Angles in a quadrilateral (beautiful and almost famous)

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Angles in a quadrilateral (beautiful and almost famous)

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Source: Problem 4, Brazil MO 1993

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Johann Peter Dirichlet

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Johann Peter Dirichlet

#1 h Mar 18, 2006, 4:50 AM • 2 Y

Y by Adventure10, Mango247

$ABCD$ is a convex quadrilateral with
$\angle BAC = 30^\circ$ $\angle CAD = 20^\circ$ $\angle ABD = 50^\circ$ $\angle DBC = 30^\circ$
If the diagonals intersect at $P$ , show that $PC = PD$ .

This post has been edited 2 times. Last edited by Amir Hossein, Aug 16, 2012, 1:40 PM
Reason: Edited.

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Kunihiko_Chikaya

14512 posts

Kunihiko_Chikaya

#2 h Mar 18, 2006, 4:52 AM • 1 Y

Y by Adventure10

These angles reminds us of Franklin's kite.

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Johann Peter Dirichlet

375 posts

Johann Peter Dirichlet

#3 h Mar 18, 2006, 5:32 AM • 2 Y

Y by Adventure10, Mango247

How I can give some info about this Franklin's Kite?

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388 posts

Jutaro

#4 h Mar 19, 2006, 6:12 AM • 1 Y

Y by Adventure10

By easy angle-chasing we have $\angle ACB=70^{\circ}$ , $\angle ADB= 80^{\circ}$ . Now we apply the sine law to the triangles $ABC$ , $ABD$ , $BCP$ and $ADP$ :

$\frac{BC}{AB}=\frac{\sin 30^{\circ}}{\sin 70^{\circ}}=\frac{1}{2\sin 70^{\circ}}$

\[ \frac{AD}{AB}=\frac{\sen 50^{\circ}}{\sin 80^{\circ}} \]

$PC=\frac{BC \sin 30^{\circ}}{\sin 80^{\circ}}=\frac{BC}{2\sin 80^{\circ}}$
$PD=\frac{AD \sin 20^{\circ}}{\sin 80^{\circ}}$

Dividing the first two equations we get $\frac{BC}{AD}=\frac{\sin 80^{\circ}}{2\sin 50^{\circ} \sin 70^{\circ}}$
and dividing the last two equations: $\frac{PC}{PD}=\frac{BC}{AD}\cdot\frac{1}{2\sin 20^{\circ}}$
Therefore $\frac{PC}{PD}=\frac{\sin 80^{\circ}}{4\sin 20^{\circ}\sin 50^{\circ} \sin 70^{\circ}}$

But
$4\sin 20^{\circ}\sin 50^{\circ} \sin 70^{\circ}=(2\sin 70^{\circ})(2\sin 20^{\circ} \sin 50^{\circ})=(2 \sin 70^{\circ})(\cos 30^{\circ}-\cos 70^{\circ})=2\sin 70^{\circ} \cos 30^{\circ}-2 \sin 70^{\circ} \cos 70^{\circ}=\sin 100^{\circ}+\sin 40^{\circ}-\sin 140^{\circ}=\sin 80^{\circ}+ \sin 40^{\circ}-\sin 40^{\circ}=\sin 80^{\circ}$
Hence $PC=PD$

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7054 posts

#5 h Mar 19, 2006, 8:44 AM • 1 Y

Y by Adventure10

Here is a short trigonometrical proof of this easily and nice proposed problem.

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388 posts

Jutaro

#6 h Mar 20, 2006, 3:27 PM • 2 Y

Y by Adventure10, Mango247

Virgil Nicula wrote:

$\sin \widehat {BAC}\cdot \sin \widehat {ADB}\cdot \sin \widehat {DCA}\cdot \sin \widehat {CBD}=$ $\sin \widehat {ABD}\cdot \sin \widehat {DAC}\cdot \sin \widehat {CDB}\cdot \sin \widehat {BCA}$

Hey, nice solution, Mr. Virgil

By the way, is the equality above a known theorem or something?

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#7 h May 2, 2011, 7:06 PM • 1 Y

Y by Adventure10

Jutaro wrote:

Virgil Nicula wrote:

$\sin \widehat {BAC}\cdot \sin \widehat {ADB}\cdot \sin \widehat {DCA}\cdot \sin \widehat {CBD}=$ $\sin \widehat {ABD}\cdot \sin \widehat {DAC}\cdot \sin \widehat {CDB}\cdot \sin \widehat {BCA}$

Hey, nice solution, Mr. Virgil

By the way, is the equality above a known theorem or something?

It is not necessary a theorem. It is more a useful lemma. The proof: Aply law of sines in $\triangle{ABC},\triangle{ABD},\triangle{DCA},\triangle{CDB}$ , and you can rewrite the equality as: $\frac{BC}{BA}\frac{AB}{AD}\frac{DA}{DC}\frac{CD}{CB}=1$ .

