Let the side length of the regular tetrahedron be a. Link and extend AP to meet the plane containing triangle BCD at E; link AQ and extend it to meet the same plane at F. We know that E and F are inside triangle BCD and that ∠PAQ = ∠EAF Now let’s look at the plane containing triangle BCD with points E and F inside the triangle. Link and extend EF on both sides to meet the sides of the triangle BCD at I and J, I on BC and J on DC. We have ∠EAF < ∠IAJ. But since E and F are interior of the tetrahedron, points I and J cannot be both at the vertices and IJ < a, ∠IAJ < ∠BAD = 60°. Therefore, ∠PAQ < 60°. Solution with graphic posted at http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1

Vo Duc Dien wrote: Let the side length of the regular tetrahedron be a. Link and extend AP to meet the plane containing triangle BCD at E; link AQ and extend it to meet the same plane at F. We know that E and F are inside triangle BCD and that ∠PAQ = ∠EAF Now let’s look at the plane containing triangle BCD with points E and F inside the triangle. Link and extend EF on both sides to meet the sides of the triangle BCD at I and J, I on BC and J on DC. We have ∠EAF < ∠IAJ. But since E and F are interior of the tetrahedron, points I and J cannot be both at the vertices and IJ < a, ∠IAJ < ∠BAD = 60°. Why we have ∠IAJ < ∠BAD from IJ < a?

Suppose $AP$ and $AQ$ intersect the plane $BCD$ at points $X$ and $Y$. It is clear that $X$ and $Y$ lie inside the triangle $\bigtriangleup BCD$. Suppose that the line $XY$ intersects the sides of $\bigtriangleup BCD$ at $S$ and $T$. WLOG, $S$ is on $BC$ and $T$ is on $BD$. It is clear that $\angle XAY< \angle SAT$, and equality cannot occur as the points $P$ and $Q$ cannot lie on the faces of the tetrahedron. Let $M$ and $N$ be the midpoints of $BC$ and $BD$ respectively, and suppose $S$ and $T$ are a distance $p$ and $q$ from $M$ and $N$ respectively. [asy][asy] size(8cm); pair B = dir(90); pair C = dir(210); pair D = dir(330); pair M = (B+C)/2; pair N = (B+D)/2; pair P = 0.7*B+0.3*C; pair PP = B+C-P; pair Q = 0.4*B+0.6*D; draw(P--Q--PP); draw(B--C--D--B--cycle); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$S$",P,dir(P)); dot("$S'$",PP,dir(PP)); dot("$T$",Q,dir(Q)); [/asy][/asy] Suppose that $S$ is in segment $BM$. Then noting that $\cos\angle SAT=\frac{AS^2+AT^2-ST^2}{2AS\cdot AT}$, we may reflect $S$ over $M$ to get $S'$. The quantity $ST$ increases (since $S$ is on the same side as $T$ on the perpendicular bisector of $BC$), and the angle $\angle SAT$ also increases. Similarly, we may assume that $T$ lies in segment $BN$. Letting the tetrahedron have side length 2, we have $$ST^2=BS^2+BT^2-2BS\cdot BT\cos\angle SBT =(1+p)^2+(1+q)^2-(1+p)(1+q)=(1+2p+p^2+1+2q+q^2)-(1+p+q+pq) =p^2+q^2+pq+p+q+1$$Thus, $$\cos\angle SAT=\frac{3+p^2+3+q^2-(p^2+q^2+pq+p+q+1)}{2\sqrt{(3+p^2)(3+q^2)}}=\frac{5-pq-p-q}{2\sqrt{(3+p^2)(3+q^2)}}$$As $p$ increases, the numerator is strictly decreasing while the denominator is strictly increasing, thus $\cos \angle SAT$ is decreasing and $\angle SAT$ is increasing. Thus we may position $S$ at vertex $C$ to maximise $\angle SAT$. Similarly, $T$ is at vertex $D$ to maximise $\angle SAT$. Thus, $\angle PAQ < \angle SAT \leq 60^{\circ}$.

