hmm, do you mean that the fraction is maximal out of the 3 combinations given P,Q,R or it's maximal considering all points R? If it's the latter, then wouldn't R have to be points infinitely close to Q, so then the fraction would be infinitely large? I'm confused...

al.M.V. wrote: hmm, do you mean that the fraction is maximal out of the 3 combinations given P,Q,R or it's maximal considering all points R? Maximal considering all points R on the plane p. The points P and Q are fixed. al.M.V. wrote: If it's the latter, then wouldn't R have to be points infinitely close to Q, so then the fraction would be infinitely large? I'm confused... The point R cannot come infinitely close to Q, since the point Q doesn't lie on the plane p (this condition was formerly invisible because of a LaTeX error in Orl's post; now I corrected it), while the point R should lie on p. Anyway, here is my solution of the problem: Let T be the orthogonal projection of the point Q on the plane p. Then, the line PT is the orthogonal projection of the line PQ on the plane p, and thus forms the least angle with the line PQ among all lines through the point P which lie in the plane p; hence, $\measuredangle RPQ\geq\measuredangle TPQ$, and equality holds if and only if the point R lies on the ray PT (the only exception is when $PQ\perp p$; in this case, P = T, so the ray PT is undefined, and equality holds for all points R in the plane p, since we always have < RPQ = < TPQ = 90°). From $\measuredangle RPQ\geq\measuredangle TPQ$, it follows that $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$; also, since the angles < RPQ and < TPQ are < 180°, their half-angles $\frac{\measuredangle RPQ}{2}$ and $\frac{\measuredangle TPQ}{2}$ are < 90°, so that from $\frac{\measuredangle RPQ}{2}\geq\frac{\measuredangle TPQ}{2}$ we can conclude that $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$. Equality holds, as in the above, if and only if the point R lies on the ray PT (and, respectively, for all points R in the plane p if $PQ\perp p$). On the other hand, we obviously have $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ with equality if and only if $\frac{\measuredangle QRP-\measuredangle PQR}{2}=0^{\circ}$, i. e. if and only if < QRP = < PQR, i. e. if and only if triangle PQR is isosceles with base QR, i. e. if and only if PR = PQ, i. e. if and only if the point R lies on the sphere with center P and radius PQ. Now, applying the Mollweide theorem in triangle QPR, we get $\frac{QP+PR}{QR}=\frac{\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}}{\sin\frac{\measuredangle RPQ}{2}}$ $\leq\frac{1}{\sin\frac{\measuredangle RPQ}{2}}$ (since $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$) $\leq\frac{1}{\sin\frac{\measuredangle TPQ}{2}}$ (since $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$), and equality holds here if and only if equality holds in both of the inequalities $\sin\frac{\measuredangle RPQ}{2}\geq\sin\frac{\measuredangle TPQ}{2}$ and $\cos\frac{\measuredangle QRP-\measuredangle PQR}{2}\leq 1$ that we have used, i. e. if and only if the point R lies both on the ray PT (this condition should be ignored if $PQ\perp p$) and on the sphere with center P and radius PQ. Hence, the point R for which the ratio $\frac{QP+PR}{QR}$ is maximum is the point of intersection of the ray PT with the sphere with center P and radius PQ (or, respectively, it can be any arbitrary point on the intersection of the plane p with the sphere with center P and radius PQ if $PQ\perp p$). Darij

Calculus is not supposed to be used, but let me 'borrow' it anyway: Assume $R'$ is the point we are looking for instead of $R$. Let $H$ be the foot of $Q$ on plane $[i][b]p[/b][/i]$. Draw ray $P$y that passes through $H$. Let a = $PQ$, b = $QR'$, c = $PH$, x = $PR'$, α = ∠$QPR'$, β = ∠$QPH$ and $[i]r[/i]$ = ($QP + PR'$)/$QR'$ = (a + x)/b. The ratio r is maximum when r² = (a + x)²/b² is also maximum. However, the law of cosines gives us (a + x)²/b² = (a² + x² + 2ax)/(a² + x² – 2ax×cosα). Observe that angle α is independent of variable x, and for the ratio r² to be maximum, cosα has to be maximum which occurs when α = β. Furthermore cosβ = c/a, and (a² + x² + 2ax)/(a² + x² – 2ax×cosβ) = (a² + x² + 2ax)/(a² + x² – 2cx). Note that c is also a constant, and the only variable in ratio r² is x. To get its maximum, take the derivative of the ratio r with respect to x and set it to zero to get x = a, and point $R$ is on ray $P$y with $PR = PQ$.

darij grinberg wrote: Anyway, here is my solution of the problem: Let T be the orthogonal projection of the point Q on the plane p. Then, the line PT is the orthogonal projection of the line PQ on the plane p, and thus forms the least angle with the line PQ among all lines through the point P which lie in the plane p; hence, $\measuredangle RPQ\geq\measuredangle TPQ$, and equality holds if and only if the point R lies on the ray PT (the only exception is when $PQ\perp p$; in this case, P = T, so the ray PT is undefined, and equality holds for all points R in the plane p, since we always have < RPQ = < TPQ = 90°). then: Quote: Now, applying the Mollweide theorem in triangle QPR, we get: $\frac{QP+PR}{QR}=\frac{cos \frac{\angle QRP - \angle PQR}{2}}{sin \frac{\angle RPQ}{2}}$ Because R lies on the ray of PT, so $\angle RPQ = \angle TPQ$. If T=P, $\angle RPQ=90^O$. Also in both cases $\angle QPR$ is fixed. So: $\frac{QP+PR}{QR}$ is max, if and only if $\angle QRP = \angle PQR \iff PR = PQ$. The answer: * R is on the ray of PT, where T is the orthogonal projection of the point Q on the plane p, and |PR|=|PQ| *Special case: T=P --> R is on the circumference of the circle on plane p with center P and radius |PQ|

Here is different approach to prove that $R$ lies on line $PT$, when $T\ne P$ (I took $T$ as projection of $Q$ onto the plane $p$ as others). Let $f(X)=\frac{QP+PX}{QX}$. Assume $R$ doesnt lie on $PT$. Take point $R_0$ on ray $PT$ such that $R_0$ doesn't lie on segment $PT$ and $PR-PT<R_0T<RT$. We can choose such $R_0$ because $PR-PT<RT$ from triangle inequality on $\triangle PRT$. $$QR^2=QT^2+TR^2>QT^2+TR_0^2=QR_0^2 \implies QR>QR_0$$$$PR_0=TR_0+PT>PR \implies PR_0>PR$$So we get $f(R_0)>f(R)$ which is contradiction. So $R$ should lie on line $PT$. Note: Since the rest of my sol is almost same with others, I don't post it.