Kvant Math Problem 4
Fix the segment $AB$ and let its length be $d$.
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Problem
Given a segment $AB$. Find the set of points $C$ in the plane such that in the triangle $ABC$ the median drawn from vertex $A$ is equal to the altitude drawn from vertex $B$.
Moscow Mathematical Olympiad (XXXII)
Exploration
Fix the segment $AB$ and let its length be $d$.
The condition says that the median from $A$ equals the altitude from $B$. Let $M$ be the midpoint of $BC$. Then the median from $A$ has length $AM$. Let $h_B$ denote the altitude from $B$ to the line $AC$.
The altitude from $B$ to $AC$ is the distance from $B$ to the line $AC$. Since $A$ lies on that line, this distance equals
$$h_B=AB\sin\angle BAC=d\sin\angle A.$$
Thus the condition becomes
$$AM=d\sin\angle A.$$
A natural coordinate choice is
$$A=(0,0),\qquad B=(d,0),\qquad C=(x,y).$$
Then
$$M=\left(\frac{d+x}{2},\frac y2\right),$$
hence
$$AM^2=\frac{(d+x)^2+y^2}{4}.$$
Also,
$$\sin\angle A=\frac{|y|}{\sqrt{x^2+y^2}},$$
so
$$h_B=\frac{d|y|}{\sqrt{x^2+y^2}}.$$
Equating the squares gives
$$\frac{(d+x)^2+y^2}{4} = \frac{d^2y^2}{x^2+y^2}.$$
The algebra is the delicate part. Multiplying by $4(x^2+y^2)$ yields
$$\bigl((d+x)^2+y^2\bigr)(x^2+y^2)=4d^2y^2.$$
Since
$$(d+x)^2+y^2=d^2+2dx+x^2+y^2,$$
the left-hand side becomes
$$(x^2+y^2)\bigl(d^2+2dx+x^2+y^2\bigr).$$
Introducing polar coordinates centered at $A$,
$$x=r\cos\theta,\qquad y=r\sin\theta,$$
gives
$$r^2\bigl(d^2+2dr\cos\theta+r^2\bigr) = 4d^2r^2\sin^2\theta.$$
For $C\neq A$, division by $r^2$ gives
$$r^2+2dr\cos\theta+d^2(4\cos^2\theta-3)=0.$$
This is a quadratic in $r$. Its discriminant equals
$$4d^2\cos^2\theta-4d^2(4\cos^2\theta-3) = 12d^2\sin^2\theta.$$
Hence
$$r=d\bigl(-\cos\theta\pm\sqrt3,|\sin\theta|\bigr).$$
The expressions
$$r=-d(\cos\theta-\sqrt3\sin\theta), \qquad r=-d(\cos\theta+\sqrt3\sin\theta)$$
suggest the polar equations of circles through $A$. Indeed,
$$r=2d\sin!\left(\theta-\frac{\pi}{6}\right), \qquad r=-2d\sin!\left(\theta+\frac{\pi}{6}\right).$$
A circle through the pole has equation
$$r=2R\sin(\theta-\varphi).$$
Thus the locus should be the union of two circles through $A$ and $B$. Converting to Cartesian coordinates will identify them precisely.
Problem Understanding
We are given a fixed segment $AB$. For a variable point $C$, consider triangle $ABC$. The median drawn from $A$ must have the same length as the altitude drawn from $B$. We must determine all points $C$ satisfying this condition.
This is a Type A problem. We must find the complete locus and prove both that every point on the claimed locus satisfies the condition and that every solution belongs to it.
The core difficulty is transforming the metric condition involving a median and an altitude into an explicit geometric equation and recognizing the resulting locus.
The answer should be the union of two circles passing through $A$ and $B$, symmetric with respect to the line $AB$.
Proof Architecture
Lemma 1. In coordinates $A=(0,0)$, $B=(d,0)$, $C=(x,y)$, the condition is equivalent to
$$\bigl((d+x)^2+y^2\bigr)(x^2+y^2)=4d^2y^2.$$
This follows from explicit formulas for the median length and the altitude length.
Lemma 2. After substituting $x=r\cos\theta$, $y=r\sin\theta$, every solution with $r\ne0$ satisfies
$$r^2+2dr\cos\theta+d^2(4\cos^2\theta-3)=0.$$
This is obtained by dividing the equation of Lemma 1 by $r^2$.
Lemma 3. The preceding quadratic factors into the two polar equations
$$r=2d\sin!\left(\theta-\frac{\pi}{6}\right), \qquad r=-2d\sin!\left(\theta+\frac{\pi}{6}\right).$$
This comes from solving the quadratic and using trigonometric identities.
Lemma 4. These two polar equations represent the circles
$$x^2+y^2-dx-\sqrt3,dy=0,$$
and
$$x^2+y^2-dx+\sqrt3,dy=0.$$
This is obtained by converting the polar equations to Cartesian form.
The hardest direction is proving that no other points occur. The critical step is showing that the original condition is exactly equivalent to the quadratic equation in $r$, whose solutions are precisely the two circles.
