Difference between revisions of "Sparrow’s lemmas"

Revision as of 12:20, 21 September 2025

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 $E.$

Sparrow’s Lemma 3A

Let lines $\ell$ and $\ell'$ be given, $O = \ell \cap \ell'.$

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

Prove that center $Q$ of the circle $\Omega = \odot OAB$ moves along fixed line with fixed speed.

Proof

$\frac{OA}{OB} = const \implies \angle OAB = \alpha = const.$ So direction of line $AB$ is fixed for given motion.

Let $m'$ be the tangent to $\Omega$ at point $O.$ Angle between $m'$ and $l'$ is $\alpha$ so $m'$ is the fixed line which is antiparallel to line $AB.$

$QO \perp m'$ so line $m \perp m'$ is the locus $Q.$

$\angle QOA$ is fixed, so $\frac{OQ}{OA}$ is fixed and $Q$ moves with fixed speed.

vladimir.shelomovskii@gmail.com, vvsss

Russian Math Olympiad 2011

Let triangle $ABC$ with circumcircle $\Omega$ be given. Let $K \in \Omega$ be the midpoint of the arc $BC$ which contain the point $A, M$ be the midpoint of $BC, I_B$ be incenter of $\triangle ABM, I_C$ be incenter of $\triangle ACM.$

Prove that points $A, I_B, I_C,$ and $K$ are concyclic.

Proof

Denote $\omega = \odot AI_BI_C, D = \omega \cap AB,$ $E = \omega \cap AC, F = \omega \cap AM.$

We use Lemma 2 for $\triangle ABM$ and get $BD + FM = BM.$

We use Lemma 2 for $\triangle ACM$ and get $CE + FM = CM = BM \implies BD = CE.$

We use Lemma 1 for $\triangle ABC$ and get result: points $A,D,E,$ and $K$ are concyclic $\blacksquare$

Russian Math Olympiad 2005

Let triangle $ABC$ with circumcircle $\Omega$ and incircle $\omega$ be given. Let $A', B',$ and $C'$ be the tangent points of the excircles of $\triangle ABC$ with the corresponding sides. Let $D, E,$ and $F$ be the tangent points of the incircle of $\triangle ABC.$ The circumscribed circles of triangles $\triangle A'B'C, \triangle AB'C',$ and $\triangle A'BC'$ intersect $\Omega$ a second time at points $D', E',$ and $F',$ respectively.

Prove that $\triangle DEF \sim \triangle D'E'F'.$

Proof

$AF = BC' = B'C.$ We use Sparrow’s Lemma 1 and get that point $D'$ is the midpoint of arc $BC$ of $\Omega$ which contain the point $A.$

So $D'O || DI, \frac {D'O}{DI} = \frac{R}{r} = \frac{OG}{GI} = \frac{D'G}{DG}.$

Similarly, $\frac{E'G}{EG} = \frac{F'G}{FG} = \frac{R}{r} \blacksquare$

Russian Math Olympiad 1999

Let triangle $\triangle ABC$ with points $D \in AB$ and $E \in BC, EC = AD$ be given.

Let $M$ and $M'$ be midpoints $AC$ and $DE,$ respectively.

Let $I$ be the incenter of $\triangle ABC.$

Prove that $BI || MM'.$

Proof

$AD = CE, \Omega = \odot ABC.$ We use Sparrow’s Lemma 1 for circle $\omega = \odot BDE$ and get that point $F = \Omega \cap \omega$ is the midpoint of arc $AC$ of $\Omega$ which contain the point $B.$

Let $G$ be the antipode of $F$ on $\Omega, G'$ be the antipode of $F$ on $\omega \implies M \in GF, M' \in G'F.$

$\angle ABG = \beta = \angle DFG' = \angle AFG,$ $\angle FDG' = 90^\circ = \angle FAG \implies$ $\frac{FM'}{FG'} = \frac{FM}{FG} \implies$ $GG' || MM' \implies BI || MM' \blacksquare$

IMO 1985 5 Sparrow

A circle with center $O$ passes through the vertices $A$ and $C$ of the triangle $ABC$ and intersects the segments $AB$ and $BC$ again at distinct points $K$ and $N$ respectively. Let $M$ be the point of intersection of the circumcircles of triangles $ABC$ and $KBN$ (apart from $B$ ). Prove that $\angle OMB = 90^{\circ}$ .

