Let be a triangle $\triangle ABC$ with $m(\angle ABC) = 75^{\circ}$ and $m(\angle ACB) = 45^{\circ}$. The angle bisector of $\angle CAB$ intersects $CB$ at point $D$. We consider the point $E \in (AB)$, such that $DE = DC$. Let $P$ be the intersection of lines $AD$ and $CE$. Prove that $P$ is the midpoint of segment $AD$.
Problem
Source: BMO SL 2019, G2
Tags: geometry
Functional_equation
08.11.2020 10:29
GorgonMathDota wrote: Let be a triangle $\triangle ABC$ with $m(\angle ABC) = 75^{\circ}$ and $m(\angle ACB) = 45^{\circ}$. The angle bisector of $\angle CAB$ intersects $CB$ at point $D$. We consider the point $E \in (AB)$, such that $DE = DC$. Let $P$ be the intersection of lines $AD$ and $CE$. Prove that $P$ is the midpoint of segment $AD$. My Solution Claim 1.$AEDC$ is a cyclic.
$\angle AED>\angle ACD$
$\frac{AD}{\sin \angle AED}=\frac{ED}{\sin \angle EAD}\implies AD=\frac{ED\cdot \sin \angle AED}{\sin 30};\frac{AD}{\sin \angle ACD}=\frac{CD}{\sin \angle CAD}\implies AD=\frac{CD\cdot \sin 45}{\sin 30};ED=CD\implies \sin \angle AED=\sin 45\implies \angle AED=180-45=135$
$\frac{AP}{\sin \angle ACP}=\frac{PC}{\sin \angle CAP}\implies AP=\frac{PC\cdot \sin 15}{\sin 30}$ $\frac{PD}{\sin \angle DCP}=\frac{PC}{\sin \angle CDP}\implies PD=\frac{PC\cdot \sin 30}{\sin 75}$ $\frac{AP}{PD}=\frac{\frac{PC\cdot \sin 15}{\sin 30}}{\frac{PC\cdot \sin 30}{\sin 75}}=\frac{\sin 15\cdot \sin 75}{(\sin 30)^2}=\frac{2\cdot \sin 15\cdot \cos 15}{\sin 30}=\frac{\sin 30}{\sin 30}=1$
Infinityfun
26.08.2021 22:16
My solution is probably similar to above, but I am gonna write anyway. Claim 1 $ACDE$ is cyclic. Proof: $\angle CDP = \angle DAE$ and $ |CD| = |DE|$ implies either $\triangle ACD \simeq \triangle AED$ or $ACDE$ is cyclic. But former is impossible, since while$\angle ACD = 45^{\circ}$, $75^{\circ}<\angle DEA <150^{\circ}$. So $\angle ACD \neq \angle DEA$ Thus, $ACDE$ is cyclic. Name the circle $\omega$ and the raidus $r$ $$\frac{CD}{AE} = \frac{\sin \angle 30}{r} \cdot \frac{r}{\sin \angle 15} = \frac{\sin \angle 30}{\sin \angle 15}$$Since $ACDE$ is cyclic $\angle CED = \angle CAD = 30^{\circ}$ , $\angle CEB = \angle CED + \angle DEB =75^{\circ}$ $$\frac{BE}{CB} = \frac{\sin \angle 30}{\sin \angle 75} $$By Menelaus $\frac{CD \cdot BE \cdot AP }{CB\cdot EA \cdot PD} =1$ Thus $\frac{PD}{AP}= \frac{CD}{AE} \cdot \frac{BE}{CB} = \frac{\sin 30}{\sin 15} \cdot \frac{\sin 30}{\sin75}= \frac{\sin 30}{\sin 15} \cdot \frac{\sin 30}{\cos 15}=\frac{\sin 30 \cdot \sin 30}{\frac{\sin 30}{2}} = 2\sin 30 = 1$ Which means $AP = AD $
OlympusHero
27.08.2021 01:50
