if incenter I in (KDE), prove that BD+CE=BC, symmetric lines, perpendicular

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if incenter I in (KDE), prove that BD+CE=BC, symmetric lines, perpendicular

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Source: 2016 Saudi Arabia GMO TST level 4, II p3

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#1 h Aug 1, 2020, 8:14 PM

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Let $ABC$ be a triangle with incenter $I$ . Let $CI, BI$ intersect $AB, AC$ at $D, E$ respectively. Denote by $\Delta_b,\Delta_c$ the lines symmetric to the lines $AB, AC$ with respect to $CD, BE$ correspondingly. Suppose that $\Delta_b,\Delta_c$ meet at $K$ .
a) Prove that $IK \perp BC$ .
b) If $I \in (K DE)$ , prove that $BD + C E = BC$ .

This post has been edited 2 times. Last edited by parmenides51, Aug 4, 2020, 2:41 PM
Reason: typo corrected, see #3

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#2 h Aug 4, 2020, 1:07 PM • 1 Y

Y by parmenides51

I think the problem statement is wrong - the points $D$ and $E$ should be defined as $CI\cap AB$ and $BI\cap AC$ respectively. Anyway, if my formulation is right, here is the solution.

(a) We use directed angles modulo $180^{\circ}$ .
$\measuredangle(DK,BC)=\measuredangle(DK,DC)+\measuredangle(DC,BC)=\measuredangle(DC,DA)+\measuredangle(AC,DC)=\measuredangle(AC,DA)=\measuredangle CAB$ Similarly, $\measuredangle(KE,BC)=\measuredangle BAC$ . It thus suffices to show that $IK$ bisects $\angle DKE$ .

$\frac{\sin\angle DKI}{\sin\angle IKE}=\frac{DI}{IE}\cdot\frac{\sin\angle KDI}{\sin\angle KEI}=\frac{DI}{IE}\cdot\frac{\sin\angle BDI}{\sin\angle CEI}=\frac{DI}{IE}\cdot\frac{\frac{\sin\angle DBI}{DI}\cdot BI}{\frac{\sin\angle ECI}{EI}\cdot CI}=\frac{\sin\angle IBC}{\sin\angle ICB}\cdot\frac{BI}{CI}=1.$ Clearly we have $\angle DKI\neq 180^{\circ}-\angle IKE$ . Thus $\angle DKI=\angle IKE$ and we're done.

(b) If $IDKE$ is cyclic, then $\angle DKE=180^{\circ}-2\angle A$ , and also equal to $180^{\circ}-\angle DIE=90^{\circ}-\frac{1}{2}\angle A$ . Hence, $2\angle A=90^{\circ}+\frac{1}{2}\angle A$ , which means that $\angle A=60^{\circ}$ .

Now
$BD+CE=\frac{BC}{AC+BC}\cdot AB+\frac{BC}{AB+BC}\cdot AC=BC\left(\frac{c}{a+b}+\frac{b}{a+c}\right).$ Now it suffices to show that $\frac{c}{a+b}+\frac{b}{a+c}=1$ , or equivalently, $c(a+c)+b(a+b)=(a+b)(a+c)$ . This is equivalent to showing that $b^2+c^2=a^2+bc$ , which is true by the cosine rule.

This post has been edited 1 time. Last edited by ACGNmath, Aug 4, 2020, 1:07 PM

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#3 h Aug 4, 2020, 2:18 PM • 1 Y

Y by Mango247

Quote:

Let $ABC$ be a triangle whose incircle $(I)$ is tangent to $AB, AC$ at $D, E$ respectively. Denote by $\Delta_b,\Delta_c$ the lines symmetric to the lines $AB, AC$ with respect to $CD, BE$ correspondingly. Suppose that $\Delta_b,\Delta_c$ meet at $K$ .
a) Prove that $IK \perp BC$ .
b) If $I \in (K DE)$ , prove that $BD + C E = BC$ .

that above was the official wrong statement, thanks for noticing
the official solution, proves you correct, I just edited it to

Quote:

Let $ABC$ be a triangle with incenter $I$ . Let $CI, BI$ intersect $AB, AC$ at $D, E$ respectively. Denote by $\Delta_b,\Delta_c$ the lines symmetric to the lines $AB, AC$ with respect to $CD, BE$ correspondingly. Suppose that $\Delta_b,\Delta_c$ meet at $K$ .
a) Prove that $IK \perp BC$ .
b) If $I \in (K DE)$ , prove that $BD + C E = BC$ .

This post has been edited 2 times. Last edited by parmenides51, Aug 5, 2020, 11:50 AM

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