is $(*)$.