Subfield: Geometry


Question: There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that

$$
\angle P A B+\angle P D C=\angle P B C+\angle P A D=\angle P C D+\angle P B A=\angle P D A+\angle P C B=90^{\circ} \tag{1}
$$

if and only if the diagonals $A C$ and $B D$ are perpendicular.

Solution:
For a point $P$ in $A B C D$ which satisfies (1), let $K, L, M, N$ be the feet of perpendiculars from $P$ to lines $A B, B C, C D, D A$, respectively. Note that $K, L, M, N$ are interior to the sides as all angles in (1) are acute. The cyclic quadrilaterals $A K P N$ and $D N P M$ give

$$
\angle P A B+\angle P D C=\angle P N K+\angle P N M=\angle K N M .
$$

Analogously, $\angle P B C+\angle P A D=\angle L K N$ and $\angle P C D+\angle P B A=\angle M L K$. Hence the equalities (1) imply $\angle K N M=\angle L K N=\angle M L K=90^{\circ}$, so that $K L M N$ is a rectangle. The converse also holds true, provided that $K, L, M, N$ are interior to sides $A B, B C, C D, D A$.

(i) Suppose that there exists a point $P$ in $A B C D$ such that $K L M N$ is a rectangle. We show that $A C$ and $B D$ are parallel to the respective sides of $K L M N$.

Let $O_{A}$ and $O_{C}$ be the circumcentres of the cyclic quadrilaterals $A K P N$ and $C M P L$. Line $O_{A} O_{C}$ is the common perpendicular bisector of $L M$ and $K N$, therefore $O_{A} O_{C}$ is parallel to $K L$ and $M N$. On the other hand, $O_{A} O_{C}$ is the midline in the triangle $A C P$ that is parallel to $A C$. Therefore the diagonal $A C$ is parallel to the sides $K L$ and $M N$ of the rectangle. Likewise, $B D$ is parallel to $K N$ and $L M$. Hence $A C$ and $B D$ are perpendicular.

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(ii) Suppose that $A C$ and $B D$ are perpendicular and meet at $R$. If $A B C D$ is a rhombus, $P$ can be chosen to be its centre. So assume that $A B C D$ is not a rhombus, and let $B R<D R$ without loss of generality.

Denote by $U_{A}$ and $U_{C}$ the circumcentres of the triangles $A B D$ and $C D B$, respectively. Let $A V_{A}$ and $C V_{C}$ be the diameters through $A$ and $C$ of the two circumcircles. Since $A R$ is an altitude in triangle $A D B$, lines $A C$ and $A V_{A}$ are isogonal conjugates, i. e. $\angle D A V_{A}=\angle B A C$. Now $B R<D R$ implies that ray $A U_{A}$ lies in $\angle D A C$. Similarly, ray $C U_{C}$ lies in $\angle D C A$. Both diameters $A V_{A}$ and $C V_{C}$ intersect $B D$ as the angles at $B$ and $D$ of both triangles are acute. Also $U_{A} U_{C}$ is parallel to $A C$ as it is the perpendicular bisector of $B D$. Hence $V_{A} V_{C}$ is parallel to $A C$, too. We infer that $A V_{A}$ and $C V_{C}$ intersect at a point $P$ inside triangle $A C D$, hence inside $A B C D$.



Construct points $K, L, M, N, O_{A}$ and $O_{C}$ in the same way as in the introduction. It follows from the previous paragraph that $K, L, M, N$ are interior to the respective sides. Now $O_{A} O_{C}$ is a midline in triangle $A C P$ again. Therefore lines $A C, O_{A} O_{C}$ and $U_{A} U_{C}$ are parallel.

The cyclic quadrilateral $A K P N$ yields $\angle N K P=\angle N A P$. Since $\angle N A P=\angle D A U_{A}=$ $\angle B A C$, as specified above, we obtain $\angle N K P=\angle B A C$. Because $P K$ is perpendicular to $A B$, it follows that $N K$ is perpendicular to $A C$, hence parallel to $B D$. Likewise, $L M$ is parallel to $B D$.

Consider the two homotheties with centres $A$ and $C$ which transform triangles $A B D$ and $C D B$ into triangles $A K N$ and $C M L$, respectively. The images of points $U_{A}$ and $U_{C}$ are $O_{A}$ and $O_{C}$, respectively. Since $U_{A} U_{C}$ and $O_{A} O_{C}$ are parallel to $A C$, the two ratios of homothety are the same, equal to $\lambda=A N / A D=A K / A B=A O_{A} / A U_{A}=C O_{C} / C U_{C}=C M / C D=C L / C B$. It is now straightforward that $D N / D A=D M / D C=B K / B A=B L / B C=1-\lambda$. Hence $K L$ and $M N$ are parallel to $A C$, implying that $K L M N$ is a rectangle and completing the proof.