Solution
Let
$$A=(0,0),\qquad B=(d,0),$$
where $d=AB>0$, and let
$$C=(x,y).$$
Let $M$ be the midpoint of $BC$. Then
$$M=\left(\frac{d+x}{2},\frac y2\right),$$
so the median from $A$ has length
$$AM=\frac12\sqrt{(d+x)^2+y^2}.$$
The altitude from $B$ is the distance from $B$ to the line $AC$. Since the angle at $A$ is $\angle BAC$,
$$h_B=AB\sin\angle BAC.$$
Because
$$\sin\angle BAC=\frac{|y|}{\sqrt{x^2+y^2}},$$
we obtain
$$h_B=\frac{d|y|}{\sqrt{x^2+y^2}}.$$
The required condition $AM=h_B$ is therefore equivalent to
$$\frac14\bigl((d+x)^2+y^2\bigr) = \frac{d^2y^2}{x^2+y^2}.$$
Multiplying by $4(x^2+y^2)$ gives
$$\bigl((d+x)^2+y^2\bigr)(x^2+y^2)=4d^2y^2. \tag{1}$$
Now write
$$x=r\cos\theta,\qquad y=r\sin\theta.$$
The point $C=A$ is not a vertex of a triangle, so $r>0$. Substituting into (1) and dividing by $r^2$ yields
$$d^2+2dr\cos\theta+r^2 = 4d^2\sin^2\theta.$$
Using
$$\sin^2\theta=1-\cos^2\theta,$$
we obtain
$$r^2+2dr\cos\theta+d^2(4\cos^2\theta-3)=0. \tag{2}$$
Solving (2) as a quadratic in $r$,
$$r = d\bigl(-\cos\theta\pm\sqrt3,|\sin\theta|\bigr).$$
This yields the two families
$$r=-d(\cos\theta-\sqrt3\sin\theta), \qquad r=-d(\cos\theta+\sqrt3\sin\theta).$$
Using
$$2\sin!\left(\theta-\frac{\pi}{6}\right) = \sqrt3\sin\theta-\cos\theta,$$
and
$$-2\sin!\left(\theta+\frac{\pi}{6}\right) = -\cos\theta-\sqrt3\sin\theta,$$
we obtain
$$r=2d\sin!\left(\theta-\frac{\pi}{6}\right), \tag{3}$$
or
$$r=-2d\sin!\left(\theta+\frac{\pi}{6}\right). \tag{4}$$
Convert (3) to Cartesian coordinates:
$$r^2 = 2dr\sin!\left(\theta-\frac{\pi}{6}\right).$$
Since
$$r\sin!\left(\theta-\frac{\pi}{6}\right) = y\cos\frac{\pi}{6}-x\sin\frac{\pi}{6} = \frac{\sqrt3}{2}y-\frac12x,$$
equation (3) becomes
$$x^2+y^2 = d(\sqrt3,y-x),$$
or
$$x^2+y^2+dx-\sqrt3,dy=0. \tag{5}$$
Similarly, (4) becomes
$$x^2+y^2+dx+\sqrt3,dy=0. \tag{6}$$
Each of (5) and (6) is the equation of a circle. Completing squares gives
$$\left(x+\frac d2\right)^2+\left(y-\frac{\sqrt3,d}{2}\right)^2=d^2,$$
and
$$\left(x+\frac d2\right)^2+\left(y+\frac{\sqrt3,d}{2}\right)^2=d^2.$$
Thus the locus is the union of two circles of radius $d$ with centers
$$\left(-\frac d2,\frac{\sqrt3,d}{2}\right), \qquad \left(-\frac d2,-\frac{\sqrt3,d}{2}\right).$$
Since every solution of the original condition led to one of these circles, and every point on either circle satisfies equations (5) or (6), hence satisfies (1), hence $AM=h_B$, these circles give exactly all solutions.
Therefore the set of all admissible points $C$ is the union of the two circles of radius $AB$ whose centers are obtained from the midpoint of $AB$ by moving a distance $\frac{\sqrt3}{2}AB$ perpendicular to $AB$, equivalently the circles with equations
$$\left(x+\frac d2\right)^2+\left(y-\frac{\sqrt3,d}{2}\right)^2=d^2,$$
and
$$\left(x+\frac d2\right)^2+\left(y+\frac{\sqrt3,d}{2}\right)^2=d^2.$$
$$\boxed{\text{The locus is the union of these two circles.}}$$
Verification of Key Steps
The first delicate step is the altitude formula. The altitude from $B$ is the distance from $B$ to the line $AC$. In triangle $ABC$,
$$[ABC]=\frac12,AC\cdot h_B =\frac12,AB\cdot AC\sin\angle BAC.$$
Cancelling $\frac12 AC$ gives
$$h_B=AB\sin\angle BAC.$$
With coordinates,
$$\sin\angle BAC=\frac{|y|}{\sqrt{x^2+y^2}},$$
hence
$$h_B=\frac{d|y|}{\sqrt{x^2+y^2}}.$$
A common mistake is to use $d\sin\angle C$ instead.
The second delicate step is solving equation (2). Its discriminant is
$$(2d\cos\theta)^2 - 4d^2(4\cos^2\theta-3) = 12d^2\sin^2\theta.$$
No solutions are lost because the discriminant is always nonnegative. Taking the square root gives
$$2\sqrt3,d,|\sin\theta|,$$
and the absolute value must be retained. Dropping it prematurely can miss one branch of the locus.
The third delicate step is recognizing the polar equations as circles. Starting from
$$r=2d\sin!\left(\theta-\frac{\pi}{6}\right),$$
multiplication by $r$ gives
$$r^2=d(\sqrt3,y-x).$$
Replacing $r^2$ by $x^2+y^2$ produces a quadratic equation in $x,y$, which is a circle. This confirms the geometric interpretation directly.
Alternative Approaches
A synthetic approach begins with the midpoint $M$ of $BC$. The condition is
$$AM=h_B.$$
Express the altitude as twice the area divided by $AC$:
$$h_B=\frac{2[ABC]}{AC}.$$
Since
$$[ABC]=[AMC],$$
one can rewrite the condition in terms of the distance from $M$ to the line $AC$. After introducing suitable coordinates along $AC$, the relation becomes a quadratic constraint describing a circle. Carrying this out for both orientations yields the same pair of circles.
The coordinate method is preferable because the median and altitude admit immediate explicit formulas, and the locus emerges from a single algebraic equation whose factorization directly reveals the two circles.