Y by samrocksnature, megarnie, son7, HWenslawski, Adventure10, Mango247, and 1 other user
Let 
be a non-equilateral, acute triangle with 
, and let 
and 
denote the circumcenter and orthocenter of , respectively.
(a) Prove that line
intersects both segments 
and 
.
(b) Line intersects segments and at 
and 
, respectively. Denote by 
and 
the respective areas of triangle 
and quadrilateral 
. Determine the range of possible values for 
.
be a non-equilateral, acute triangle with 
, and let 
and 
denote the circumcenter and orthocenter of , respectively.(a) Prove that line
intersects both segments 
and 
.(b) Line
intersects segments and at 
and 
, respectively. Denote by 
and 
the respective areas of triangle 
and quadrilateral 
. Determine the range of possible values for 
.This post has been edited 1 time. Last edited by djmathman, Apr 6, 2015, 6:43 PM
is cyclic. This means 
and get 
for 
.
.
,
,
is cyclic.
,
,
,
.
,
,
,
equalaterial. (Actually 
and 
are congruent.)
is 
,
be points on 
.
be the feet of the altitudes from 
,
,
with 
,
and 
.
lies on the perpendicular bisector of 
.
.
,
are collinear.
is a rhombus formed by equilateral triangles 
and 
.
,
.
,
,
is between 
and 
.
.
and 
.
gives us 
.
and 
.
.
is 
.
,
.
,
is 
.![\[t=\sqrt{3}(1+3x)/2-\sqrt{3}(1+2x+x^2)/4=\sqrt{3}(2+6x-1-2x-x^2)/4=\sqrt{3}(1+4x-x^2)/4\]](http://latex.artofproblemsolving.com/2/4/f/24f8026bcb9ae2633b62546f840d8f8f16fa2a8b.png)
![\[\frac{s}{t}=\frac{1+2x+x^2}{1+4x-x^2}\]](http://latex.artofproblemsolving.com/0/6/1/0616d21733ac0a17e043cf2b7de6134fb71c6c70.png)
.
to get 
,
.
,
.
to get 
,
.
since 
.
.
.
because otherwise, 
by 30-60-90 triangles, 
.
is a 30-60-90 triangle (Consider quadrilateral 
.
and 
)
.
is also 
,
.
.
.
.
.
if and only if 
.
.
if and only if 
.
has no roots in the range of 
is continuous. 
when 
,
when 
.
.
to vary (some work out better than others), but in the end 
and you know what domain you need so you get the range.
and it worked out very nicely.
.
,
,
and 
for part 2a?
.
due to 
and the perpendicular bisector of 
.
.
and the altitude to 
,
.
,
at 
and 
at 
.
which is evident by 
.
,
and the area of 
will be 
.
Let us substitute 
.
We need to find the minimum and maximum of this expression on the range 
.
which is a quadratic in 
with negative first coefficient. Therefore the maximum of this expression is assumed at 
or 
,
.
we get that the minimum is 
.
and the maximum is 
,
.
is equalaterial (as we get two more 
angles there),
,
.
and 
,
as 
,
,
is the midpoint of 
.
,
than 
intersects 
.
and the foot from 
.
is equilateral.
.
because 
is a 
.
is a 
.
,
.
is the radius of 
,
,
.
,
,
and in particular, 
,
.![$[APQ]=\frac{(b+c)^2\sqrt{3}}{36}$](http://latex.artofproblemsolving.com/f/6/6/f66b5c7c0d43208c2930c84b9943179f8f1e55bb.png)
,![$[ABC]=\frac{bc\sqrt{3}}{4}$](http://latex.artofproblemsolving.com/3/d/5/3d5c9c9629e52c806cc6c5779e4651be64ef1f2b.png)
,![$\frac{[ABC]}{[APQ]}=\frac{9bc}{(b+c)^2}$](http://latex.artofproblemsolving.com/f/e/f/fef78081e48f272f07bb732b69233c3d423c357e.png)
.
,
and 
.![$\frac{[ABC]}{[APQ]}<\frac{9}{4}$](http://latex.artofproblemsolving.com/9/6/7/967bbe25fde2cc98214af571bc63a0493f4cd55c.png)
.![$\frac{[ABC]}{[APQ]}$](http://latex.artofproblemsolving.com/a/6/b/a6b2076775b10a2f35bb7be20d0d30b0a7ab64f0.png)
become arbitrarily close to 
by considering 
approaching 
,
.
,
.
