Kvant Math Problem 1351
Let the right angle of triangle $ABC$ be at $A$.
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Problem
Let $AB$ and $AC$ be the legs of a right triangle, with $AC\gt AB$. A point $E$ is chosen on $AC$, and a point $D$ on $BC$, so that $AB=AE=BD$. Prove that the triangle $ADE$ is right-angled if and only if the sides of the triangle $ABC$ are in the ratio $3:4:5$.
A. Parovyan
Tournament of Towns
Exploration
Let the right angle of triangle $ABC$ be at $A$. Put
$AB=b,\qquad AC=c,\qquad c>b.$
Since $AE=AB=b$, the point $E$ lies on $AC$ and
$CE=c-b.$
Since $BD=AB=b$, the point $D$ lies on $BC$ with
$\frac{BD}{BC}=\frac{b}{\sqrt{b^2+c^2}}.$
A coordinate model seems natural. Take
$A=(0,0),\qquad B=(b,0),\qquad C=(0,c).$
Then $E=(0,b)$.
The point $D$ divides $BC$ in the ratio $BD:BC=b:\sqrt{b^2+c^2}$, hence
$D=B+\frac{b}{\sqrt{b^2+c^2}}(C-B).$
Writing $s=\sqrt{b^2+c^2}$ gives
$D=\left(b-\frac{b^2}{s},,\frac{bc}{s}\right).$
The condition that $ADE$ be right-angled must now be translated into an algebraic equation. Since the right angle could occur at $A$, $D$, or $E$, all three possibilities must be checked.
At $A$ we would need $AD\perp AE$. Since $AE$ is vertical, this would force $D$ to lie on the $x$-axis, impossible because $D$ is an interior point of $BC$.
At $E$ we require
$\overrightarrow{EA}\cdot\overrightarrow{ED}=0.$
This gives $\frac{bc}{s}=b$, hence $c=s$, impossible.
Thus the only possible right angle is at $D$.
Computing
$\overrightarrow{DA}\cdot\overrightarrow{DE}=0$
produces a single equation in $b,c$. After simplification it becomes
$c^2=\frac{16}{9}b^2,$
which yields the ratio $3:4:5$.
The delicate point is the algebra leading from the orthogonality condition at $D$ to this ratio.
Problem Understanding
We are given a right triangle $ABC$ with legs $AB$ and $AC$, where $AC>AB$. Points $E\in AC$ and $D\in BC$ satisfy
$AB=AE=BD.$
We must prove that triangle $ADE$ is right-angled if and only if triangle $ABC$ has side lengths proportional to $3:4:5$.
This is a Type B problem. The statement itself is to be proved.
The core difficulty is identifying where the right angle of triangle $ADE$ can occur and converting that geometric condition into an equation relating the legs $AB$ and $AC$.
Proof Architecture
Let $b=AB$ and $c=AC$, and place the triangle in coordinates with $A=(0,0)$, $B=(b,0)$, $C=(0,c)$.
The first claim is that
$E=(0,b),\qquad D=\left(b-\frac{b^2}{s},,\frac{bc}{s}\right),\qquad s=\sqrt{b^2+c^2},$
because $AE=b$ and $BD=b$.
The second claim is that the right angle of triangle $ADE$ cannot be at $A$ or at $E$; this follows from direct orthogonality tests.
The third claim is that $ADE$ is right-angled exactly when
$\overrightarrow{DA}\cdot\overrightarrow{DE}=0,$
which simplifies to
$9c^2=16b^2.$
The final claim is that $9c^2=16b^2$ is equivalent to
$AB:AC:BC=3:4:5.$
The hardest direction is deriving the ratio from the orthogonality condition at $D$. That is the step most likely to fail under scrutiny.