We will first show $AEDC$ is cyclic. Note that LoS gives $\frac{AD}{\sin \angle AED}=\frac{ED}{\sin \angle EAD}$, or $AD=\frac{ED \sin \angle AED}{\sin 30}$. But we have $\frac{DA}{\sin \angle ACD}=\frac{DC}{\sin \angle CAD}$ from LoS, so $AD=\frac{CD \sin 45}{\sin 30}$. However $CD=ED$, meaning $\sin \angle AED = \sin 45$. We cannot have $\angle AED = 45$, so $\angle AED = 135$. Now note that $\frac{AP}{\sin 15} = \frac{PC}{\sin 30}$ from LoS, so $AP = \frac{PC \sin 15}{\sin 30}$. Similarly, $\frac{PD}{\sin 30} = \frac{PC}{\sin 105}$ from LoS, so $PD=\frac{PC \sin 30}{\sin 105}$. It hence suffices to show $\frac{\sin 15}{\sin 30}=\frac{\sin 30}{\sin 105}$, or $\sin^2{30}=\sin 15 \sin 105$. But $\sin 15 = \frac{\sqrt6-\sqrt2}{4}, \sin 105 = \frac{\sqrt6+\sqrt2}{4}, \sin 30 = \frac{1}{2}$, so upon computation this is true as desired. All steps are reversible, so we're done.
BVKRB-
28.08.2021 20:12
Very nice problem
By the angle bisector theorem we have and LoS $$\frac{\sin(\angle ABC)}{\sin(\angle ACB)}=\frac{AB}{AC}= \frac{BD}{DC}= \frac{BD}{DE}= \frac{\sin(\angle BED)}{\sin(\angle EBD)}= \frac{\sin(\angle ABC}{\angle EBD} \implies \sin(\angle ACB)=\sin(\angle EBD) \implies \angle ACB= \angle EBD$$This is because the angles can’t be supplementary (If so, sum of the angles in $\triangle EBD$ Will be greater than 180, absurd)
This means that $AEDC$ Is Cyclic
So $$\angle EAD =\angle CAD =\angle CED \implies \text{ DE tangent to circumcircle of (AEP)} \implies DE^2=DP \cdot DA$$Now by LoS on $\triangle ADC$ We get $AD = \sqrt{2}\cdot DC$ This means $$DP \cdot \sqrt{2} \cdot DC = DE^2 = DC^2 \implies DP = \sqrt{\frac{1}{2}} \cdot DP = \frac{DA}{2}$$as desired $\blacksquare$
sunken rock
28.08.2021 22:29
What about a synthetic solution? It's quite nice and easy!
sunken rock
30.08.2021 17:58
So, nobody ... take a look at https://stanfulger.blogspot.com/2021/08/aops-bmo-2019-sl.html Best regards, sunken rock
Mahdi_Mashayekhi
26.02.2022 09:50
Easy Claim1 : $CDEA$ is cyclic. Proof : we have $\angle DAC = \angle DAE$ and $DC = DE$ so $CDEA$ is either kite or cyclic. Note that if $CDEA$ is kite then $\angle DEB = \angle 135$ so in triangle $DEB$ sum of angles is more than 180 witch gives contradiction so $CDEA$ is cyclic. Now we have $\frac {AP}{DP} = \frac {CA.\sin{15}}{CD.sin{30}}$ so we need to prove $\frac {CD}{CA} = \frac {sin{15}}{sin{30}}$. we know $\frac {CD}{CA} = \frac {sin{30}}{sin{75}}$ so we need to prove $\sin^2{30} = \sin{15}.\sin{75}$. we know $2\sin{a}.\sin{b} = \cos{(a-b)} - \cos{(a+b)}$ so $\sin{15}.\sin{75} = (\cos{60} - \cos{90})/2 = 0.25 = \sin^2{30}$. we're Done.
Tellocan
23.01.2024 16:05
Basic cyclic + trigo kills