,
,
.![$\frac{[ABC]}{[APQ]}-1=\frac{[BPQC]}{[APQ]}$](http://latex.artofproblemsolving.com/7/9/4/7941d8843aca12dfdf9b3fe4044b92ffc1f54e4e.png)
,![$1<\frac{[BPQC]}{[APQ]}<\frac{5}{4}$](http://latex.artofproblemsolving.com/b/9/9/b99aafe3a862b66748e2df5a6edfeebd4023f2f9.png)
,![$\frac{4}{5}<\frac{[APQ]}{[BPQC]}<1$](http://latex.artofproblemsolving.com/e/b/a/eba326ea1e3068c4a52caaacc16c6fae908c352b.png)
,
denote the incenter, and recall (``fact 5'') that the arc midpoint 
,
.![\[ \angle BOC = \angle BIC = \angle BHC = 120^{\circ}. \]](http://latex.artofproblemsolving.com/6/a/d/6ade5192df8e758d2d9d7f91b9d81a1c02fadeaa.png)
Since all three centers lie inside 
as 
of ![[asy] pair A = dir(50); pair B = dir(210); pair C = dir(330);
pair M = dir(270); pair I = incenter(A, B, C); pair O = circumcenter(A, B, C); pair H = orthocenter(A, B, C); pair P = extension(O, H, A, B); pair Q = extension(O, H, A, C);
filldraw(A--P--Q--cycle, invisible, deepgreen); filldraw(B--P--Q--C--cycle, invisible, deepgreen); filldraw(circumcircle(A, B, C), invisible, red); filldraw(CP(M, I), invisible, dotted+blue); draw(A--M, red);
filldraw(A--O--M--H--cycle, invisible, lightred);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(M)); dot("$I$", I, dir(20)); dot("$O$", O, dir(90)); dot("$H$", H, dir(50)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q));
/* TSQ Source:
A = dir 50 B = dir 210 C = dir 330
M = dir 270 I = incenter A B C R20 O = circumcenter A B C R90 H = orthocenter A B C R50 P = extension O H A B Q = extension O H A C
A--P--Q--cycle 0.2 lightgreen / deepgreen B--P--Q--C--cycle 0.2 lightcyan / deepgreen circumcircle A B C 0.1 yellow / red CP M I 0.1 lightblue / dotted blue A--M red
A--O--M--H--cycle 0.1 orange / lightred
*/ [/asy]](http://latex.artofproblemsolving.com/8/8/0/8802a3fd8bf77eb00909bf303df563aad6f10b17.png)
.
,
,
is a rhombus. Thus 
and in this way we derive that 
,
,
is isosceles and 
.
.
as needed. ![\[ \frac{s}{s+t} = \frac{[APQ]}{[ABC]} = \frac{\frac{\sqrt3}4 \left( \frac{b+c}{3} \right)^2} {\frac{\sqrt3}4 bc} = \frac{b^2+bc+c^2}{9bc} = \frac19 \left( 1 + \frac bc + \frac cb \right). \]](http://latex.artofproblemsolving.com/0/c/c/0cc185e3f5d5058aa073d0fb862042fb70bb28e3.png)
.
the circumcircle of 
,
and 
can be achieved.![\[ \frac{1}{2} < \frac bc < 2 \implies 4/9 < \frac{s}{s+t} < 1/2 \implies 4/5 < \frac st < 1 \]](http://latex.artofproblemsolving.com/4/1/3/4133e034c0ec33e9d2a1772cf0d2866c5d84d73a.png)
as needed.
.
and 
,
.
be the reflection of 
,
,
,
in 
,![\[ \angle HOB = \tfrac12 \widehat{BH} = \tfrac12 \widehat{BH'} = \angle BAH' = 90-B.\]](http://latex.artofproblemsolving.com/4/b/d/4bda40653e168ce9fe18b7abfc315d2fd718090b.png)
Now,![\[ \angle BHO = 180-\angle HBO-\angle HOB = 180-(60-C)-(90-B) = 150, \]](http://latex.artofproblemsolving.com/0/7/1/071a35bc2639c0e95a54aa62c4cb796a1f98a27a.png)
so![\[ \angle AHO = 360-\angle BHA - 150 = 360-(180-C)-150 = 150-B.\]](http://latex.artofproblemsolving.com/4/9/e/49e547f5c72d6ba1fdccad1e8c25a4f4d36dcd80.png)
Therefore,![\[ \angle APH = 180-\angle PAH - (180-\angle AHO) = 60. \]](http://latex.artofproblemsolving.com/7/5/f/75f3500665a80e927092830b4fd7403141ec6561.png)
This means that 
,
.
