Kvant Math Problem 38
Let $AB=p$ and $AC=q$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m52s
Source on kvant.digital
Problem
Fig. 2
A circle constructed on the altitude $AD$ of the right triangle $ABC$ as a diameter intersects leg $AB$ at point $K$ and leg $AC$ at point $M$. The segment $KM$ intersects the altitude $AD$ at point $L$. It is known that the segments $AK$, $AL$, and $AM$ form a geometric progression, i.e., $\dfrac{AK}{AL}=\dfrac{AL}{AM}$. Find the acute angles of triangle $ABC$ (Fig. 2).
L. M. Lopovok
Exploration
Let $AB=p$ and $AC=q$. Since $ABC$ is a right triangle and $AD$ is its altitude, the right angle is at $A$ and $D$ lies on the hypotenuse $BC$.
A coordinate model seems natural. Put
$$A=(0,0),\qquad B=(p,0),\qquad C=(0,q).$$
The hypotenuse has equation
$$\frac{x}{p}+\frac{y}{q}=1.$$
The foot of the perpendicular from the origin to this line is
$$D=\left(\frac{pq^{2}}{p^{2}+q^{2}},\frac{p^{2}q}{p^{2}+q^{2}}\right).$$
The circle with diameter $AD$ passes through $A$ and $D$. Its equation is
$$x^{2}+y^{2}-D_xx-D_yy=0.$$
Intersecting this circle with $AB$ gives $K=(D_x,0)$, hence
$$AK=D_x=\frac{pq^{2}}{p^{2}+q^{2}}.$$
Similarly,
$$AM=D_y=\frac{p^{2}q}{p^{2}+q^{2}}.$$
The crucial point is to determine $AL$. The line through $K$ and $M$ has intercept form
$$\frac{x}{D_x}+\frac{y}{D_y}=1.$$
The altitude $AD$ consists of points $(sD_x,sD_y)$. Substituting into the equation of $KM$ yields
$$s+s=1,$$
so $s=\frac12$. Thus $L$ is the midpoint of $AD$, and
$$AL=\frac{AD}{2}.$$
The geometric progression condition becomes
$$AL^2=AK\cdot AM.$$
This should reduce to a relation between $p$ and $q$. Carrying out the algebra gives
$$p^2+q^2=4pq.$$
Writing $r=q/p$,
$$r^2-4r+1=0,$$
hence
$$r=2\pm\sqrt3.$$
Since
$$\tan 15^\circ=2-\sqrt3,\qquad \tan 75^\circ=2+\sqrt3,$$
the acute angles must be $15^\circ$ and $75^\circ$.
The step most likely to hide an error is the determination of $L$. If $L$ were not the midpoint of $AD$, the final relation would change completely.
Problem Understanding
A right triangle $ABC$ has altitude $AD$ drawn from the right angle $A$ to the hypotenuse $BC$. A circle with diameter $AD$ meets the legs $AB$ and $AC$ again at $K$ and $M$. The chord $KM$ intersects $AD$ at $L$. The lengths $AK$, $AL$, and $AM$ satisfy
$$\frac{AK}{AL}=\frac{AL}{AM}.$$
We must determine the acute angles of the triangle.
This is a Type A problem. We must determine all triangles satisfying the condition and prove that no others do.
The core difficulty is expressing the three lengths in a manageable form and exploiting the geometry of the circle. The answer should be $15^\circ$ and $75^\circ$, because the geometric progression condition leads to the ratio of the legs satisfying
$$\frac{q}{p}=2\pm\sqrt3,$$
which is exactly the tangent of $15^\circ$ or $75^\circ$.
Proof Architecture
The first lemma states that if
$$D=\left(\frac{pq^{2}}{p^{2}+q^{2}},\frac{p^{2}q}{p^{2}+q^{2}}\right),$$
then
$$AK=\frac{pq^{2}}{p^{2}+q^{2}},\qquad AM=\frac{p^{2}q}{p^{2}+q^{2}}.$$
This follows from the equation of the circle with diameter $AD$.
The second lemma states that $L$ is the midpoint of $AD$.
This follows from solving simultaneously the equations of the lines $AD$ and $KM$.
The third lemma states that the geometric progression condition is equivalent to
$$p^{2}+q^{2}=4pq.$$
This follows by substituting the expressions for $AK$, $AL$, and $AM$.
The final step solves the resulting quadratic equation for $q/p$ and converts the result into angles.
The lemma most likely to fail under insufficient scrutiny is the second one, because the midpoint property is not visually obvious and must be proved algebraically.
Solution
Place the triangle in the coordinate plane:
$$A=(0,0),\qquad B=(p,0),\qquad C=(0,q),$$
where $p,q>0$.
