Let $AA_1$, $CC_1$ be the altitudes of $\Delta ABC$, and $P$ be an arbitrary point of side $BC$. Point $Q$ on the line $AB$ is such that $QP = PC_1$, and point $R$ on the line $AC$ is such that $RP = CP$. Prove that $QA_1RA$ is a cyclic quadrilateral.
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Tags: geometry
06.03.2019 16:09
Construct a parallel to $CC_1$ through $Q$ and Let it intersect $BC$ at $E$ and then see the magic
06.03.2019 16:13
I complex bashed it BTW I haad constructed the parallel line , but didn't notice one obvious thing
07.03.2019 01:48
I solved P14 and thought this was impossible >.>
16.03.2019 04:37
AlastorMoody wrote: Construct a parallel to $CC_1$ through $Q$ and Let it intersect $BC$ at $E$ and then see the magic Can I ask what motivates this?
13.08.2019 03:49
Jesus Christ this took me forever [asy][asy] size(250); defaultpen(linewidth(0.7)); pair A = origin, B = (5,12), C = (14,0), Ap = foot(A,B,C), Cp = foot(C,A,B), P = 0.7*B + 0.3*C, M = foot(P,A,C), N = foot(P,A,B), Q = 2*N - Cp, R = 2*M-C, D = 2*P - C; draw(A--B--C--A,lightblue); draw(A--Ap^^C--Cp^^B--Q--P--R,heavycyan); draw(Q--D--B^^R--D^^N--P--M^^rightanglemark(P,M,C,16)^^rightanglemark(D,R,A,16)^^rightanglemark(P,N,A,16),orange); draw(circumcircle(A,R,Ap),red+linetype("3 3")); dot("$A$",A,SW); dot("$B$",B,W); dot("$C$",C,SE); dot("$A_1$",Ap,dir(A--Ap)); dot("$C_1$",Cp,dir(C--Cp)); dot("$P$",P,dir(A--P)); dot("$Q$",Q,dir(A--Q)); dot("$R$",R,S); dot("$M$",M,dir(P--M)); dot("$N$",N,dir(P--N)); dot("$D$",D,dir(C--D)); [/asy][/asy] Let $D$ be the reflection of $C$ across $P$, and let $M$ and $N$ denote the midpoints of $\overline{RC}$ and $\overline{QC_1}$ respectively. Observe that by similar triangles, $PM\parallel DR$, so $\angle DRA = \angle DA_1A = 90^\circ$; this implies quadrilateral $DA_1RA$ is cyclic. Similarly, since $PN\perp CC_1$, we have $DQ\perp CC_1$ as well, so $\angle DQA = \angle DA_1A = 90^\circ$ and quadrilateral $DQA_1A$ is cyclic. Combining both of these results proves the claim. Kagebaka wrote: AlastorMoody wrote: Construct a parallel to $CC_1$ through $Q$ and Let it intersect $BC$ at $E$ and then see the magic Can I ask what motivates this? Luck, mostly. Okay, actually it wasn't pure luck. I wanted to introduce the point $D$ to find some point on the circumcircle of $\triangle ARA_1$. Since $AMA_1P$ is cyclic, we know that $CM\cdot CA = CA_1\cdot CP$. Now in order to get the length $CR$ we multiply the length $CM$ by $2$; this motivates constructing the point $D$ so that we get $CR\cdot CA = CA_1\cdot CD$, implying the cyclicity. (Only later did I realize I didn't need to do any Power of a Point.)
20.08.2019 05:32
SHREYAS333 wrote: I complex bashed it BTW I haad constructed the parallel line , but didn't notice one obvious thing Is it bash able??I can't find anyway to complex bash it. How to find $p,q,r$?
10.01.2020 18:53
I got it. I kinda guessed the construction and then worked it out.
17.05.2020 05:19
Kagebaka wrote: AlastorMoody wrote: Construct a parallel to $CC_1$ through $Q$ and Let it intersect $BC$ at $E$ and then see the magic Can I ask what motivates this? Interesting. I solved the easier side of the problem first(related to R) and then noticed that the intersection of the circle with center P through R and C intersects BC at the point opposite A on (A,R,A(1),Q). Then it took me way too long to notice that Q$A_{opp}$ is parallel to P$M_{idpoint of QC_1}$