Kvant Math Problem 851
Place the square in a coordinate system so that computations can be expressed in terms of two parameters.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m53s
Source on kvant.digital
Problem
The side length of square $ABCD$ is 1. On the sides $AB$ and $AD$, points $P$ and $Q$ are chosen such that the perimeter of triangle $APQ$ is equal to 2. Prove that $\widehat{PCQ}=45^\circ$.
A. B. Khodulyev
Exploration
Place the square in a coordinate system so that computations can be expressed in terms of two parameters. Let $A=(0,0)$, $B=(1,0)$, $D=(0,1)$, $C=(1,1)$. Let $P=(x,0)$ with $0\le x\le 1$ and $Q=(0,y)$ with $0\le y\le 1$. The condition on the perimeter becomes $x+y+\sqrt{x^2+y^2}=2$, which couples $x$ and $y$ in a nonlinear way.
The target angle is $\angle PCQ$, which is the angle between vectors $(x-1,-1)$ and $(-1,y-1)$. A direct cosine computation reduces the geometric statement to an algebraic identity involving $x$ and $y$. The main obstacle is eliminating the square root in the perimeter condition in a way that produces a symmetric relation between $x$ and $y$. Introducing $s=x+y$ and $t=xy$ suggests itself, since the perimeter condition involves $x^2+y^2$, which can be expressed through $s$ and $t$.
The key point is whether the perimeter constraint forces a rigid algebraic relation strong enough to make the cosine simplify to a constant independent of the specific positions of $P$ and $Q$. The computation indicates that the expression for $\cos \angle PCQ$ simplifies exactly to $1/\sqrt{2}$ once the constraint is fully translated into symmetric form.
Problem Understanding
This is a Type B problem. One must prove that the angle $\angle PCQ$ is constant and equal to $45^\circ$ under the condition that $P$ lies on $AB$, $Q$ lies on $AD$, and $AP+AQ+PQ=2$ in a unit square.
The difficulty lies in showing that a nonlinear geometric constraint forces a fixed angular value at a vertex not directly involved in the constraint. The expectation is that the perimeter condition encodes a hidden quadratic relation between $x$ and $y$ that rigidifies the configuration.
Proof Architecture
The proof begins by introducing coordinates for the square and parameterizing $P$ and $Q$ by $x$ and $y$. The first lemma translates the perimeter condition into a relation between $x+y$ and $xy$.
The second lemma expresses $\cos \angle PCQ$ in terms of $x$ and $y$, then in terms of symmetric quantities.
The third lemma shows that after substitution of the perimeter relation, the cosine simplifies to $1/\sqrt{2}$.
The hardest step is the algebraic elimination of the square root and correct handling of symmetric expressions, where cancellation must be exact and no hidden sign ambiguity may remain.
Solution
Let $A=(0,0)$, $B=(1,0)$, $D=(0,1)$, $C=(1,1)$. Let $P=(x,0)$ and $Q=(0,y)$ with $0\le x,y\le 1$. Then $AP=x$, $AQ=y$, and $PQ=\sqrt{x^2+y^2}$. The condition on the perimeter gives
$$x+y+\sqrt{x^2+y^2}=2.$$
Set $s=x+y$. Then $\sqrt{x^2+y^2}=2-s$, and squaring yields
$$x^2+y^2=4-4s+s^2.$$
Since $x^2+y^2=s^2-2xy$, substitution gives
$$s^2-2xy=4-4s+s^2,$$
hence
$$xy=2s-2.$$
Now consider vectors
$$\overrightarrow{CP}=(x-1,-1),\quad \overrightarrow{CQ}=(-1,y-1).$$
Their dot product is
$$\overrightarrow{CP}\cdot\overrightarrow{CQ}=(x-1)(-1)+(-1)(y-1)=2-x-y=2-s.$$
Their squared lengths are
$$|\overrightarrow{CP}|^2=(x-1)^2+1=x^2-2x+2,$$
$$|\overrightarrow{CQ}|^2=(y-1)^2+1=y^2-2y+2.$$
Hence
$$\cos \angle PCQ=\frac{2-s}{\sqrt{(x^2-2x+2)(y^2-2y+2)}}.$$
To evaluate the denominator, expand
$$(x^2-2x+2)(y^2-2y+2).$$
This equals
$$x^2y^2 -2x^2y +2x^2 -2xy^2 +4xy -4x +2y^2 -4y +4.$$
Group symmetric expressions:
$$x^2y^2=t^2,\quad x^2y+xy^2=xy(x+y)=ts,\quad x^2+y^2=s^2-2t.$$
Rewriting the expression yields
$$t^2 -2ts + 2(s^2-2t) + 4t -4s +4.$$
Substituting $t=2s-2$ gives
$$(x^2-2x+2)(y^2-2y+2)=2(s-2)^2.$$
Therefore
$$\cos \angle PCQ=\frac{2-s}{\sqrt{2}(2-s)}=\frac{1}{\sqrt{2}},$$
since $s=x+y\le 2$, and equality $s=2$ would force $x=y=1$, which is incompatible with $PQ>0$ in the given perimeter condition, hence $2-s>0$.
Thus $\angle PCQ=45^\circ$, completing the proof. ∎
Verification of Key Steps
The critical point is the elimination of the square root in the perimeter constraint. Squaring $x+y+\sqrt{x^2+y^2}=2$ is valid because all terms are nonnegative, so no extraneous sign ambiguity arises in replacing $\sqrt{x^2+y^2}$ with $2-s$.
The second delicate step is the reduction of $(x^2-2x+2)(y^2-2y+2)$ to a symmetric expression. Each mixed term $x^2y$ and $xy^2$ is correctly grouped into $xy(x+y)$, ensuring that no asymmetric residue remains. Substitution of $t=2s-2$ is consistent with the earlier derived constraint and yields an expression depending only on $s$.
Finally, cancellation in the cosine expression is justified because $2-s>0$ under the perimeter condition with $P\ne A$ and $Q\ne A$, ensuring no division by zero occurs.
Alternative Approaches
A synthetic approach replaces coordinates with an affine transformation sending $A$ to the origin and $AB,AD$ to coordinate axes, then interprets the perimeter condition as a level set of a norm function. The result can then be derived using homogeneity properties of quadratic forms, leading to the same constant angle without explicit expansion, though the algebraic structure remains implicitly identical.