Let $\ell$ be a line in the plane and let $90^\circ<\theta<180^\circ$. Consider any distinct points $P,Q,R$ that satisfy the following: (i) $P$ lies on $\ell$ and $PQ$ is perpendicular to $\ell$ (ii) $R$ lies on the same side of $\ell$ as $Q$, and $R$ doesn’t lie on $\ell$ (iii) for any points $A,B$ on $\ell$, if $\angle ARB=\theta$ then $\angle AQB \geq \theta$. Find the minimum value of $\angle PQR$.
Problem
Source: Thailand MO 2023 Day 1 P5
Tags: geometry
demmy
11.05.2023 12:03
I think this is weirdest geo in my opinion. This year is first year that has 2 geo in day 1
DaRKst0Rm
11.05.2023 16:13
$270-\theta$
We fix point $R$ and construct line $l'$ passing $R$ such that $l'$ parallel to $l$. Consider a point $X, Y$ on $l$ such that $\measuredangle (l,XR) = \measuredangle (YR,l) = \theta$.$Q$ must be inside $\triangle XRY$ (include the boundary)
gghx
12.05.2023 02:55
We claim that the answer is $270-\theta$. You'll probably need diagrams to understand this since I'm too lazy to draw any. Suppose $\angle PQR<270-\theta$. Consider a point $X$ on $l$ such that $X$ is on the opposite side of $PQ$ as $R$, $\triangle PXR$ does not contain $Q$, and $\angle PXR<180-\theta$. This is possible by making line $RX$ arbitrarily close to $Q$. Let $\angle PXR=180-\theta-2\alpha$. Consider the point $A$ such that $\angle PAR=180-\theta-\alpha$ and $B$ on the other side such that $\angle RBA=\alpha$. The key point is that $XR$ is tangent to $(ARB)$, and $\angle ARB=\theta$. By construction $\triangle XRB$ does not contain $Q$. Thus $Q$ lies outside $(ARB)$ on the same side of $AB$ as $R$, which means $\angle ARB>\angle AQB$. This violates the condition. Now take $R$ such that $\angle PQR=270-\theta$. Suppose $A,B$ exist such that $\angle AQB<\theta$. Let $RQ$ meet $l$ at $Y$. Then WLOG if $A$ is on the opposite side of $PQ$ than $R$ then $A$ must lie on the opposite side of $Y$ than $P$ (else $B$ cannot exist). Thus triangle $ARB$ must contain $Q$, which means $\angle AQB>\angle ARB$, a contradiction.