[/quote]
,
be a positive integer, and let its positive divisors be![\[
d_1 < d_2 < \cdots < d_k.
\]](http://latex.artofproblemsolving.com/6/3/e/63ec82aba0d6c4eece89af6a1247a89498616281.png)
Define 
to be the number of ordered pairs 
with 
such that 
.
.![\[
n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},
\]](http://latex.artofproblemsolving.com/2/7/2/272dc2cc33ebbf695311cb61567fa0297e1299b6.png)
where the 
are distinct primes and the 
are positive integers.
be an interior point for it such that 
.
When does equality hold?
denote the set of all positive integers less than 
that are relatively prime to 
.
reducible over the rational numbers, given that it has integer coefficients?
blocks, each weighing at least 
;
.
with 
you can choose a subset of the blocks whose total weight is at least 
.
such that 
.
chessboard. Initially, Azer colors all cells of this chessboard with some colors. Then, Babek takes 
rows and 
cells in the intersection. Babek wants to have all these 
be non-negative real numbers such that 
Prove that
and 
Find the minimum 
be a positive integer and 
be real numbers such that ![\[a_1+a_2+\dots+a_n=0.\]](http://latex.artofproblemsolving.com/0/f/d/0fde8d75cebc35103ae9acf8ebe393653d336d60.png)
Define the set 
by![\[A=\left\{(i, j)\,|\,1 \leqslant i<j \leqslant n,\left|a_{i}-a_{j}\right| \geqslant 1\right\}\]](http://latex.artofproblemsolving.com/6/9/7/6974ebf20c44bcb4d52d85adc0805e28aafa5d6e.png)
Prove that, if ![\[\sum_{(i, j) \in A} a_{i} a_{j}<0.\]](http://latex.artofproblemsolving.com/b/e/3/be3e65a633dc2ac092a264968f1cf8514d3d6124.png)
is defined by 
and for all positive integers 
,
.
,
such that 
is a multiple of 
and incenter 
.
at 
respectively. Circumcircle of triangle 
intersects 
,
intersects 
distinct from 
.
are concyclic.
.
is a parallelogram. Prove that 
.
Y by Sihakmath, Davi-8191, Durjoy1729, MarkBcc168, Inconsistent, HamstPan38825, jhu08, Adventure10
Let 
be an acute triangle with altitudes 
, 
, and 
, and let 
be the center of its circumcircle. Show that the segments 
, 
, 
, 
, 
, 
dissect the triangle into three pairs of triangles that have equal areas.
be an acute triangle with altitudes 
, 
, and 
, and let 
be the center of its circumcircle. Show that the segments 
, 
, 
, 
, 
, 
dissect the triangle into three pairs of triangles that have equal areas.
,
and 
.
are midpoints of 
and 
then 
is cyclic so 
(
so 
and 
(
)
and 
similarly we get the other equalities.
and 
and 
.
:![$[X]$](http://latex.artofproblemsolving.com/f/6/3/f631f7a53d834de52b718ed6b496689e9658eadf.png)
denote the area of the region 
.
.
and 
.![$\frac{[BOD]}{[AOE]}=\frac{BD\cdot OM}{AE\cdot ON}=\frac{BD\cdot AH}{AE\cdot BH}.
(AH=2OM;BH=2ON)$](http://latex.artofproblemsolving.com/f/b/8/fb80f6bad8b1659cabde498979310b9b705d4623.png)
is cyclic. So,
.
.
are similar
.![$\frac{[BOD]}{[AOE]}=\frac{AC\cdot BE}{AD\cdot BC}=1$](http://latex.artofproblemsolving.com/7/6/5/76511464c50afba95992873c3761d664dd9409e3.png)
.![$[BOD]=[AOE]$](http://latex.artofproblemsolving.com/5/a/4/5a4f55f2b535fc5112c110b6c89332e83aed8f75.png)
.![$[DOC]=[AOF]$](http://latex.artofproblemsolving.com/6/f/0/6f0cfdd697de79412eb907d053c116fa7dba508d.png)
,![$[COE]=[BOF]$](http://latex.artofproblemsolving.com/8/c/7/8c75d2c95b9374b4e308a722fb55eebcdd032a13.png)
.
and 
,
is a parallelogram and hence 
.
.
be 
and let 
be the circumradius. We will find the area of 
and the others will follow by symmetry. Notice that 
.
,
.
.![$2[OCD] = bR\cos{A}\cos{C}$](http://latex.artofproblemsolving.com/d/a/a/daa58e1ef90c2f9787a56000316205c1980d896d.png)
.![$2[OAF] = bR\cos{C}\cos{A}$](http://latex.artofproblemsolving.com/e/5/d/e5d624464785579d426bebe85b5bba4ee9ba6a81.png)
,![$[OCD] = [OAF]$](http://latex.artofproblemsolving.com/e/c/e/ece386570a1a9e0390df7573c0f0f69e52454971.png)
.
and 
have equal areas as well, so done.