,![\[ \frac{AQ}{\sin \angle AHQ} = \frac{AH}{\sin 60} \implies AQ = R\frac{\sin(150-B)}{\sqrt3/2}=\frac{2R}{\sqrt3}\sin(30+B). \]](http://latex.artofproblemsolving.com/4/d/3/4d364ecf58df10f819c7c4646848d4fa791b2c2b.png)
Now, we find the areas. We have![\begin{align*}
[APQ] &= \frac{\sqrt3}{4} AQ^2 \\
&= \frac{\sqrt3}{4} \cdot \left(\frac{2R\sin (30+B)}{\sqrt3}\right)^2 \\
&= \frac{R^2\sin^2(30+B)}{\sqrt3}.
\end{align*}](http://latex.artofproblemsolving.com/5/c/7/5c70dde99329fa2643f6b956f66e04a1888ab6e4.png)
And we know![\[ [ABC] = \frac12 bc\sin 60 = \frac{\sqrt3}{4}(2R\sin B)(2R\sin C) = R^2 \sqrt3 \sin B \sin(120-B). \]](http://latex.artofproblemsolving.com/2/3/a/23a2da04db2a3acf7b07bce59509d2fe7f4a4901.png)
Therefore,![\begin{align*}
\frac{s}{s+t} &= \frac{[APQ]}{[ABC]} \\
&= \frac{\sin^2(30+B)/\sqrt3}{\sqrt3 \sin B \sin(120-B)} \\
&= \frac{\sin^2(30+B)}{3\sin B\sin (60+B) } \\
&= \frac{\sin^2x}{3\sin (x-30) \sin (x+30)}
\end{align*}](http://latex.artofproblemsolving.com/e/8/5/e857fb582966409c374ffd5550a442bd613f4444.png)
where we let 
,
and 
.![\[ \frac{\sin^2x}{3(\sin^2x-1/4)} = \frac{a}{3(a-1/4)}\]](http://latex.artofproblemsolving.com/d/e/6/de64bdda26857dd3ca5eac666143cb8bbccbfc80.png)
for 
and 
.
,![\[ \frac{s}{s+t} \in \left(\frac12, \frac49 \right) \implies \frac{s}{t} \in \left(\frac45,1\right). \]](http://latex.artofproblemsolving.com/e/f/1/ef1d5534b0e11c83ff2f02b1e257d4f0941f32f5.png)
then 
(
,but also : 
. Thus 
, so 
, done.
We claim that 
let 
and 
Then note 
and 
Since 
is a parallelogram. Since 
it follows that 
and 
is the perpendicular bisector of 
intersects 
Note 
and 
Since 
Since 
and 
Note that 
and 
Without loss of generality, let 
Then note 
implying that 
Thus 
![$[APQ]=\frac{AD^2\sqrt{3}}{12}=\frac{\sin^2(C+30^{\circ})\sqrt{3}}{3}$](http://latex.artofproblemsolving.com/5/f/4/5f49f7d05da614ab1efdbb234b5f4fe99b14d7c7.png)
and ![$[ABC]=\frac{1}{2}\cdot AC\cdot BC\cdot \sin C=\frac{\sqrt{3}}{2}\cdot AC\sin C=\sqrt{3}\sin C\sin B,$](http://latex.artofproblemsolving.com/b/5/3/b534e995faadadc374ac2d7c09862a7b07f51144.png)
where the last step follows from the Extended Law of Sines. Also note 
Let 
Then![\[\frac{[APQ]}{[ABC]}=\frac{\frac{\sqrt{3}}{3}\sin^2\theta}{\sqrt{3}\sin (\theta-30)\sin(\theta+30)}=\frac{1}{3}\cdot \frac{\sin^2\theta}{(\sin\theta\cos 30^{\circ})^2-(\cos\theta\sin 30^{\circ})^2}=\frac{1}{3}\cdot \frac{\sin^2\theta}{\frac{3}{4}\sin^2\theta-\frac{1}{4}\cos^2\theta}=\frac{1}{3}\cdot \frac{1}{\frac{3}{4}-\frac{1}{4}\cot^2\theta},\]](http://latex.artofproblemsolving.com/0/8/4/08465c9b47b3287302a8d986722f776ecf4471a1.png)
where 
since 
The minimum is 
and the maximum is 
This gives us ![$\frac{[APQ]}{[ABC]}\in (\frac{4}{9},\frac{1}{2}),$](http://latex.artofproblemsolving.com/0/a/2/0a235186ae11e9c68862e0f414fe56b65086c87f.png)
giving us 
as desired.