ACGNmath wrote: Suppose $AP$ and $AQ$ intersect the plane $BCD$ at points $X$ and $Y$. It is clear that $X$ and $Y$ lie inside the triangle $\bigtriangleup BCD$. Suppose that the line $XY$ intersects the sides of $\bigtriangleup BCD$ at $S$ and $T$. WLOG, $S$ is on $BC$ and $T$ is on $BD$. It is clear that $\angle XAY< \angle SAT$, and equality cannot occur as the points $P$ and $Q$ cannot lie on the faces of the tetrahedron. Let $M$ and $N$ be the midpoints of $BC$ and $BD$ respectively, and suppose $S$ and $T$ are a distance $p$ and $q$ from $M$ and $N$ respectively. [asy][asy] size(8cm); pair B = dir(90); pair C = dir(210); pair D = dir(330); pair M = (B+C)/2; pair N = (B+D)/2; pair P = 0.7*B+0.3*C; pair PP = B+C-P; pair Q = 0.4*B+0.6*D; draw(P--Q--PP); draw(B--C--D--B--cycle); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$S$",P,dir(P)); dot("$S'$",PP,dir(PP)); dot("$T$",Q,dir(Q)); [/asy][/asy] Suppose that $S$ is in segment $BM$. Then noting that $\cos\angle SAT=\frac{AS^2+AT^2-ST^2}{2AS\cdot AT}$, we may reflect $S$ over $M$ to get $S'$. The quantity $ST$ increases (since $S$ is on the same side as $T$ on the perpendicular bisector of $BC$), and the angle $\angle SAT$ also increases. Similarly, we may assume that $T$ lies in segment $BN$. Letting the tetrahedron have side length 2, we have $$ST^2=BS^2+BT^2-2BS\cdot BT\cos\angle SBT =(1+p)^2+(1+q)^2-(1+p)(1+q)=(1+2p+p^2+1+2q+q^2)-(1+p+q+pq) =p^2+q^2+pq+p+q+1$$Thus, $$\cos\angle SAT=\frac{3+p^2+3+q^2-(p^2+q^2+pq+p+q+1)}{2\sqrt{(3+p^2)(3+q^2)}}=\frac{5-pq-p-q}{2\sqrt{(3+p^2)(3+q^2)}}$$As $p$ increases, the numerator is strictly decreasing while the denominator is strictly increasing, thus $\cos \angle SAT$ is decreasing and $\angle SAT$ is increasing. Thus we may position $S$ at vertex $C$ to maximise $\angle SAT$. Similarly, $T$ is at vertex $D$ to maximise $\angle SAT$. Thus, $\angle PAQ < \angle SAT \leq 60^{\circ}$. In your last step for simplifying the expression for $ST^2$, you flipped the sign of $pq$. It should be \[ST^2 = p^2 + q^2 - pq + p + q + 1. \] Thus, unfortunately, the last step of your proof is also not true. Or rather it needs to be fixed: \[cos SAT = \frac{4 + (p-1)(q-1)}{2\sqrt{(3+p^2)(3+q^2)}}.\] Since half the edge length of the tetrahedron in your proof is 1, and $q < 1$ and $p < 1$ since they are distances to the midpoints. both the quantities $p-1$ and $q-1$ are negative. I'll see if this proof is fixable later. Edit: Since in $(p-1)(q-1),$ $p$ and $q$ both lie in the interval $(0,1)$, we may rewrite as $(1-p)(1-q)$. As $p$ or $q$ increases, the numerator decreases strictly. Then the proof is complete as above.

Are there any proofs by contradiction? Also, just by intuition, the points are bounded inside of the figure, meaning maximum angle would occur if our points are at the border of the tetrahedron, which is precisely 60 degrees, but since it needs to be inside of that, $<60^o$.

huashiliao2020 wrote: Are there any proofs by contradiction? Also, just by intuition, the points are bounded inside of the figure, meaning maximum angle would occur if our points are at the border of the tetrahedron, which is precisely 60 degrees, but since it needs to be inside of that, $<60^o$. I believe so. Wish problems were still this easy Assume for the sake of contradiction that $\angle PAQ > 60^{\circ}$. By the sine area formula we have $[PAQ]=\frac{1}{2}\cdot PA \cdot QA \sin{\angle PAQ}$. We now note that since $P$ and $Q$ are inside $ABCD$, the lengths $AP$ and $AQ$ must be less than the side length of the tetrahedron $AB$. Also since $\angle PAQ > 60, \sin{\angle PAQ}> \frac{\sqrt{3}}{4}$. Thus $[PAQ]>AB^2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = AB^2 \cdot \frac{\sqrt{3}}{4}$. However we note that this is exactly the area of $[ABC]$. But $[PAQ]$ cannot be greater than $[ABC]$ because of the restriction of $P$ and $Q$ being inside $ABCD$. Thus we have a contradiction, and $\angle PAQ < 60^{\circ}$.