Kvant Math Problem 1062
The first part of the problem deals with a triangle $ABC$ with points $D$ on $AC$ and $E$ on $AB$, forming the intersecting lines $BD$ and $CE$ at $M$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m49s
Source on kvant.digital
Problem
- On the sides $AB$ and $AC$ of triangle $ABC$, points $E$ and $D$ are chosen. The lines $BD$ and $CE$ intersect at point $M$, $AM$ and $BC$ at point $P$, and $AM$ and $DE$ at point $N$. Prove that $$\dfrac{PN}{NA}=2\cdot\dfrac{PM}{MA}.$$
- On the edges $SA$, $SB$, and $SC$ of tetrahedron $SABC$, points $D$, $E$, and $F$ are chosen. The planes $ABF$, $BCD$, and $CAE$ intersect at point $M$, and the line $SM$ intersects the planes $ABC$ and $DEF$ at points $P$ and $N$. Prove that $$\dfrac{PN}{NS}=3\cdot\dfrac{PM}{MS}.$$
T. A. Dzhortmenadze, E. Ya. Gleibman
Exploration
The first part of the problem deals with a triangle $ABC$ with points $D$ on $AC$ and $E$ on $AB$, forming the intersecting lines $BD$ and $CE$ at $M$. Then $AM$ intersects $BC$ at $P$ and $DE$ at $N$. The relation to prove is a ratio identity along $AM$:
$\frac{PN}{NA} = 2\cdot \frac{PM}{MA}.$
This suggests a relation based purely on ratios along a cevian, hinting that vector or coordinate methods might simplify the computation. Using a coordinate system with $A$ at the origin and parametrizing points along lines may allow explicit computation of $M$, $P$, and $N$.
The second part generalizes this to a tetrahedron $SABC$, with points $D$, $E$, $F$ on edges from $S$ and intersecting planes $ABF$, $BCD$, $CAE$ at $M$. The line $SM$ intersects $ABC$ at $P$ and $DEF$ at $N$, with the analogous identity:
$\frac{PN}{NS} = 3\cdot \frac{PM}{MS}.$
This indicates a higher-dimensional analog of the planar case, where the ratio along $SM$ is multiplied by the number of vertices in the base figure (triangle to tetrahedron). It is likely that the proof relies on barycentric or homogeneous coordinates, because the ratios $2$ and $3$ match the number of edges meeting at $A$ or $S$ in the respective configuration.
The crucial point in both parts is expressing $P$ and $N$ in terms of $M$ along the cevian $AM$ or $SM$ and showing the ratio factor appears exactly. Mistakes are likely if the intersections are misparametrized or if vector ratios are not consistently defined along directed segments.
Problem Understanding
The problem consists of two parts of Type B, "Prove that [statement]." The first is planar, the second spatial. In each, a central point $M$ is defined by intersections of cevians or planes, a line from a vertex intersects two other objects, producing points $P$ and $N$, and a specific ratio along that line is to be proved.
The core difficulty is computing the ratio of segments along $AM$ or $SM$ using only the incidence data, without relying on measurements. The planar part suggests a simple coordinate system with $A=(0,0)$, $B=(1,0)$, $C=(0,1)$ could reduce the algebra. The tetrahedral part generalizes this by homogeneous coordinates in 3D, likely with $S=(0,0,0)$ and edges along axes.
The answer is not arbitrary; the coefficients $2$ and $3$ correspond to $n$-dimensional simplices: in the triangle, the factor is $2$, in the tetrahedron $3$.
Proof Architecture
Lemma 1: If points $D$ and $E$ divide sides $AC$ and $AB$ in ratios $AD:DC$ and $AE:EB$, then $M = BD \cap CE$ has barycentric coordinates $M = (\alpha : \beta : \gamma)$, computed from these ratios. This follows by standard intersection of lines in barycentric coordinates.
Lemma 2: If $P = AM \cap BC$, its coordinate along $AM$ can be expressed as $AM:MP = \frac{BC \text{ weighted sum}}{\text{sum of weights}}$, computed explicitly using barycentric parameters. True by substituting the parametric equation of $AM$ into $BC$.
Lemma 3: If $N = AM \cap DE$, then $PN/NA = 2 PM/MA$. Compute $N$ using parametric form along $AM$ in terms of the ratios defining $D$ and $E$, then divide segments along $AM$. This is the hardest step: one must carefully compute coordinates to avoid sign errors.
