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Difference between revisions of "Sparrow’s lemmas"
(→‎Sparrow’s Lemma 1A)
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<math>\angle BKD = \angle CKE  \implies \triangle BKD = \triangle CKE \implies BD = CE.</math>
 
<math>\angle BKD = \angle CKE  \implies \triangle BKD = \triangle CKE \implies BD = CE.</math>
−
==Sparrow’s Lemma 1A==
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==Sparrow’s Lemma 2==
 
[[File:Sparrow 1A.png|300px|right]]
 
[[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 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.
Line 41: Line 41:
 
<cmath>\triangle CIF = \triangle CIE, \triangle BIF = \triangle BID \implies </cmath>
 
<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>
 
<cmath>180^\circ = \angle BFI + \angle CFI =  \angle BDI + \angle CEI =  \angle ADI + \angle AEI \blacksquare</cmath>
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==Sparrow’s Lemma 3==
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[[File:Sparrow 3.png|300px|right]]
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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>
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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>
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Prove that circle <math>\Omega = \odot OAB</math> contain fixed point (<math>P</math>).
 +
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<i><b>Proof</b></i>
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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>
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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>

Revision as of 09:22, 17 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

Sparrow 1.png

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

Sparrow 1A.png

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

Let $M$ be the midpoint of $BC, I$ be the incenter.

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

Proof

1. Let points $M, D, E,$ 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

Sparrow 3.png

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 $P \blacksquare$

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