Kvant Math Problem 124
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Verdicts: SKIP + SKIP
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Problem
Given a triangle $ABC$. Find a point $O$ inside it with the following property: for any line passing through the point $O$ and intersecting the sides of the triangle $AB$ at point $K$ and $BC$ at point $L$, the equality $$\dfrac{AK}{KB}+\dfrac{CL}{LB}=1.$$ holds.
More generally, prove that if $p$ and $q$ are arbitrary positive numbers, then inside the triangle $ABC$ one can specify a point $O$ such that for any line $KL$ passing through this point ($K$ lies on $AB$, $L$ on $BC$), $$p\dfrac{AK}{KB}+q\dfrac{CL}{LB}=1.$$.
G. Noten, 10th-grade student
Exploration
Let
$$x=\frac{AK}{KB},\qquad y=\frac{CL}{LB}.$$
The condition asks for a point $O$ such that every line $KL$ through $O$ with $K\in AB$, $L\in BC$, satisfies
$$px+qy=1.$$
The expression involves ratios on two sides of the triangle. Ratios on sides suggest using mass points, Menelaus, or coordinates.
Choose coordinates adapted to the two sides. Let
$$B=(0,0),\qquad A=(1,0),\qquad C=(0,1).$$
Then a point $K\in AB$ has coordinates $(t,0)$. Since
$$\frac{AK}{KB}=\frac{1-t}{t},$$
we obtain
$$t=\frac1{1+x}.$$
Similarly, if $L=(0,s)$, then
$$\frac{CL}{LB}=\frac{1-s}{s},$$
hence
$$s=\frac1{1+y}.$$
The line through $K=(t,0)$ and $L=(0,s)$ has intercept form
$$\frac{X}{t}+\frac{Y}{s}=1.$$
Substituting $t=1/(1+x)$ and $s=1/(1+y)$,
$$(1+x)X+(1+y)Y=1.$$
The required relation $px+qy=1$ gives
$$x=\frac{1-qy}{p}.$$
Substituting into the line equation,
$$\left(1+\frac1p-\frac q p,y\right)X+(1+y)Y=1.$$
For all admissible lines satisfying $px+qy=1$, the parameter $y$ varies. A common point $O=(X,Y)$ exists precisely when the coefficient of $y$ vanishes:
$$-\frac q p X+Y=0.$$
Thus
$$Y=\frac q p X.$$
The constant term then yields
$$\left(1+\frac1p\right)X+Y=1.$$
Using $Y=\frac q p X$,
$$\left(1+\frac{1+q}{p}\right)X=1,$$
so
$$X=\frac{p}{p+q+1},\qquad Y=\frac{q}{p+q+1}.$$
This gives a candidate
$$O=\left(\frac{p}{p+q+1},\frac{q}{p+q+1}\right).$$
The crucial point is to prove the converse: every line through this point indeed satisfies $px+qy=1$. That must be checked directly from the intercept equation.
For the original problem $p=q=1$, the point becomes
$$O=\left(\frac13,\frac13\right),$$
which is the centroid of the reference triangle. Since affine transformations preserve ratios on a side and collinearity, this corresponds to the centroid in an arbitrary triangle.
Problem Understanding
We must construct a point $O$ inside triangle $ABC$ such that every line through $O$ meeting $AB$ at $K$ and $BC$ at $L$ satisfies
$$p\frac{AK}{KB}+q\frac{CL}{LB}=1.$$
The special case $p=q=1$ asks for the corresponding point when the coefficients are equal.
This is a Type D problem. We must explicitly construct a point and verify that every line through it has the required property.
The difficulty is to find a point through which exactly the family of lines satisfying a linear relation between the side ratios passes. The natural approach is to express those ratios in coordinates and identify the common point of all such lines.
The answer is the point whose barycentric coordinates with respect to $ABC$ are proportional to
$$(p,;1,;q).$$
In the coordinate system $A=(1,0)$, $B=(0,0)$, $C=(0,1)$, it is
$$\left(\frac{p}{p+q+1},\frac{q}{p+q+1}\right).$$
For $p=q=1$, this is the centroid.
Proof Architecture
We place the triangle in coordinates $A=(1,0)$, $B=(0,0)$, $C=(0,1)$.
A point $K\in AB$ satisfies $K=(1/(1+x),0)$, where $x=AK/KB$, because solving the ratio condition on the segment gives this coordinate.
A point $L\in BC$ satisfies $L=(0,1/(1+y))$, where $y=CL/LB$, by the same computation.
The line $KL$ has equation $\frac{X}{t}+\frac{Y}{s}=1$, where $t$ and $s$ are its intercepts.
Substituting $t=1/(1+x)$ and $s=1/(1+y)$ converts the line equation into $(1+x)X+(1+y)Y=1$.
Imposing $px+qy=1$ and eliminating one parameter shows that all such lines pass through the fixed point
$$\left(\frac{p}{p+q+1},\frac{q}{p+q+1}\right).$$
Conversely, substituting this point into the equation of an arbitrary line $KL$ yields $px+qy=1$, proving the required property.