The hypotenuse $BC$ has equation
$$\frac{x}{p}+\frac{y}{q}=1.$$
Since $AD$ is perpendicular to $BC$, the foot of the perpendicular from the origin to this line is
$$D=\left(\frac{pq^{2}}{p^{2}+q^{2}},\frac{p^{2}q}{p^{2}+q^{2}}\right).$$
Write
$$D=(D_x,D_y).$$
The circle with diameter $AD$ passes through $A$ and $D$, hence its equation is
$$x^{2}+y^{2}-D_xx-D_yy=0.$$
To find $K$, set $y=0$. Then
$$x^{2}-D_xx=0,$$
so the intersections with $AB$ are $A$ and
$$K=(D_x,0).$$
Therefore
$$AK=D_x=\frac{pq^{2}}{p^{2}+q^{2}}.$$
To find $M$, set $x=0$. Then
$$y^{2}-D_yy=0,$$
so
$$M=(0,D_y),$$
and hence
$$AM=D_y=\frac{p^{2}q}{p^{2}+q^{2}}.$$
The line through $K=(D_x,0)$ and $M=(0,D_y)$ has equation
$$\frac{x}{D_x}+\frac{y}{D_y}=1.$$
A point of the line $AD$ has the form
$$(x,y)=(tD_x,tD_y).$$
Substituting into the equation of $KM$ gives
$$t+t=1,$$
hence
$$t=\frac12.$$
Thus
$$L=\left(\frac{D_x}{2},\frac{D_y}{2}\right),$$
so $L$ is the midpoint of $AD$ and
$$AL=\frac{AD}{2}.$$
The length of the altitude is
$$AD=\sqrt{D_x^2+D_y^2} =\frac{pq}{\sqrt{p^{2}+q^{2}}},$$
therefore
$$AL=\frac{pq}{2\sqrt{p^{2}+q^{2}}}.$$
Since $AK$, $AL$, and $AM$ form a geometric progression,
$$AL^2=AK\cdot AM.$$
Substituting the expressions found above,
$$\frac{p^{2}q^{2}}{4(p^{2}+q^{2})} = \frac{pq^{2}}{p^{2}+q^{2}} \cdot \frac{p^{2}q}{p^{2}+q^{2}}.$$
After cancellation of the common factor $p^{2}q^{2}$,
$$\frac1{4(p^{2}+q^{2})} = \frac{pq}{(p^{2}+q^{2})^{2}}.$$
Multiplying by $(p^{2}+q^{2})^{2}$ yields
$$p^{2}+q^{2}=4pq.$$
Let
$$r=\frac{q}{p}.$$
Then
$$r^{2}-4r+1=0,$$
so
$$r=2\pm\sqrt3.$$
In the right triangle,
$$\tan\angle B=\frac{AC}{AB}=\frac{q}{p}=r.$$
Hence
$$\tan\angle B=2-\sqrt3 \quad\text{or}\quad \tan\angle B=2+\sqrt3.$$
Using
$$\tan 15^\circ=2-\sqrt3, \qquad \tan 75^\circ=2+\sqrt3,$$
we obtain
$$\angle B=15^\circ,\quad \angle C=75^\circ,$$
or the reverse.
Thus the acute angles of the triangle are
$$\boxed{15^\circ \text{ and } 75^\circ}.$$
Verification of Key Steps
The midpoint property of $L$ deserves an independent check. The points $K$ and $M$ have coordinates $(D_x,0)$ and $(0,D_y)$. Their connecting line is
$$\frac{x}{D_x}+\frac{y}{D_y}=1.$$
The midpoint of $AD$ is
$$\left(\frac{D_x}{2},\frac{D_y}{2}\right).$$
Substituting into the equation of $KM$ gives
$$\frac12+\frac12=1.$$
Hence the midpoint lies on $KM$, and since it also lies on $AD$, it is exactly $L$.
The formula for $AD$ can also be checked directly:
$$AD^{2} = \frac{p^{2}q^{4}+p^{4}q^{2}}{(p^{2}+q^{2})^{2}} = \frac{p^{2}q^{2}(p^{2}+q^{2})}{(p^{2}+q^{2})^{2}} = \frac{p^{2}q^{2}}{p^{2}+q^{2}}.$$
Taking square roots gives
$$AD=\frac{pq}{\sqrt{p^{2}+q^{2}}}.$$
The final angle determination can be verified numerically. If
$$r=2+\sqrt3\approx3.732,$$
then
$$\arctan r=75^\circ.$$
If
$$r=2-\sqrt3\approx0.268,$$
then
$$\arctan r=15^\circ.$$
Since the acute angles of a right triangle sum to $90^\circ$, these two values indeed form the required pair.
Alternative Approaches
A synthetic solution can be built from the fact that the circle has diameter $AD$. Since $\angle AKD=\angle AMD=90^\circ$, the quadrilaterals arising from the construction contain several similar right triangles. One can derive
$$AK=\frac{AD^{2}}{AB}, \qquad AM=\frac{AD^{2}}{AC},$$
using projections on the legs. The line $KM$ is the polar of $A$ with respect to the circle, which yields that $L$ is the midpoint of $AD$. Substituting into the geometric progression condition again leads to
$$AB^{2}+AC^{2}=4AB\cdot AC.$$
The coordinate approach is preferable because every length appears directly from a simple equation, and the midpoint property of $L$ emerges from a single linear substitution.