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Source: 2020 Dutch IMO TST 1.1

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#1 h Nov 21, 2020, 9:53 PM • 1 Y

Y by Ha_ha_ha

In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$ . Prove that $\angle BAC = 2 \angle ABC$ .

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#2 h Nov 22, 2020, 8:24 PM • 4 Y

Y by Ha_ha_ha, cryptographer, abc9, z81699854321

Extend $AC$ past $A$ to $P$ so that $AP = AI$ . Then $PC = PA + AC = AI + AC = BC$ , so triangle $PBC$ is isosceles.

Since $\angle ACI = \angle BCI$ , triangles $BIC$ and $PIC$ are congruent, so $\angle IPC = \angle IBC = \frac{B}{2}$ . Then $\angle AIP = \angle API = \frac{B}{2}$ . Also, $\angle BAI = \frac{A}{2}$ and $\angle PAB = 180^\circ - A$ , so adding up all the angles in triangle $PAI$ , we get
$\frac{B}{2} + 180^\circ - A + \frac{A}{2} + \frac{B}{2} = 180^\circ.$ This simplifies to $A = 2B$ .

$[asy] unitsize(1 cm); pair A, B, C, I, P; B = (0,0); C = (5,0); A = extension(B, B + dir(50), C, C + dir(180 - 30)); I = incenter(A,B,C); P = C + 5*dir(180 - 30); draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(A--P--B); draw(B--I--C); draw(A--I--P); label("$A$", A, NE); label("$B$", B, SW); label("$C$", C, SE); dot("$I$", I, S); label("$P$", P, N); [/asy]$

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#3 h Dec 24, 2022, 11:40 AM

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Let $K$ be a point on $BC$ such that $KC=AC$ . Note that
$AI=BC-AC=BC-KC=BK.$ Moreover,
$\angle AIB=90^{\circ}+\frac{1}{2}\angle ACB=180^{\circ}-\angle AKC=\angle AKB.$ Thus, $AIKB$ is an isosceles trapezoid, so $\angle IAB=\angle KBA\implies \frac{1}{2}\angle BAC=\angle ABC\implies BAC=2\angle ABC.$

This post has been edited 1 time. Last edited by Quidditch, Dec 24, 2022, 11:41 AM

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