Kvant Math Problem 976
Place the square in coordinates:
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m22s
Source on kvant.digital
Problem
From vertex $A$ of the square $ABCD$, two rays are drawn forming an angle of $45^\circ$ between them. One ray intersects side $BC$ at point $E$ and diagonal $BD$ at point $P$; the other intersects side $CD$ at point $F$ and diagonal $BD$ at point $O$. Prove that the area of triangle $AEF$ is twice the area of triangle $APO$.
E. G. Gotman
Exploration
Place the square in coordinates:
$$A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1).$$
The diagonal $BD$ has equation
$$x+y=1.$$
Let the ray through $E$ and $P$ have slope $m>0$. Its equation is
$$y=mx.$$
Since it meets $BC$, we must have $m\le 1$.
The second ray forms an angle $45^\circ$ with the first. If its slope is $n$, then
$$\tan 45^\circ=\frac{n-m}{1+mn},$$
hence
$$n=\frac{m+1}{1-m}.$$
Since $m\le 1$, this ray indeed meets $CD$.
Compute the intersection points:
$$E=(1,m),\qquad P=\left(\frac1{1+m},\frac{m}{1+m}\right),$$
$$F=\left(\frac1n,1\right),\qquad O=\left(\frac1{1+n},\frac{n}{1+n}\right).$$
The area of a triangle with one vertex at the origin equals half the absolute value of the determinant of the other two vertices. Thus
$$[AEF] =\frac12\left| \begin{vmatrix} 1&m\[2mm] 1/n&1 \end{vmatrix} \right| =\frac12\left(1-\frac mn\right).$$
Likewise
$$[APO] =\frac12 \left| \begin{vmatrix} \dfrac1{1+m}&\dfrac{m}{1+m}\[3mm] \dfrac1{1+n}&\dfrac{n}{1+n} \end{vmatrix} \right| = \frac12\cdot \frac{n-m}{(1+m)(1+n)}.$$
The relation $n=(m+1)/(1-m)$ should convert both expressions to the same parameter. Carrying this out gives
$$1-\frac mn=\frac{1+m^2}{1+m},$$
while
$$\frac{n-m}{(1+m)(1+n)} =\frac{1+m^2}{2(1+m)}.$$
Hence
$$[AEF]=2[APO].$$
The step most likely to hide an error is the algebraic simplification after substituting
$$n=\frac{m+1}{1-m}.$$
That computation must be checked carefully.
Problem Understanding
A square $ABCD$ is given. From $A$ two rays are drawn making an angle of $45^\circ$. One ray meets side $BC$ at $E$ and diagonal $BD$ at $P$. The other meets side $CD$ at $F$ and diagonal $BD$ at $O$. The goal is to prove that
$$[AEF]=2[APO],$$
where brackets denote area.
This is a Type B problem, a pure proof.
The core difficulty is relating the geometry of the two rays to the positions of their intersections with the diagonal $BD$. A coordinate model turns the $45^\circ$ condition into an explicit relation between the slopes of the rays, after which both areas can be computed directly.
Proof Architecture
Let the square have coordinates $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$.
The first lemma states that if the first ray has slope $m$, then the second ray has slope
$$n=\frac{m+1}{1-m}.$$
This follows from the tangent formula for the angle between two lines.
The second lemma computes the coordinates of $E,F,P,O$:
$$E=(1,m),\quad F=\left(\frac1n,1\right),\quad P=\left(\frac1{1+m},\frac{m}{1+m}\right),\quad O=\left(\frac1{1+n},\frac{n}{1+n}\right).$$
These come from intersecting the rays with the sides of the square and with the diagonal $x+y=1$.
The third lemma computes the two areas:
$$[AEF]=\frac12\left(1-\frac mn\right),$$
$$[APO] = \frac12\cdot \frac{n-m}{(1+m)(1+n)}.$$
This uses the determinant formula for the area of a triangle with one vertex at the origin.
The final step substitutes the expression for $n$ and verifies
$$[AEF]=2[APO].$$
The most delicate point is the algebraic reduction after substituting
$$n=\frac{m+1}{1-m}.$$
Solution
Let
$$A=(0,0),\quad B=(1,0),\quad C=(1,1),\quad D=(0,1).$$
Then the diagonal $BD$ has equation
$$x+y=1.$$
Let the ray passing through $E$ and $P$ have slope $m$. Since it intersects side $BC$, its equation is
$$y=mx,$$
with $0<m\le1$.
