Sparrow’s lemmas

Sparrow’s lemmas have been known to Russian Olympiad participants since at least 2016. Page was made by vladimir.shelomovskii@gmail.com, vvsss

Sparrow's Lemma 1

Let triangle $ABC$ with circumcircle $\Omega$ and points $D$ and $E$ on the sides $AB$ and $AC,$ respectively be given.

Let $K \in \Omega$ be the midpoint of the arc $BC$ which contain the point $A.$

Prove that $BD = CE$ iff points $A, D, E,$ and $K$ are concyclic.

Proof

$BK = CK, \angle ABK = \angle ACK.$

Let $BD = CE \implies \triangle BKD = \triangle CKE \implies$

$\angle KDA = \angle KEA \implies A, D, E,$ and $K$ are concyclic.

Let $A, D, E,$ and $K$ are concyclic $\implies \angle KDA = \angle KEA \implies$

$\angle BKD = \angle CKE \implies \triangle BKD = \triangle CKE \implies BD = CE.$

Sparrow’s Lemma 2

Let triangle $ABC$ with circumcircle $\Omega$ and points $D$ and $E$ on the sides $AB$ and $AC,$ respectively be given.

Let $I$ be the incenter.

Prove that $BD + CE = BC$ iff points $A, D, E,$ and $I$ are concyclic.

Proof

1. Let points $A, D, E,$ and $I$ are concyclic.

Denote $F \in BC$ such $BD = BF, \varphi = \angle ADI.$

So point $F$ is symmetric to $D$ with respect to $BI \implies \angle BDI = 180^\circ - \varphi, \angle IEC = \varphi.$ $\triangle BDI = \triangle BFI \implies DI = FI, \angle CFI = \varphi.$ $\angle DAI = \angle EAI \implies DI = EI = FI.$ $\triangle CIF = \triangle CIE \implies CE = CF \implies BD + CE = BC \blacksquare$ 2. Let $BD + CE = BC \implies$ there is point $F$ such that $BF = BD, CF = CE \implies$ $\triangle CIF = \triangle CIE, \triangle BIF = \triangle BID \implies$ $180^\circ = \angle BFI + \angle CFI = \angle BDI + \angle CEI = \angle ADI + \angle AEI \blacksquare$

Sparrow’s Lemma 3

Let lines $\ell$ and $\ell'$ and points $A_0 \in \ell$ and $B_0 \in \ell'$ be given, $O = \ell \cap \ell'.$

Points $A$ and $B$ moves along $\ell$ and $\ell',$ respectively with fixed speeds. At moment $t = 0,$ $A = A_0, B = O$ , at moment $t_0$ $A = O, B = B_0.$

Prove that circle $\Omega = \odot OAB$ contain fixed point ( $P$ ).

Proof

Let $\omega$ be the circle contains $O$ and $B_0$ and tangent to $\ell.$ Let $\omega'$ be the circle contains $O$ and $A_0$ and tangent to $\ell'.$ $P = \omega \cap \omega' \ne O.$ It is known that $P$ is the spiral center of spiral similarity $T$ mapping segment $A_0O$ to $OB_0.$ The ratio of the speeds of points $A$ and $B$ is $\frac{AA_0}{BO} = \frac{OA_0}{B_0O},$ so $T$ mapping segment $AA_0$ to $BO.$ Therefore $\Omega$ contain the spiral center $P \blacksquare$

Corollary 1

Lemma 1 is partial case of Lemma 3 with spiral center $K,$ equal speeds and two positions of the pare moving points - $D,E$ and $B,C.$

Corollary 2

Lemma 2 is partial case of Lemma 3 with spiral center $I,$ and equal speeds (from $D$ to $A$ and from $E$ to $C$ ). Start positions of these points are $D$ and $E.$

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Sparrow’s lemmas

Sparrow's Lemma 1

Sparrow’s Lemma 2

Sparrow’s Lemma 3