Let $I$ be the incenter of an acute-angled triangle $ABC$. Let $P$, $Q$, $R$ be points on sides $AB$, $BC$, $CA$ respectively, such that $AP=AR$, $BP=BQ$ and $\angle PIQ = \angle BAC$. Prove that $QR \perp AC$.
Problem
Source: III Caucasus Mathematical Olympiad
Tags: geometry
sharp0407
23.03.2018 09:53
AC와 QR이 수직이 됨을 보이는 것은 각ARP가90도가 됨을 보이면 충분하다. 삼각형IAP와 삼각형 IAR은 합동이고(SAS), 삼각형IBP와 삼각형IBQ도 합동이므로(SAS) IP=IQ=IR이 되고 즉 I는 삼각형 PQR의 외심이 된다. 조건(A=PIQ)과 중심각 원주각 사이의 관계를 이용하면 각PRQ=각A/2이고 삼각형 APR은 이등변삼각형이므로 각ARP=90-각A/2이므로 각ARP=(90-A/2)+A/2=90이다.
BSWJY
23.03.2018 11:39
i think sharp0407 is Korean I guess....
integrated_JRC
23.03.2018 11:44
$\textbf{English Version :}$ It is sufficient to show that $AC$ and $QR$ are perpendicular to each $ARP$ to be $90$ degrees. Triangle $IAP$ and triangle $IAR$ are joined $(SAS)$, triangle $IBP$ and triangle $IBQ$ are joined $(SAS)$ $IP = IQ = IR$, that is, $I$ becomes the outer radius of the triangular $PQR$. $ARP = (90-A / 2) + (2)$ Since each triangle $APR$ is an isosceles triangle, each $ARP = 90-A / 2$ is obtained by using the relationship between the condition $(A = PIQ) A / 2 = 90$. ** I don't know Korean. I simply used $\textit{Google Translate}$.
BSWJY
23.03.2018 11:45
Oh thanks!!
sharp0407
23.03.2018 17:24
thanks....i am korean....sorry...i cannot write english....
Mr.Bash
23.03.2018 17:50
bigant146 wrote: Let $I$ be the incenter of an acute-angled triangle $ABC$. Let $P$, $Q$, $R$ be points on sides $AB$, $BC$, $CA$ respectively, such that $AP=AR$, $BP=BQ$ and $\angle PIQ = \angle BAC$. Prove that $QR \perp AC$.
Let $DEF$ be the contact triangle of $ABC$, with $D,E,F$ on $BC,CA,AB$ respectively. Note that $IFP\cong IDQ$, as $IF=ID, QD=PF$ (as $BF=BD; BQ=BP),\angle IDQ=\angle IFP=90^\circ$. So, $IP=IQ$ and similarly, $IP=IR$. So, $I$ is the center of circle $(PQR)$. Note that then, $\angle PRQ = \frac{\angle PIQ}{2}=\frac{\angle BAC}{2}$. Note that as triangle $APR$ is isosceles with $AP=AR, \angle ARP=90^\circ - \angle BAC$. So, $\angle QRA= \angle QRP + \angle PRA = 90^\circ$. Hence, $QR\perp AC$.
$\blacksquare$.
e_plus_pi
23.03.2018 17:58
Nice solution Mr. Bash The Key Point Is to show that I is the circumcenter of $\triangle PQR$. From That By an easy angle chase we get that $\angle QRA= \dfrac {\pi}{2}$.
Mr.Chem-Mathy
24.03.2018 01:10
Too lazy to write Latex so I'll just use words Angle will be denoted by < <ARP=<RPA=90-(<A)/2| PRQ= <A/2 since it subtends the same arc as <IPQ=A Thus, <QRP+<PRA=90 - <A/2 + <A/2 = 90 = <QRA. Important note! I am an idiot so this solution is probably wrong.
Math-Ninja
24.03.2018 03:16
Mr.Chem-Mathy wrote: Too lazy to write $\textbf{\LaTeX}$ so I'll just use words: Angle will be denoted by $\angle$: $\angle ARP=\angle RPA=90-(<A)/2|$ $PRQ= \angle A/2$ since it subtends the same arc as $\angle IPQ=A$ Thus, $\angle QRP+\angle PRA=90 - \angle A/2 + \angle A/2 = 90 = \angle QRA$. Important note! I am an idiot so this solution is probably wrong.
DimPlak
18.04.2018 10:20
How can we construct the figure with ruler and compass?
dikhendzab
16.04.2020 21:20
$\angle CIB=90^\circ+\frac{\alpha}{2}, \angle IPA=\angle IQC=\angle IRA=180^\circ-\angle IRC \implies IQCR$ is cyclic quadrilateral. Since $\angle PIQ=\alpha \implies \angle PIB=\angle QIB=\frac{\alpha}{2}$. Now, using that $IQCR$ is cyclic, we obtain: $\angle CRQ=\angle CIB-\angle BIQ=90^\circ+\frac{\alpha}{2}-\frac{\alpha}{2}=90^\circ \implies QR \perp AC$