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For a point $P$ distinct from $A, B, C, D$, let circles $(A P D)$ and $(B P C)$ intersect again at $Q(Q=P$ if the circles are tangent). Next, let circles $(A Q B)$ and $(C Q D)$ intersect again at $R$. We show that if $P$ lies in $A B C D$ and satisfies (1) then $A C$ and $B D$ intersect at $R$ and are perpendicular; the converse is also true. It is convenient to use directed angles. Let $\measuredangle(U V, X Y)$ denote the angle of counterclockwise rotation that makes line $U V$ parallel to line $X Y$. Recall that four noncollinear points $U, V, X, Y$ are concyclic if and only if $\measuredangle(U X, V X)=\measuredangle(U Y, V Y)$.

The definitions of points $P, Q$ and $R$ imply

$$
\begin{aligned}
\measuredangle(A R, B R) & =\measuredangle(A Q, B Q)=\measuredangle(A Q, P Q)+\measuredangle(P Q, B Q)=\measuredangle(A D, P D)+\measuredangle(P C, B C), \\
\measuredangle(C R, D R) & \measuredangle(C Q, D Q)=\measuredangle(C Q, P Q)+\measuredangle(P Q, D Q)=\measuredangle(C B, P B)+\measuredangle(P A, D A), \\
\measuredangle(B R, C R) & =\measuredangle(B R, R Q)+\measuredangle(R Q, C R)=\measuredangle(B A, A Q)+\measuredangle(D Q, C D) \\
& =\measuredangle(B A, A P)+\measuredangle(A P, A Q)+\measuredangle(D Q, D P)+\measuredangle(D P, C D) \\
& =\measuredangle(B A, A P)+\measuredangle(D P, C D) .
\end{aligned}
$$

Observe that the whole construction is reversible. One may start with point $R$, define $Q$ as the second intersection of circles $(A R B)$ and $(C R D)$, and then define $P$ as the second intersection of circles $(A Q D)$ and $(B Q C)$. The equalities above will still hold true.



Assume in addition that $P$ is interior to $A B C D$. Then

$$
\begin{gathered}
\measuredangle(A D, P D)=\angle P D A, \measuredangle(P C, B C)=\angle P C B, \measuredangle(C B, P B)=\angle P B C, \measuredangle(P A, D A)=\angle P A D \\
\measuredangle(B A, A P)=\angle P A B, \measuredangle(D P, C D)=\angle P D C .
\end{gathered}
$$

(i) Suppose that $P$ lies in $A B C D$ and satisfies (1). Then $\measuredangle(A R, B R)=\angle P D A+\angle P C B=90^{\circ}$ and similarly $\measuredangle(B R, C R)=\measuredangle(C R, D R)=90^{\circ}$. It follows that $R$ is the common point of lines $A C$ and $B D$, and that these lines are perpendicular.

(ii) Suppose that $A C$ and $B D$ are perpendicular and intersect at $R$. We show that the point $P$ defined by the reverse construction (starting with $R$ and ending with $P$ ) lies in $A B C D$. This is enough to finish the solution, because then the angle equalities above will imply (1).

One can assume that $Q$, the second common point of circles $(A B R)$ and $(C D R)$, lies in $\angle A R D$. Then in fact $Q$ lies in triangle $A D R$ as angles $A Q R$ and $D Q R$ are obtuse. Hence $\angle A Q D$ is obtuse, too, so that $B$ and $C$ are outside circle $(A D Q)(\angle A B D$ and $\angle A C D$ are acute).

Now $\angle C A B+\angle C D B=\angle B Q R+\angle C Q R=\angle C Q B$ implies $\angle C A B<\angle C Q B$ and $\angle C D B<$ $\angle C Q B$. Hence $A$ and $D$ are outside circle $(B C Q)$. In conclusion, the second common point $P$ of circles $(A D Q)$ and $(B C Q)$ lies on their arcs $A D Q$ and $B C Q$.

We can assume that $P$ lies in $\angle C Q D$. Since

$$
\begin{gathered}
\angle Q P C+\angle Q P D=\left(180^{\circ}-\angle Q B C\right)+\left(180^{\circ}-\angle Q A D\right)= \\
=360^{\circ}-(\angle R B C+\angle Q B R)-(\angle R A D-\angle Q A R)=360^{\circ}-\angle R B C-\angle R A D>180^{\circ},
\end{gathered}
$$

point $P$ lies in triangle $C D Q$, and hence in $A B C D$. The proof is complete.

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