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429 posts

#8 h May 2, 2011, 7:15 PM • 1 Y

Y by Adventure10

By the way, my solution. We have(easy angle chasing) $\triangle{BPC}\sim \triangle{ABC}$ , which gives $\frac{PC}{BC}=\frac{PB}{AB}$ , so $\frac{PC}{PB}=\frac{BC}{AB}=\frac{sin30}{sin70}$ (law of sines in $\triangle{BPC}$ ), In $\triangle{ABD}, \frac{PC}{BP}=\frac{sin20}{sin30}\frac{sin80}{sin50}$ (using and the law of sines).
Finally, we have to prove that $sin20 cos20 sin50=sin30 sin30 sin80$ , (using $2sinacosa=sin2a$ ), $sin40 sin 50=\frac{sin80}{2}$ , obvios true using $sina sinb=\frac{-1}{2}(cos(a+b)-cos(a-b)$

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2643 posts

yetti

#9 h May 3, 2011, 1:28 AM • 5 Y

Y by mastermind.hk16, Pirkuliyev Rovsen, Polynom_Efendi, myh2910, Adventure10

WITHOUT TRIGONOMETRY:
Take regular 18-gon $P_1P_2...P_{18}$ with circumcircle $(O)$ and let $X \equiv P_1P_2 \cap P_7P_8.$ $\triangle P_2XP_7$ is equilateral, having $60^\circ$ angles at the base $P_2P_7$ $\Longrightarrow$ $P_7X = P_2P_7 = P_7P_{12}$ $\Longrightarrow$ $\triangle XP_7P_{12}$ is isosceles with $140^\circ$ angle at the vertex $P_7$ $\Longrightarrow$ $\angle XP_{12}P_7 = 20^\circ$ $\Longrightarrow$ $P_5 \in XP_{12}.$ Likewise, $P_4 \in XP_{15}.$ It follows that the 18-gon diagonals $P_1P_7, P_2P_8, P_4P_{12}, P_5P_{15}$ are concurrent at $Y$ on the polar of $X$ WRT $(O).$ Take $\triangle P_2P_4P_7$ with concurrent cevians $P_2YP_8, P_4YP_{12}, P_7YP_1.$ Their isogonals $P_2P_3, P_4P_{15}, P_7P_5$ WRT this triangle are concurrent at the isogonal conjugate $Z$ of $Y.$ Re-labeling $A \equiv P_{15}, B \equiv P_7, C \equiv Z, D \equiv P_2$ yields quadrilateral $ABCD$ with the given angles $\angle BAC = 30^\circ, \angle CAD = 20^\circ, \angle ABD = 50^\circ, \angle DBC = 30^\circ$ and diagonal intersection $P \equiv AC \cap BD.$ $\angle PCD = \angle ACD = 40^\circ = \angle BDC = \angle PDC$ $\Longrightarrow$ $\triangle PCD$ is isosceles with $PC = PD.$

EVEN SIMPLER:
$\angle BAD = 50^\circ = \angle ABD$ $\Longrightarrow$ $\triangle ABD$ is isosceles with $DA = DB.$ Let $\triangle BXD$ be equilateral, with $X, A$ on the opposite sides of $BD.$ $\triangle AXD$ is isosceles with $DA = DX$ and $\angle XDA = 140^\circ$ $\Longrightarrow$ $\angle XAD = 20^\circ = \angle CAD$ $\Longrightarrow$ $C \in AX.$ $\angle DBC = 30^\circ$ $\Longrightarrow$ $BC$ is perpendicular bisector of $DX$ $\Longrightarrow$ $\triangle DCX$ is isosceles with $CX = CD$ and base angles $\angle CDX = \angle CXD = 20^\circ$ $\Longrightarrow$ $\angle PDC = \angle BDC = 40^\circ.$ Since $\angle CPD = \angle APB = 100^\circ,$ $\triangle CPD$ is isosceles with $PC = PD.$

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756 posts

#10 h Feb 15, 2018, 2:11 PM • 3 Y

Y by Dr_Vex, Adventure10, Mango247

We Use Quadrilateral Ceva's Theorem
Solution:
Since $AC \cap BD = P:$
$\implies \dfrac {Sin \angle PAD}{Sin \angle PAB} * \dfrac {Sin \angle PBA}{Sin \angle PBC}* \dfrac {Sin \angle PCB}{Sin \angle PCD}*\dfrac {Sin \angle PDC}{Sin \angle PDA}=1$

$\therefore \dfrac{Sin 20^\circ}{Sin 30^\circ} *\dfrac {Sin 50^\circ}{Sin 30^\circ} * \dfrac {Sin 70^\circ}{Sin \angle PCD}* \dfrac {Sin \angle PDC}{Sin 80^\circ} =1$

Simplifying We Get :

$\dfrac {Sin 140^\circ}{Sin 40^\circ}= \dfrac { Sin \angle PCD}{Sin \angle PDC}$

$\implies Sin \angle PDC = Sin \angle PCD$

Thus, Either $\angle PCD = \angle PDC$ or $\angle PDC + \angle PCD =180^\circ$

The later is not possible as both angles are angles of triangle .

Thus $\angle PCD = \angle PDC$
Hence $\triangle PCD$ is Isosceles

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#11 h Feb 15, 2018, 2:49 PM • 3 Y

Y by Drunken_Master, Adventure10, Mango247

By law of sines,

$AP=2BP\sin50=\sin80 \cdot \frac{DP}{\sin20}$

Let $\angle PCD=x$ ,

$CP=\frac{BP}{2\sin70}=\sin(80-x)\cdot\frac{DP}{\sin x}$

Dividing, $\frac{AP}{CP}=4\sin50\sin70=\frac{\sin80\sin x}{\sin20\sin(80-x)}$ $\implies \frac{\sin x}{\sin (80-x)}=\frac{ 2 \sin 20 \cos 20 \cdot 2 \cos 40}{\sin 80}=\frac{2 \sin 40 \cos 40}{\sin 80}=1$
Hence, $x=80-x=40^{\circ} \implies PC=PD$ , as desired. $\square$

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