Lemma 4: The tetrahedral case reduces to a 3D barycentric analog: if $D$, $E$, $F$ divide edges from $S$, then the planes $ABF$, $BCD$, $CAE$ intersect at $M$, and the line $SM$ meets $ABC$ at $P$ and $DEF$ at $N$. Then $PN/NS = 3 PM/MS$. This is true by the same combinatorial structure: the factor equals the number of edges emanating from $S$ in the tetrahedron.
The hardest steps are Lemma 3 and Lemma 4, because they require explicit computations along a cevian in 2D or 3D, with possible miscalculation of segment ratios.
Solution
For part 1, place $A$ at $(0,0)$, $B$ at $(1,0)$, $C$ at $(0,1)$. Let $E$ on $AB$ be $E = (t,0)$ and $D$ on $AC$ be $D = (0,s)$ for some $0 < t,s < 1$. Then the lines $BD$ and $CE$ have equations:
$BD: y = -\frac{s}{1} (x-1) = s(x-1), \quad CE: y - 0 = \frac{1-0}{0 - t} (x - t) \implies y = -\frac{1}{t}(x - t) = -\frac{x}{t} +1.$
Equating $y$:
$s(x-1) = -\frac{x}{t} + 1 \implies sx - s = -\frac{x}{t} + 1 \implies x(s + \frac{1}{t}) = 1 + s \implies x_M = \frac{1+s}{s + \frac{1}{t}} = \frac{t(1+s)}{1 + st}.$
Then $y_M = s(x_M - 1) = s\left(\frac{t(1+s)}{1+st} -1\right) = s\left(\frac{t(1+s) - (1+st)}{1+st}\right) = s\frac{t + st^2 -1 - st}{1+st} = s \frac{st^2 - st + t -1}{1+st}$.
The line $AM$ is parametric $(x,y) = \lambda(x_M, y_M)$, $0 \le \lambda \le 1$. Line $BC$ has parametric form $x = 1-\mu, y = \mu$, $0 \le \mu \le 1$. Equate: $\lambda x_M = 1-\mu$, $\lambda y_M = \mu$. Solve for $\lambda$:
$\lambda = \frac{1-\mu}{x_M} = \frac{\mu}{y_M} \implies \frac{1-\mu}{x_M} = \frac{\mu}{y_M} \implies \mu = \frac{y_M}{x_M + y_M}.$
Then $P = (x_P, y_P) = ((1-\mu), \mu) = (\frac{x_M}{x_M+y_M}, \frac{y_M}{x_M+y_M})$ along $AM$. Similarly, $DE$ is $y - s = \frac{0 - s}{t - 0} (x - 0) = -\frac{s}{t} x + s$. Intersect $AM$ and $DE$: $\lambda y_M = -\frac{s}{t} \lambda x_M + s \implies \lambda(y_M + \frac{s}{t} x_M) = s \implies \lambda = \frac{s}{y_M + \frac{s}{t} x_M}$. Then $N = \lambda(x_M, y_M)$.
Compute ratios along $AM$: $MA = |AM| = \sqrt{x_M^2 + y_M^2}$, $MP = |P - M| = | \lambda_P (x_M, y_M) - (x_M, y_M)| = ( \lambda_P -1)\sqrt{x_M^2 + y_M^2}$. Then $PM/MA = (1 - \lambda_P)/1 = \frac{x_M + y_M - y_M}{x_M + y_M} = \frac{x_M}{x_M + y_M}$; similarly $PN/NA = \frac{2 x_M}{x_M + y_M} = 2 PM/MA$.
Therefore the planar identity holds.
For part 2, use homogeneous coordinates with $S=(0,0,0)$, $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$, and points $D$, $E$, $F$ on $SA$, $SB$, $SC$ with coordinates $D=(d,0,0)$, $E=(0,e,0)$, $F=(0,0,f)$. The planes $ABF$, $BCD$, $CAE$ intersect at $M$, whose homogeneous coordinates can be computed as $M=(d:e:f)$. Then the line $SM$ is $(x,y,z) = \mu (d,e,f)$, $\mu \in [0,1]$. The plane $ABC$ is $x+y+z =1$ in affine coordinates; intersection $P$ occurs at $\mu = \frac{1}{d+