(DEP) tangent to BC

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(DEP) tangent to BC

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Source: 2019-2020 Winter SDPC #8

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mira74

#1 h Mar 20, 2020, 3:06 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $ABC$ be a triangle with circumcircle $\Gamma$ . If the internal angle bisector of $\angle A$ meets $BC$ and $\Gamma$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $A$ and $D$ tangent to $BC$ , let the external angle bisector of $\angle A$ meet $\Gamma$ at $F$ , and let $FO_1$ meet $\Gamma$ at some point $P \neq F$ . Show that the circumcircle of $DEP$ is tangent to $BC$ .

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#2 h Mar 20, 2020, 3:30 AM

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WLOG assume $AB < AC.$

Let $P' = (\triangle AO_1D) \cap \Gamma.$ We claim that $P' = P.$

Note that $\angle AO_1D = 2 \angle C + \angle A.$

This means that $\angle AP'D = 180 - (2 \angle C + \angle A) = \angle B - \angle C.$ Since $O_1A = O_1 D$ , we have that $\angle AP'O_1 = \angle DP'O_1 = \frac12 \angle AP'D = \frac12 (\angle B - \angle C).$ This indicates that $P' O_1$ passes through $F$ because $\angle AP'F = \frac 12 (\angle B - \angle C).$

So therefore we have that $P = P'$ , and $A' = PD \cap \Gamma$ is the point on $\Gamma$ such that $AA' || BC$ . Thus $\angle EDC = \angle C + \frac12 \angle A = \angle DPE,$ which means we're done.

$\square$

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#3 h Mar 20, 2020, 3:31 AM

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Let $\omega$ be the circle through $A$ and $D$ that's also tangent to $BC$ . Observe by $\sqrt{bc}$ that $\omega$ is also tangent to $\Omega$ at $A$ . Let $(AO_1D)$ intersect $\Omega$ again at $P' \neq A$ . Then $\measuredangle AP'O_1 = \measuredangle ADO_1 = \measuredangle AEF = \measuredangle AP'F$ since $O_1D \perp BC$ , so $P' = P$ . If $T$ is the intersection of the tangent to $\Omega$ at $A$ with $BC$ , it's clear that $T$ lies on $(AO_1DP)$ since $\angle O_1AT = \angle O_1DT = 90^{\circ}$ . This means $\angle TPF = \angle TDO_1 = 90^{\circ} = \angle FPE$ , so $T$ lies on $PE$ . Finally, we have $TD^2 = TA^2 = TP \cdot TE,$ so $(DPE)$ is tangent to $BC$ .

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DPS

#5 h Mar 20, 2020, 4:57 AM • 2 Y

Y by lilavati_2005, Mango247

Similar to Pathological’s:

Observe that $O_1D \parallel OE$ . So, the homothety centred at $A$ mapping $D$ to $E$ maps $O_1$ to $O$ . So, $A-O_1-O$ .

Let $L=AE\cap PF$

From $O_1D\parallel FE$ we get,

$\frac{LE}{LD} =\frac{LF}{LO_1}$

Also, $AL\cdot LE=PL\cdot LF$ .

From these two equations we get,

$AL\cdot LD= PL\cdot LO_1$

So, { $A,O_1,D,P$ } are concyclic.

Now, $\angle DPE= \angle FPE - \angle O_1PD=\angle FAE -\angle O_1AD=\angle FAO=\angle ACE =\angle CDE$

So, $BD$ is tangent to $(PDE)$ $\blacksquare$

This post has been edited 2 times. Last edited by DPS, Mar 20, 2020, 5:11 AM

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