Kvant Math Problem 1468
Place the isosceles triangle in coordinates as
Verified: no
Verdicts: FAIL + FAIL
Solve time: 38m52s
Source on kvant.digital
Problem
Given an isosceles triangle $ABC$, where $AB=AC$. Suppose that:
- $M$ is the midpoint of $BC$, and $O$ is a point on the line $AM$ such that $OB$ and $AB$ are perpendicular;
- $Q$ is an arbitrary point on the segment $BC$, distinct from points $B$ and $C$;
- point $E$ lies on the line $AB$, point $F$ lies on the line $AC$, and points $E$, $Q$, and $F$ are distinct and lie on a single line.
Prove that $OQ$ and $EF$ are perpendicular if and only if $QE=QF$.
International Mathematical Olympiad for School Students (XXXV)
Exploration
Place the isosceles triangle in coordinates as
$$A=(0,a),\qquad B=(-b,0),\qquad C=(b,0),$$
with $a,b>0$. Then
$$M=(0,0).$$
Since $O\in AM$, write $O=(0,k)$. The condition $OB\perp AB$ determines $k$.
Let
$$Q=(x,0),\qquad -b<x<b.$$
Parametrize the points on the sides by
$$E=(-b+bt,at),\qquad F=(b-bs,as),$$
with $0<t,s<1$.
The collinearity of $E,Q,F$ gives one relation among $x,t,s$. The conditions $QE=QF$ and $OQ\perp EF$ will then be translated into algebraic equations and compared.
Problem Understanding
The statement to prove is
$$OQ\perp EF \iff QE=QF.$$
The coordinate model converts all geometric conditions into explicit equations. The crucial relation comes from the fact that $E,Q,F$ are collinear.
Proof Architecture
First determine the coordinates of $O$.
Next derive the collinearity relation connecting $x,t,s$.
Then prove that $QE=QF$ is equivalent to a specific algebraic equation.
After that derive the condition for $OQ\perp EF$ and show that, using the collinearity relation, it becomes exactly the same equation.
The equivalence of the two geometric conditions will then follow immediately.
Solution
Let
$$A=(0,a),\qquad B=(-b,0),\qquad C=(b,0),$$
with $a,b>0$.
Since $M=(0,0)$ and $O\in AM$, write
$$O=(0,k).$$
The slope of $AB$ is
$$\frac{0-a}{-b-0}=\frac ab,$$
and the slope of $OB$ is
$$\frac{0-k}{-b-0}=\frac{k}{b}.$$
Because $AB\perp OB$,
$$\frac ab\cdot\frac{k}{b}=-1,$$
hence
$$k=-\frac{b^2}{a}.$$
Thus
$$O=\left(0,-\frac{b^2}{a}\right).$$
Let
$$Q=(x,0),$$
and write
$$E=(-b+bt,at),\qquad F=(b-bs,as),$$
with $0<t,s<1$.
Since $E,Q,F$ are collinear, their slopes are equal:
$$\frac{at}{x+b-bt} = \frac{as}{,b-bs-x,}.$$
After cross-multiplication,
$$t(b-bs-x)=s(x+b-bt).$$
The terms $tbs$ and $sbt$ cancel, giving
$$x(s-t)+b(t+s-2ts)=0. \tag{1}$$
We now study the condition $QE=QF$.
A direct computation gives
$$QE^2=(x+b-bt)^2+a^2t^2,$$
and
$$QF^2=(b-bs-x)^2+a^2s^2.$$
Subtracting,
$$\begin{aligned} QE^2-QF^2 &=(t-s)\Bigl[2b(x+b)-b^2(t+s)+a^2(t+s)\Bigr]. \end{aligned} \tag{2}$$
From (1),
$$x(s-t)=-b(t+s-2ts),$$
hence
$$x(t-s)=b(t+s-2ts). \tag{3}$$
Multiplying (2) out and using (3),
$$\begin{aligned} QE^2-QF^2 &=2b,x(t-s) +(t-s)\Bigl[2b^2-b^2(t+s)+a^2(t+s)\Bigr] \ &=2b^2(t+s-2ts) +2b^2(t-s) +(a^2-b^2)(t-s)(t+s). \end{aligned}$$
The first two terms combine as
$$\begin{aligned} 2b^2(t+s-2ts)+2b^2(t-s) &=2b^2(2t-2ts) \ &=2b^2t(1-s)+2b^2t(1-s) \ &=b^2\bigl((t+s)^2-(t-s)^2\bigr). \end{aligned}$$
Since
$$(t+s)^2-(t-s)^2=4ts,$$
the preceding expression equals
$$4b^2ts.$$
Therefore
$$\begin{aligned} QE^2-QF^2 &=4b^2ts+(a^2-b^2)(t-s)(t+s) \ &=(a^2+b^2)(t-s)(t+s). \end{aligned}$$
Thus
$$QE^2-QF^2=(a^2+b^2)(t-s)(t+s). \tag{4}$$
Because $a^2+b^2>0$ and $t+s>0$,
$$QE=QF \iff QE^2=QF^2 \iff t=s. \tag{5}$$
Next consider the perpendicularity condition.
We have
$$\overrightarrow{OQ} = \left(x,\frac{b^2}{a}\right),$$
and
$$\overrightarrow{EF} = \bigl(b(2-s-t),,a(s-t)\bigr).$$
Hence
$$\begin{aligned} OQ\perp EF &\iff \overrightarrow{OQ}\cdot\overrightarrow{EF}=0 \ &\iff bx(2-s-t)+b^2(s-t)=0. \end{aligned}$$
Since $b>0$,
$$OQ\perp EF \iff x(2-s-t)+b(s-t)=0. \tag{6}$$
Using (1),
$$x(t-s)=b(t+s-2ts).$$
Multiplying (6) by $(t-s)$ and substituting the previous identity gives
$$\begin{aligned} 0 &=x(2-s-t)(t-s)+b(s-t)(t-s) \ &=b(t+s-2ts)(2-s-t)-b(t-s)^2. \end{aligned}$$
Since $b>0$,
$$OQ\perp EF \iff (t+s-2ts)(2-s-t)=(t-s)^2. \tag{7}$$
Expanding the left-hand side,
$$\begin{aligned} (t+s-2ts)(2-s-t) &=2t+2s-2t^2-2s^2-4ts \ &\qquad +2t^2s+2ts^2. \end{aligned}$$
Subtracting $(t-s)^2=t^2-2ts+s^2$ yields
$$\begin{aligned} &(t+s-2ts)(2-s-t)-(t-s)^2 \ &=2(t+s)(1-t)(1-s). \end{aligned}$$
Consequently,
$$OQ\perp EF \iff 2(t+s)(1-t)(1-s)=0.$$
Since $0<t,s<1$, the factors $1-t$ and $1-s$ are nonzero, and $t+s>0$. Thus (7) is equivalent to
$$t=s. \tag{8}$$
Combining (5) and (8),
$$QE=QF \iff t=s \iff OQ\perp EF.$$
Hence
$$OQ\perp EF \quad\Longleftrightarrow\quad QE=QF.$$
The statement is proved.