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Difference between revisions of "Sparrow’s lemmas"

(→‎Sparrow's Lemma 1)
(→‎Sparrow's Lemma 1)
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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>
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==Sparrow’s Lemma 1A==
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[[File:Sparrow 1A.png|300px|right]]
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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.
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Let <math>M</math> be the midpoint of <math>BC, I</math> be the incenter.
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Prove that <math>BD + CE = BC</math> iff  points <math>M, D, E,</math> and <math>I</math> are concyclic.
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<i><b>Proof</b></i>
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1. Let points <math>M, D, E,</math> and <math>I</math> are concyclic.
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Denote <math>F \in BC</math> such <math>BD = BF, \varphi = \angle ADI.</math>
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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>
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<cmath>\triangle BDI = \triangle BFI \implies DI = FI, \angle CFI = \varphi.</cmath>
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<cmath>\angle DAI = \angle EAI \implies DI = EI = FI.</cmath>
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<cmath>\triangle CIF = \triangle CIE \implies CE = CF \implies BD + CE = BC \blacksquare</cmath>
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2. Let <math>BD + CE = BC \implies</math> there is point <math>F</math> such that <math>BF = BD, CF = CE \implies</math>
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<cmath>\triangle CIF = \triangle CIE, \triangle BIF = \triangle BID \implies </cmath>
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<cmath>180^\circ = \angle BFI + \angle CFI =  \angle BDI + \angle CEI =  \angle ADI + \angle AEI \blacksquare</cmath>

Revision as of 16:11, 16 September 2025

Sparrow’s lemmas have been known to Russian Olympiad participants since at least 2016.

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 1A

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 $BF = BD, CF = CE \implies$ \[\triangle CIF = \triangle CIE, \triangle BIF = \triangle BID \implies\] \[180^\circ = \angle BFI + \angle CFI = \angle BDI + \angle CEI = \angle ADI + \angle AEI \blacksquare\]

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