The most delicate step is the converse implication, because it must show that every line through the constructed point satisfies the relation, not merely that every line satisfying the relation passes through the point.
Solution
Choose affine coordinates so that
$$B=(0,0),\qquad A=(1,0),\qquad C=(0,1).$$
Let
$$x=\frac{AK}{KB},\qquad y=\frac{CL}{LB}.$$
Since $K\in AB$, write $K=(t,0)$ with $0<t<1$. Along the segment $AB$,
$$KB=t,\qquad AK=1-t,$$
hence
$$x=\frac{1-t}{t}.$$
Solving for $t$,
$$t=\frac1{1+x}.$$
Therefore
$$K=\left(\frac1{1+x},0\right).$$
Similarly, writing $L=(0,s)$ with $0<s<1$, we have
$$LB=s,\qquad CL=1-s,$$
so
$$y=\frac{1-s}{s},$$
which gives
$$s=\frac1{1+y}.$$
Hence
$$L=\left(0,\frac1{1+y}\right).$$
The equation of the line through $K$ and $L$ is
$$\frac{X}{t}+\frac{Y}{s}=1.$$
Substituting the expressions for $t$ and $s$,
$$(1+x)X+(1+y)Y=1.$$
Assume first that $K$ and $L$ satisfy
$$px+qy=1.$$
Then
$$x=\frac{1-qy}{p}.$$
Substituting into the equation of the line,
$$\left(1+\frac1p-\frac q p,y\right)X+(1+y)Y=1.$$
Rearranging,
$$\left(1+\frac1p\right)X+Y +y\left(Y-\frac q p X\right)=1.$$
Consequently every such line passes through any point $(X,Y)$ satisfying
$$Y=\frac q p X$$
and
$$\left(1+\frac1p\right)X+Y=1.$$
Solving these two equations,
$$X=\frac{p}{p+q+1},\qquad Y=\frac{q}{p+q+1}.$$
Thus every line for which $px+qy=1$ passes through the fixed point
$$O=\left(\frac{p}{p+q+1},\frac{q}{p+q+1}\right).$$
Since $p,q>0$,
$$X>0,\qquad Y>0,\qquad X+Y=\frac{p+q}{p+q+1}<1,$$
so $O$ lies inside triangle $ABC$.
It remains to prove the converse. Let $KL$ be any line through $O$. Let
$$x=\frac{AK}{KB},\qquad y=\frac{CL}{LB}.$$
The equation of $KL$ is
$$(1+x)X+(1+y)Y=1.$$
Substituting the coordinates of $O$,
$$(1+x)\frac{p}{p+q+1} + (1+y)\frac{q}{p+q+1} =1.$$
Multiplying by $p+q+1$,
$$p+px+q+qy=p+q+1.$$
Hence
$$px+qy=1.$$
Therefore every line through $O$ satisfies the required relation.
The constructed point is the desired point. In barycentric coordinates relative to $ABC$, it is
$$\boxed{O=(p:1:q)}.$$
For the original problem $p=q=1$, we obtain
$$\boxed{O=(1:1:1)},$$
the centroid of triangle $ABC$.
Verification of Key Steps
The first delicate step is converting the side ratios into coordinates. If $K=(t,0)$, then $KB=t$ and $AK=1-t$ because $A=(1,0)$ and $B=(0,0)$. Thus
$$x=\frac{1-t}{t},$$
which indeed gives
$$t=\frac1{1+x}.$$
A sign error here would change every subsequent formula.
The second delicate step is identifying the common point of all lines satisfying $px+qy=1$. After substituting $x=(1-qy)/p$, the line equation becomes
$$\left(1+\frac1p\right)X+Y +y\left(Y-\frac q p X\right)=1.$$
For a point to belong to every line of this one-parameter family, the coefficient of the free parameter $y$ must vanish. This yields
$$Y=\frac q p X.$$
Without enforcing this condition, one would obtain a point lying on only one member of the family.
The third delicate step is the converse. Substituting the coordinates of $O$ into the equation of an arbitrary line $KL$ through $O$ gives
$$p+px+q+qy=p+q+1,$$
hence
$$px+qy=1.$$
This argument proves the required property for every line through $O$, which is stronger than showing merely that the prescribed family of lines passes through $O$.
Alternative Approaches
A projective approach starts from Menelaus' theorem. Assign masses to the vertices so that the ratios on the sides satisfy
$$p\frac{AK}{KB}+q\frac{CL}{LB}=1.$$
The relation can be rewritten in terms of the side division parameters, and the resulting family of transversals forms a pencil through a fixed point. The fixed point turns out to have barycentric coordinates $(p:1:q)$.
Another approach uses barycentric coordinates from the outset. A point on $AB$ has barycentric coordinates $(\lambda,\mu,0)$, and a point on $BC$ has coordinates $(0,\nu,\rho)$. Expressing $AK/KB$ and $CL/LB$ through the barycentric coefficients, the condition $p(AK/KB)+q(CL/LB)=1$ becomes linear. The equation of the corresponding line family immediately yields the common point $(p:1:q)$. The coordinate method used above is preferable because it requires only elementary analytic geometry and makes the converse verification completely transparent.