Solution
Let
$AB=b,\qquad AC=c,\qquad c>b,$
and place the triangle in the coordinate plane by
$A=(0,0),\qquad B=(b,0),\qquad C=(0,c).$
Denote
$s=BC=\sqrt{b^2+c^2}.$
Since $AE=b$ and $E\in AC$,
$E=(0,b).$
Since $BD=b$ and $D\in BC$, the point $D$ divides the segment $BC$ in the ratio
$\frac{BD}{BC}=\frac{b}{s}.$
Hence
$$=\left(b-\frac{b^2}{s},,\frac{bc}{s}\right).$$
Assume first that triangle $ADE$ is right-angled.
We determine where its right angle can be.
If the right angle were at $A$, then $AD\perp AE$. Since $AE$ is the vertical line $x=0$, the line $AD$ would have to be horizontal. Thus the $y$-coordinate of $D$ would be $0$. But
$\frac{bc}{s}>0,$
a contradiction.
If the right angle were at $E$, then
$$$$
Now
$$$$
and
$$=\left(b-\frac{b^2}{s},,\frac{bc}{s}-b\right).$$
Therefore
$$=-b\left(\frac{bc}{s}-b\right).$$
The condition of orthogonality becomes
$$$$
hence $c=s$, impossible because
$$$$
Thus the right angle can only be at $D$.
Consequently,
$$$$
We have
$$=\left(-b+\frac{b^2}{s},-\frac{bc}{s}\right),$$
and
$$=\left(-b+\frac{b^2}{s},,b-\frac{bc}{s}\right).$$
Writing
$$$$
the orthogonality condition becomes
$$$$
Substituting $u=-b+b^2/s$ and multiplying by $s^2$ gives
$$$$
Since $b>0$,
$$$$
Expanding,
$$$$
Using $s^2=b^2+c^2$,
$$$$
Since $b^2+c^2=s^2$,
$$$$
Because $s>0$,
$$$$
Squaring,
$$$$
Hence
$$$$
so
$$$$
Since $c>0$,
$$$$
Therefore
$$$$
Then
$$=\sqrt{b^2+\frac{16}{9}b^2} =\frac53,b.$$
Thus
$$$$
This proves the necessity.
Conversely, suppose
$$$$
Write
$$$$
Then
$$=\left(\frac65k,\frac{12}5k\right),$$
and
$$$$
Therefore
$$=\left(-\frac65k,-\frac{12}5k\right),$$
and
$$=\left(-\frac65k,\frac35k\right).$$
Their dot product equals
$$+\left(-\frac{12}5k\right)\left(\frac35k\right) =\frac{36}{25}k^2-\frac{36}{25}k^2 =0.$$
Hence $DA\perp DE$, so triangle $ADE$ is right-angled.
We have proved both implications. This completes the proof.
∎
Verification of Key Steps
The first delicate step is excluding a right angle at $E$. The orthogonality condition there is
$$=-b\left(\frac{bc}{s}-b\right)=0.$$
Since $b>0$, this yields $c=s$. Because
$$$$
we have $s>c$, so such a right angle is impossible. Any argument that merely compares lengths informally risks overlooking this strict inequality.
The second delicate step is simplifying
$$$$
Expanding and substituting $s^2=b^2+c^2$ gives
$$$$
Replacing $b^2+c^2$ by $s^2$ yields
$$$$
Since $s>0$, division by $s$ is legitimate and produces
$$$$
This is the exact point where positivity of $s$ is required.
The third delicate step is passing from
$$$$
to
$$$$
After expansion, one obtains
$$$$
Subtracting $c^2$ from both sides gives
$$$$
and dividing by $c>0$ yields
$$$$
No additional solutions are lost because $c$ is a side length.
Alternative Approaches
A synthetic solution can be built from Stewart's theorem or from ratios on the hypotenuse. Since
$$$$
the point $D$ is determined by the ratio $BD:DC$. Expressing $AD$ through Stewart's theorem and using $AE=AB$ makes triangle $ADE$ depend on a single parameter, namely $AC/AB$. Applying the Pythagorean condition in triangle $ADE$ then leads to the same equation $3c=4b$.
The coordinate method is preferable because the locations of $D$ and $E$ are explicit, the possible positions of the right angle are checked directly, and the condition $DA\perp DE$ converts immediately into a single algebraic relation whose simplification is straightforward.