Juggling concurrent cevian ratios (ft. Asymptote)
by math_explorer, Nov 10, 2010, 10:57 AM
So, today we'll be playing with the standard concurrent cevian figure.
![[asy]unitsize(.6cm);
pair pa=(0,0);
pair pb=(-2, -4);
pair pc=(6, -4);
pair pd=(0, -4);
pair pe=(3, -2);
pair pf=(-1.5, -3);
pair pp=(0, -3.2);
draw(pa--pb);
draw(pb--pc);
draw(pc--pa);
draw(pa--pd);
draw(pb--pe);
draw(pc--pf);
label("$A$", pa, N, black);
label("$B$", pb, S, black);
label("$C$", pc, S, black);
label("$D$", pd, S, black);
label("$E$", pe, N, black);
label("$F$", pf, NW, black);
label("$P$", pp, NE, black);[/asy]](//latex.artofproblemsolving.com/f/7/3/f73366420bb0c15d6b2dbcca4abbb5084aa2677c.png)
Yay a diagram.
totally-well-known
Ceva's Theorem
Hopefully you know that one. Because I'm too lazy to explain it. I think Menelaus is slightly more applicable in general, though. Which, although it can be explained with the above diagram if we pretend
doesn't exist, isn't really that much about the same thing. Anyway.
Gergonne's Theorem
which is not about the Gergonne point, even though both do kinda use cevians. But the Gergonne point is basically degenerate Brianchon. This is just area bash.
Well, I've already told you how to prove it. Area bash. Like![$[PBC]/[ABC]$](//latex.artofproblemsolving.com/7/7/a/77ac13b51f3ea5b707069414e1f49524b5accc80.png)
and go cyclic.
Van Obel's Theorem
which was used in the drawing of this particular figure.
Auxiliary line time! Basically, we reflect
and 
about the center of 
, and extend 
to cut 
and 
at 
and 
.
![[asy]
unitsize(.6cm);
pair pa=(0,0);
pair pb=(-2, -4);
pair pc=(6, -4);
pair pd=(0, -4);
pair pe=(3, -2);
pair pf=(-1.5, -3);
pair pp=(0, -3.2);
pair ppp=(4, -4.8);
pair pdp=(4, -4);
pair px=(0, -4.266);
pair py=(0, -6.4);
draw(pa--pb);
draw(pb--pc);
draw(pc--pa);
draw(pa--pd--px--py);
draw(pb--pe);
draw(pc--pf);
draw(pb--ppp--pc);
draw(ppp--pdp);
draw(pc--py);
label("$A$", pa, N, black);
label("$B$", pb, S, black);
label("$C$", pc, S, black);
label("$D$", pd, NW, black);
label("$E$", pe, N, black);
label("$F$", pf, NW, black);
label("$P$", pp, NE, black);
label("$P'$", ppp, S, black);
label("$D'$", pdp, N, black);
label("$X$", px, SW, black);
label("$Y$", py, S, black);
[/asy]](//latex.artofproblemsolving.com/a/a/8/aa8407065425a65ff329749b6cfa095b9551f42f.png)
(I came up with these auxiliary lines myself, by the way, although I expect others have drawn them before.)
So
and 
.
So we want to prove that
.
Interestingly:
so applying that we get
that is to say
which is easy because
which are equal by symmetry.
Although apparently area bash suffices nevertheless.
![[asy]unitsize(.6cm);
pair pa=(0,0);
pair pb=(-2, -4);
pair pc=(6, -4);
pair pd=(0, -4);
pair pe=(3, -2);
pair pf=(-1.5, -3);
pair pp=(0, -3.2);
draw(pa--pb);
draw(pb--pc);
draw(pc--pa);
draw(pa--pd);
draw(pb--pe);
draw(pc--pf);
label("$A$", pa, N, black);
label("$B$", pb, S, black);
label("$C$", pc, S, black);
label("$D$", pd, S, black);
label("$E$", pe, N, black);
label("$F$", pf, NW, black);
label("$P$", pp, NE, black);[/asy]](http://latex.artofproblemsolving.com/f/7/3/f73366420bb0c15d6b2dbcca4abbb5084aa2677c.png)
Yay a diagram.
totally-well-known
Ceva's Theorem
Hopefully you know that one. Because I'm too lazy to explain it. I think Menelaus is slightly more applicable in general, though. Which, although it can be explained with the above diagram if we pretend
doesn't exist, isn't really that much about the same thing. Anyway.Gergonne's Theorem
which is not about the Gergonne point, even though both do kinda use cevians. But the Gergonne point is basically degenerate Brianchon. This is just area bash.
Well, I've already told you how to prove it. Area bash. Like
![$[PBC]/[ABC]$](http://latex.artofproblemsolving.com/7/7/a/77ac13b51f3ea5b707069414e1f49524b5accc80.png)
and go cyclic.Van Obel's Theorem
which was used in the drawing of this particular figure.
Auxiliary line time! Basically, we reflect
and 
about the center of 
, and extend 
to cut 
and 
at 
and 
.![[asy]
unitsize(.6cm);
pair pa=(0,0);
pair pb=(-2, -4);
pair pc=(6, -4);
pair pd=(0, -4);
pair pe=(3, -2);
pair pf=(-1.5, -3);
pair pp=(0, -3.2);
pair ppp=(4, -4.8);
pair pdp=(4, -4);
pair px=(0, -4.266);
pair py=(0, -6.4);
draw(pa--pb);
draw(pb--pc);
draw(pc--pa);
draw(pa--pd--px--py);
draw(pb--pe);
draw(pc--pf);
draw(pb--ppp--pc);
draw(ppp--pdp);
draw(pc--py);
label("$A$", pa, N, black);
label("$B$", pb, S, black);
label("$C$", pc, S, black);
label("$D$", pd, NW, black);
label("$E$", pe, N, black);
label("$F$", pf, NW, black);
label("$P$", pp, NE, black);
label("$P'$", ppp, S, black);
label("$D'$", pdp, N, black);
label("$X$", px, SW, black);
label("$Y$", py, S, black);
[/asy]](http://latex.artofproblemsolving.com/a/a/8/aa8407065425a65ff329749b6cfa095b9551f42f.png)
(I came up with these auxiliary lines myself, by the way, although I expect others have drawn them before.)
So
and 
.So we want to prove that
.Interestingly:
so applying that we get
that is to say
which is easy because
which are equal by symmetry.
Although apparently area bash suffices nevertheless.
May 2020
January 2020