Let the second ray, passing through $F$ and $O$, have slope $n$. Since the angle between the rays equals $45^\circ$,
$$\frac{n-m}{1+mn}=\tan 45^\circ=1.$$
Hence
$$n-m=1+mn,$$
and therefore
$$n=\frac{m+1}{1-m}.$$
The point $E$ lies on $BC$, whose equation is $x=1$. Thus
$$E=(1,m).$$
The point $F$ lies on $CD$, whose equation is $y=1$. Since the second ray is $y=nx$,
$$F=\left(\frac1n,1\right).$$
To find $P$, solve
$$y=mx,\qquad x+y=1.$$
Substituting gives
$$x+mx=1,$$
so
$$x=\frac1{1+m},\qquad y=\frac{m}{1+m}.$$
Hence
$$P=\left(\frac1{1+m},\frac{m}{1+m}\right).$$
Similarly, $O$ is the intersection of
$$y=nx$$
with
$$x+y=1.$$
Therefore
$$O=\left(\frac1{1+n},\frac{n}{1+n}\right).$$
Since $A$ is the origin, the area of a triangle with vertices $A,X,Y$ equals
$$\frac12\left| \begin{vmatrix} x_X&y_X\ x_Y&y_Y \end{vmatrix} \right|.$$
Applying this to $AEF$,
$$[AEF] = \frac12 \left| \begin{vmatrix} 1&m\[2mm] 1/n&1 \end{vmatrix} \right| = \frac12\left(1-\frac mn\right).$$
Applying the same formula to $APO$,
$$[APO] = \frac12 \left| \begin{vmatrix} \dfrac1{1+m}&\dfrac{m}{1+m}\[3mm] \dfrac1{1+n}&\dfrac{n}{1+n} \end{vmatrix} \right|.$$
The determinant equals
$$\frac{n}{(1+m)(1+n)} - \frac{m}{(1+m)(1+n)} = \frac{n-m}{(1+m)(1+n)}.$$
Hence
$$[APO] = \frac12\cdot \frac{n-m}{(1+m)(1+n)}.$$
Now substitute
$$n=\frac{m+1}{1-m}.$$
First,
$$1-\frac mn = 1-\frac{m(1-m)}{1+m} = \frac{1+m-m+m^2}{1+m} = \frac{1+m^2}{1+m}.$$
Next,
$$n-m = \frac{m+1}{1-m}-m = \frac{1+m^2}{1-m},$$
and
$$1+n = 1+\frac{m+1}{1-m} = \frac2{1-m}.$$
Therefore
$$\frac{n-m}{(1+m)(1+n)} = \frac{\frac{1+m^2}{1-m}} {(1+m)\frac2{1-m}} = \frac{1+m^2}{2(1+m)}.$$
Thus
$$[APO] = \frac12\cdot \frac{1+m^2}{2(1+m)} = \frac{1+m^2}{4(1+m)},$$
while
$$[AEF] = \frac12\cdot \frac{1+m^2}{1+m} = \frac{1+m^2}{2(1+m)}.$$
Comparing the two expressions,
$$[AEF]=2[APO].$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the slope relation. For lines of slopes $m$ and $n$,
$$\tan\theta=\frac{n-m}{1+mn}.$$
Since $\theta=45^\circ$, the right-hand side equals $1$. Solving
$$n-m=1+mn$$
gives
$$n(1-m)=1+m,$$
hence
$$n=\frac{1+m}{1-m}.$$
No sign ambiguity occurs because both rays lie inside the first quadrant.
The second delicate step is the determinant for $[APO]$. Using the coordinates directly,
$$\det(P,O) = \frac1{1+m}\cdot\frac n{1+n} - \frac m{1+m}\cdot\frac1{1+n}.$$
Factoring out $(1+m)^{-1}(1+n)^{-1}$ leaves $n-m$, producing
$$\frac{n-m}{(1+m)(1+n)}.$$
A common mistake is to forget one denominator factor.
The third delicate step is the substitution of $n$. Computing separately,
$$n-m=\frac{1+m^2}{1-m},$$
and
$$1+n=\frac2{1-m}.$$
Substituting these values yields
$$\frac{n-m}{(1+m)(1+n)} = \frac{1+m^2}{2(1+m)},$$
which gives exactly the factor $2$ between the two areas.
Alternative Approaches
A synthetic proof can be obtained by projecting points on the diagonal $BD$. The diagonal divides the square into two congruent right isosceles triangles, and the condition that the rays differ by $45^\circ$ allows the use of angle-chasing inside triangle $ABD$. Expressing the areas through distances from points on $BD$ to the sides of the square leads to proportional segments on the diagonal and eventually to the same factor $2$.
The coordinate method is preferable because the geometry is encoded by a single parameter $m$. The $45^\circ$ condition becomes one algebraic identity, every relevant point has an explicit coordinate, and both areas reduce to straightforward determinant computations.