lie on the same side of 
,![$[ABC]=\frac{1}{2} \sin(60)\sin(B)\sin(120-B)$](http://latex.artofproblemsolving.com/3/2/7/327f79827ca556b2cd18094746315112f6343711.png)
.
and 
,
. Thus we compute![\begin{align*}
\frac{[APQ]}{[ABC]} &= \frac{AP^2 \cdot \sqrt{3}/4}{\sin(B)\sin(120-B) \cdot \sqrt{3}/4} \\
&=\frac{1}{3} \cdot \frac{\sin^2(B+30)}{\sin(B)\sin(120-B)} \\
&=\frac{1}{3} \cdot \frac{\sin^2(x)}{\sin(x-30)\sin(x+30)} \\
&=\frac{1}{3} \cdot \frac{\sin^2(x)}{\frac{3}{4}\sin^2(x)-\frac{1}{4}\cos^2(x)} \\
&=\frac{1}{3} \cdot \left( 1+ \frac{1}{4\sin^2(x)-1} \right) \\
\end{align*}](http://latex.artofproblemsolving.com/9/4/5/9451fb4577bfeae4d1e0ee21ccaab6c9cc36c60c.png)
where we set 
.
we have 
,![$$\frac{[APQ]}{[ABC]} \in (\frac{4}{3},\frac{3}{2})$$](http://latex.artofproblemsolving.com/8/3/3/833421401c62865fca832ae4a9eb2b4d6e659a6e.png)
. It's easy to see that every value in this range can indeed be achieved by varying 
acute (since 
)![$\frac{[APQ]}{[BCQP]} \in (\frac{4}{5},1)$](http://latex.artofproblemsolving.com/b/d/6/bd6a184ef7fab6348c614c5477c697423ad86467.png)
.
,
.
and this is obviously a contradiction, because we get that 
forcing 
meaning BHOC is cyclic. We have 
so 
meaning 
.
then 
.
is isosceles so 
.
.
(
)
to 
and since quadratics are continuous, this is a continuous range and we are done.
be the foot of altitude from 
.
.
is cyclic
.
is convex. 
.
By symmetry, 
thus 
.![$$\frac{[APQ]}{[BPQC]} = \frac{AP \cdot AQ}{AB \cdot AC - AP \cdot AQ} = \frac{\left(\frac{b+3}{3}\right)^2}{bc-\left(\frac{b+3}{3}\right)^2} = \frac{1}{b(3-b)-1} \ge \frac{4}{5}$$](http://latex.artofproblemsolving.com/4/2/6/426832e954a65debb66d5b708a741007b34d57ff.png)
WLOG 
.
by law of cosines. 
.
.
.
doesn't intersect 
and 
as any line passing through the interior of 
and (abusing notation),
hence 
,
,
.
,
,
,
,
Since any valid choice of 
with 
yields a valid 
(which gives 
(which gives 
by AM-GM, it follows that
,
,
.
lie on 
,
meet 
,
which means 
,
so ![$$[APQ] = \frac{AP^2 \sqrt{3}}{4} = \sin^2 (B + 30^{\circ}) \cdot \frac{R^2}{\sqrt{3}}.$$](http://latex.artofproblemsolving.com/9/1/7/917e22b65751860ffa0222f41e32be32f231c10e.png)
Now, the Extended LoS gives ![$$[ABC] = \frac{abc}{4R} = \frac{(2R \cdot \sin 60^{\circ}) \cdot (2R \cdot \sin B) \cdot (2R \cdot \sin C)}{4R}$$](http://latex.artofproblemsolving.com/d/8/8/d88ebd90c2a2bbe618557763bedbbf8181066d72.png)
Hence, ![$$\frac{s}{t} = \frac{[APQ]}{[BPQC]} = \frac{[APQ]}{[ABC] - [APQ]} = \frac{\sin^2 (B + 30^{\circ})}{3 \cdot \sin B \cdot \sin (120^{\circ} - B) - \sin^2 (B + 30^{\circ})}.$$](http://latex.artofproblemsolving.com/2/5/4/2547e3d220be877ee69e1971b25fda4fb543d707.png)
as desired. 
Now, let 
.
Because 
,
.
is strictly decreasing on the open interval 
,
or 
which implies 
is strictly decreasing for all 
.
or 
and 
,
and
Which is between 
and 
.
