View source for 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==
[[File:Sparrow 1.png|300px|right]]
Let triangle <math>ABC</math> with circumcircle <math>\Omega</math> and points <math>D</math> and <math>E</math> on the sides <math>AB</math> and <math>AC,</math> respectively be given. 

Let <math>K \in \Omega</math> be the midpoint of the arc <math>BC</math> which contain the point <math>A.</math>

Prove that <math>BD = CE</math> iff  points <math>A, D, E,</math> and <math>K</math> are concyclic.

<i><b>Proof</b></i>

<math>BK = CK, \angle ABK = \angle ACK.</math>

Let <math>BD = CE \implies \triangle BKD = \triangle CKE \implies</math>

<math>\angle KDA = \angle KEA \implies A, D, E,</math> and <math>K</math> are concyclic.

Let <math>A, D, E,</math> and <math>K</math> are concyclic <math>\implies \angle KDA = \angle KEA  \implies</math>

<math>\angle BKD = \angle CKE  \implies \triangle BKD = \triangle CKE \implies BD = CE.</math>
==Sparrow’s Lemma 2==
[[File:Sparrow 1A.png|300px|right]]
Let triangle <math>ABC</math> with circumcircle <math>\Omega</math> and points <math>D</math> and <math>E</math> on the sides <math>AB</math> and <math>AC,</math> respectively be given.
 
Let <math>I</math> be the incenter.

Prove that <math>BD + CE = BC</math> iff  points <math>A, D, E,</math> and <math>I</math> are concyclic.

<i><b>Proof</b></i>

1. Let points <math>A, D, E,</math> and <math>I</math> are concyclic.
 
Denote <math>F \in BC</math> such <math>BD = BF, \varphi = \angle ADI.</math>

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

==Sparrow’s Lemma 3==
[[File:Sparrow 3.png|300px|right]]
Let lines <math>\ell</math> and <math>\ell'</math> and points <math>A_0 \in \ell</math> and <math>B_0 \in \ell'</math> be given, <math>O = \ell \cap \ell'.</math>

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

Prove that circle <math>\Omega = \odot OAB</math> contain fixed point (<math>P</math>).

<i><b>Proof</b></i>

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

<i><b>Corollary 1</b></i>

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

<i><b>Corollary 2</b></i>

Lemma 2 is partial case of Lemma 3 with spiral center <math>I,</math> and equal speeds (from <math>D</math> to <math>A</math> and from <math>E</math> to <math>C</math>). Start positions of these points are <math>D</math> and <math>E.</math>
==Russian Math Olympiad 2011==
[[File:Sparrow problem 1.png|300px|right]]
Let triangle <math>ABC</math> with circumcircle <math>\Omega</math> be given. 
Let <math>K \in \Omega</math> be the midpoint of the arc <math>BC</math> which contain the point <math>A, M</math> be the midpoint of <math>BC, I_B</math> be incenter of <math>\triangle ABM, I_C</math> be incenter of <math>\triangle ACM.</math>

Prove that points <math>A, I_B, I_C,</math> and <math>K</math> are concyclic.

<i><b>Proof</b></i>

Denote <math>\omega = \odot AI_BI_C, D = \omega \cap AB,</math>
<math>E = \omega \cap AC, F = \omega \cap AM.</math>

We use Lemma 2 for <math>\triangle ABM</math> and get <math>BD + FM = BM.</math>

We use Lemma 2 for <math>\triangle ACM</math> and get <math>CE + FM = CM = BM \implies BD = CE.</math>

We use Lemma 1 for <math>\triangle ABC</math> and get result: points <math>A,D,E,</math> and <math>K</math> are concyclic <math>\blacksquare</math>
==Russian Math Olympiad 2005==
[[File:Sparrow problem 2.png|300px|right]]
Let triangle <math>ABC</math> with circumcircle <math>\Omega</math> and incircle <math>\omega</math> be given. Let <math>A', B',</math> and <math>C'</math> be the tangent points of the excircles of <math>\triangle ABC</math> with the corresponding sides. Let <math>D, E,</math> and <math>F</math> be the tangent points of the incircle of <math>\triangle ABC.</math> The circumscribed circles of triangles <math>\triangle A'B'C, \triangle AB'C',</math> and <math>\triangle A'BC'</math> intersect <math>\Omega</math> a second time at points <math>D', E',</math> and <math>F',</math> respectively.

Prove that <math>\triangle DEF \sim \triangle D'E'F'.</math>

<i><b>Proof</b></i>

<math>AF = BC' = B'C.</math> We use [[Sparrow’s lemmas | Sparrow’s Lemma 1]] and get that point <math>D'</math> is the midpoint of arc <math>BC</math> of <math>\Omega</math> which contain the point <math>A.</math>

So <math>D'O || DI, \frac {D'O}{DI} = \frac{R}{r} = \frac{OG}{GI} = \frac{D'G}{DG}.</math>

Similarly, <math>\frac{E'G}{EG} = \frac{F'G}{FG} = \frac{R}{r} \blacksquare</math>