Proof

Let $\Omega,Q,$ and $R$ be the circumcircle of $\triangle ABC,$ circumcenter and radius of $\Omega.$

Let $\omega,O',$ and $r$ be the circumcircle of $\triangle KBN,$ circumcenter and radius of $\omega.$ $BO' = MO' = r, BQ = MQ = R \implies QO' \perp MB, \angle BQO' = \angle MQO'.$

$AKNC$ is cyclic, so $KN$ is antiparallel $AC, O'O \perp KN.$

We use Sparrow’s Lemma 3A for circle $\omega$ and get that point $O'$ lies on altitude of $\triangle ABC \implies BO' \perp AC.$

Let $D$ be the point on $\omega$ opposite $B.$

$BQ$ is isogonal to $BO' \implies OO' || BQ.$

$OQ$ lies on bisector $AC \implies BO' || QO \implies BO'OQ$ is parallelogram $\implies OO' = BQ = R, BO' = QO = r = O'D \implies BO'OQ$ is parallelogram.

Let $\angle BO'M = 2 \varphi \implies \angle O'MD = \angle O'DM = \varphi.$

$\angle DO'Q = 180^\circ - \varphi - (180^\circ - 2 \varphi) = \varphi \implies MD || O'Q \perp MB \blacksquare$

@@ Line 134: / Line 134: @@
 <cmath>\frac{FM'}{FG'} = \frac{FM}{FG} \implies</cmath>
 <cmath>GG' || MM' \implies BI || MM' \blacksquare</cmath>
+==IMO 1985 5 Sparrow==
+[[File:IMO 1985 5 Sparrow.png|300px|right]]
+A circle with center <math>O</math> passes through the vertices <math>A</math> and <math>C</math> of the triangle <math>ABC</math> and intersects the segments <math>AB</math> and <math>BC</math> again at distinct points <math>K</math> and <math>N</math> respectively. Let <math>M</math> be the point of intersection of the circumcircles of triangles <math>ABC</math> and <math>KBN</math> (apart from <math>B</math>). Prove that <math>\angle OMB = 90^{\circ}</math>.
+<i><b>Proof</b></i>
+Let <math>\Omega,Q,</math> and <math>R</math> be the circumcircle of <math>\triangle ABC,</math> circumcenter and radius of <math>\Omega.</math>
+Let <math>\omega,O',</math> and <math>r</math> be the circumcircle of <math>\triangle KBN,</math> circumcenter and radius of <math>\omega.</math>
+<cmath>BO' = MO' = r, BQ = MQ = R \implies QO' \perp MB, \angle BQO'  = \angle MQO'.</cmath>
+<math>AKNC</math> is cyclic, so <math>KN</math> is antiparallel <math>AC, O'O \perp KN.</math>
+We use [[Sparrow’s lemmas | Sparrow’s Lemma 3A]] for circle <math>\omega</math> and get that point <math>O'</math> lies on altitude of <math>\triangle ABC \implies BO' \perp AC.</math>
+Let <math>D</math> be the point on <math>\omega</math> opposite <math>B.</math>
+<math>BQ</math> is isogonal to <math>BO' \implies OO' || BQ.</math>
+<math>OQ</math> lies on bisector <math>AC \implies BO' || QO \implies BO'OQ</math> is parallelogram <math>\implies OO' = BQ = R, BO' = QO = r = O'D \implies BO'OQ</math> is parallelogram.
+Let <math>\angle BO'M = 2 \varphi \implies \angle O'MD = \angle O'DM = \varphi.</math>
+<math>\angle DO'Q = 180^\circ - \varphi - (180^\circ - 2 \varphi) = \varphi \implies MD || O'Q \perp MB \blacksquare</math>

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Difference between revisions of "Sparrow’s lemmas"

Revision as of 12:20, 21 September 2025

Contents

Sparrow's Lemma 1

Sparrow’s Lemma 2

Sparrow’s Lemma 3

Sparrow’s Lemma 3A

Russian Math Olympiad 2011

Russian Math Olympiad 2005

Russian Math Olympiad 1999

IMO 1985 5 Sparrow