5AB = 4BC, <B = 60

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5AB = 4BC, <B = 60

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Source: Philippines MO 2018/1

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#1 h Aug 4, 2018, 6:30 AM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ with $\angle ABC = 60^{\circ}$ and $5AB = 4BC$ , points $D$ and $E$ are the feet of the altitudes from $B$ and $C$ , respectively. $M$ is the midpoint of $BD$ and the circumcircle of triangle $BMC$ meets line $AC$ again at $N$ . Lines $BN$ and $CM$ meet at $P$ . Prove that $\angle EDP = 90^{\circ}$ .

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#2 h Aug 4, 2018, 7:39 AM • 2 Y

Y by Adventure10, Mango247

We can assume that, $AB=4$ and $BC=5$ .
By Cosine Law, $AC=\sqrt{21}$ , and by Pythagorean Theorem, $BD=\frac{10}{\sqrt7}$ , which implies that $MD=\frac{5}{\sqrt7}$ . Also, $AD=\frac{2\sqrt3}{\sqrt7},CD=\frac{5\sqrt3}{\sqrt7}$ . From $\frac{MD}{CD}=\frac{1}{\sqrt{3}}$ , we have, $\angle DMC=60^\circ$ .

Because $BMNC$ is cyclic, we have $\angle MBN=\angle MCN\implies\angle DBN=\angle DCM\implies\Delta DBN\sim\Delta DCM$ .
From this, we have $\frac{DB}{DN}=\frac{DC}{DM}\implies DN=\frac{10}{\sqrt{21}}$ and $\angle DNB=60^\circ$ .

Claim 1: $ED||BN$

Proof: It suffices to show that $\Delta AED\sim\Delta ABN$ .

It is well-known that $\Delta AED\sim\Delta ACB$ , and because $\angle ABC=\angle DNB=60^\circ$ , we now have $\Delta ACB\sim\Delta ABN$ .

Claim 2: $DF\perp BN$

Proof: We know that $BM:MD=1:1$ and $CN:ND=1:2$ . Using mass points tells us that $MF:FC=1:1$ .

This means $F$ is the midpoint of $MC$ , and because $MDC$ is a right triangle, $DF=MF=\frac{1}{2}MC=\frac{5}{\sqrt7}$

We know that $\angle DBF=30^\circ$ , and by using Sine Law, $\angle DFB=90^\circ$ .

From the two claims, we have $\angle EDP=90^\circ$ .

This post has been edited 1 time. Last edited by NikoIsLife, Aug 4, 2018, 7:39 AM

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#3 h Dec 4, 2020, 4:01 PM

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Let $AB=4$ and $BC=5$ , since we can resize triangle to right size.

By the law of cosines,
$AC=AB^2+BC^2-2BC\cdot AB\cdot \cos{\angle ABC}=16+25-4\cdot 5=21\implies AC=\sqrt{21}$ .
By area calculation,
$AC\cdot BD=AB\cdot CB\cdot \sin{\angle ABC}\implies BD=\frac{\sqrt{3}\cdot 2\cdot 5}{4\sqrt{21}}=\frac{10\sqrt{7}}{7}$ .
$DM\cdot DB=DN\cdot DC\implies DN=\frac{DM\cdot DB}{DC}=\frac{DM\cdot DB}{DC}$
$\cos{\angle BCA}=\frac{CB^2+AC^2-AB^2}{2 \cdot CA\cdot CB}=\frac{25+21-16}{2 \cdot \sqrt{21}\cdot 5}=\frac{\sqrt 3}{\sqrt 7}$ Thus, $\cos{\angle BCA}=\frac{CD}{BC}\implies CD=\cos{\angle BCA}\cdot BC=\frac{\sqrt 3}{\sqrt 7}\cdot 5=\frac{\sqrt{3}\cdot 5}{\sqrt 7}$ Thus, $DN=\frac{\frac{5\sqrt{7}}{7}\cdot \frac{10\sqrt{7}}{7}}{\frac{\sqrt{3}\cdot 5}{\sqrt 7}}=\frac{10\cdot \sqrt{7}}{7\sqrt{3}}$ Thus, $\tan{\angle DBN}=\frac{DN}{BD}=\frac{\frac{10\cdot \sqrt{7}}{7\sqrt{3}}}{\frac{10\sqrt{7}}{7}}=\frac{1}{\sqrt{3}}\implies \angle DBN=30^{\circ}.$ Hence, $\angle DBN=\angle MBN=\angle PCN=30^{\circ}$ and $\angle BND=60^{\circ}$ , thus $NC=PN$ .
We have $PN=NC=DC-DN=\frac{\sqrt{3}\cdot 5}{\sqrt 7}-\frac{10}{\sqrt{7}\sqrt{3}}=\frac{1}{\sqrt{7}}\cdot \frac{5}{\sqrt{3}}$ Hence, we easily have that $\triangle DPN$ is right triangle. Hence, $\triangle DPM$ is equilateral.
Also, $\cos{\angle BAC}=\frac{AB^2+AC^2-BC^2}{2 \cdot AB\cdot AC}=\frac{16+21-25}{2 \cdot \sqrt{21}\cdot 4}=\frac{3}{2\sqrt 21}$ Hence, $AE=\cos{\angle BAC}\cdot AC=\frac{3}{2}$ and $BE=\cos{\angle ABC}\cdot BC=\frac{1}{2}\cdot 5=\frac{5}{2}$ and $AD=AC-CD=\sqrt{21}-\frac{\sqrt{3}\cdot 5}{\sqrt{7}}=\frac{7{\sqrt{3}}-\sqrt{3}\cdot 5}{\sqrt{7}}=\frac{2{\sqrt{3}}}{\sqrt{7}}.$ Thus, $\frac{3}{5}=\frac{\frac{2{\sqrt{3}}}{\sqrt{7}}}{\frac{10\cdot \sqrt{7}}{7\sqrt{3}}}\Longleftrightarrow \frac{\frac{3}{2}}{\frac{5}{2}}=\frac{\frac{2{\sqrt{3}}}{\sqrt{7}}}{\frac{10\cdot \sqrt{7}}{7\sqrt{3}}}\Longleftrightarrow \frac{AE}{EB}=\frac{AD}{DN}\Longleftrightarrow \text{ $DE$ and $BN$ are parallel.}$ Hence, $\angle EDP=\angle DPN=90^{\circ}.\qquad \square$

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