,
,
,
,
,
,
![\begin{align*}[AOE] &= [ABC]\cdot\frac{1}{(a^2S_A+b^2S_B+c^2S_C)\cdot (S_BS_C+S_AS_B)}\begin{vmatrix}1 & 0 & 0\\ a^2S_A & b^2S_B & c^2S_C\\ S_BS_C & 0 & S_AS_B\end{vmatrix}\\ &= [ABC]\cdot\frac{b^2S_AS_B^2}{(a^2S_A+b^2S_B+c^2S_C)\cdot (S_BS_C+S_AS_B)}\\ &= [ABC]\cdot\frac{b^2S_AS_B}{(a^2S_A+b^2S_B+c^2S_C)\cdot (S_C+S_A)} \\
&= [ABC] \cdot \frac{S_AS_B}{a^2S_A + b^2S_B + c^2S_C}
\end{align*}](http://latex.artofproblemsolving.com/3/8/b/38b98aad9b4369fb2e77c4cf65ec99ffda6deb27.png)
![\begin{align*}
[BDO] &= [ABC] \cdot \frac{1}{(a^2S_A + b^2S_B + c^2S_C) \cdot (S_CS_A + S_AS_B)} \cdot \begin{vmatrix}
0 & 1 & 0\\
0 & S_CS_A & S_AS_B \\
a^2S_A & b^2S_B & c^2S_C\\
\end{vmatrix} \\
&= [ABC] \cdot \frac{a^2S_A^2S_B}{(a^2S_A + b^2S_B + c^2S_C) \cdot (S_CS_A + S_AS_B)} \\
&= [ABC] \cdot \frac{a^2S_AS_B}{(a^2S_A + b^2S_B + c^2S_C) \cdot (S_C + S_B)} \\
&= [ABC] \cdot \frac{S_AS_B}{a^2S_A + b^2S_B + c^2S_C}
\end{align*}](http://latex.artofproblemsolving.com/5/8/7/587d0835acbc490e060f0bb10a6dfad0b85269c3.png)
![$[AOE] = [BDO]$](http://latex.artofproblemsolving.com/5/0/8/508af026077b0de6f7a1cb115998ec9b99bb2142.png)
.
such that 
and 
such that 
.
,
.
and 
.
![$[OBD]=[OAE]$](http://latex.artofproblemsolving.com/8/7/3/873c783a657d833b69575d48f83bac1e9e12fe37.png)
.
,
.
,
.![$[OBD]=\frac{1}{2}c\cos{B}\cdot AM=\frac{cR^2\sin{2A}\cos{B}}{2a}$](http://latex.artofproblemsolving.com/1/9/0/1901edd50c85fa19d6460148fb43beed3c2fd22a.png)
.
,
is the foot of the perpendicular from 
,
.
,![$[OAE]=\frac{1}{2}ON\cdot AE=\frac{R^2c\sin{2B}\cos{A}}{2b}$](http://latex.artofproblemsolving.com/0/8/5/08597822fd6181d682a3fb1b3b65ed439372e970.png)
.
,
,
is similar to 
![$[OFB]=[OEC]$](http://latex.artofproblemsolving.com/6/d/2/6d2b640f7a79f2c70831b00505870be9ef3ef27d.png)
. Then by symmetry other parts will follow .So ,![$[OFB]=\frac{1}{2} \cdot BF \cdot OT=\frac{1}{2} \left(acosB \right) \left(RcosC \right) = \frac{1}{2} \left(acosC \right) \left(RcosB \right) = \frac{1}{2} \cdot CE \cdot OT'=[OEC]$](http://latex.artofproblemsolving.com/b/8/2/b82f6ceb6d359166b8b8c416885bf52985afffa9.png)
. Note that 
is midpt. of 
is midpt. of 
![$[OAF] = [OCD]$](http://latex.artofproblemsolving.com/2/4/6/246e4d739bc730302466dfc241a2afd0e8975302.png)
.
be the midpoints of 
.
.
.
,
done.![$[OCE]=[OBF]$](http://latex.artofproblemsolving.com/d/a/f/daf894b1b9dc23b73ee7f61b4ade810543fa03a3.png)
and ![$[OAE]=[OBD]$](http://latex.artofproblemsolving.com/d/c/8/dc885e52bf04878381f3bca989b1470b6252e71f.png)
so the problem is finished.
![$[AFO]=[DCO]$](http://latex.artofproblemsolving.com/9/8/5/985cbe0ee75b0fc88ca0cb6d30f5c75f652a1b58.png)
.
and 
,![\begin{align*} [AFO] &= \frac12 \cdot |AF| \cdot R \cdot \sin{(90^{\circ}-\gamma)}
\\ &= \frac12 \cdot |AF| \cdot R \cdot \frac{|CD|}{|AC|}
\\ &= \frac12 \cdot |CD| \cdot R \cdot \frac{|AF|}{|AC|}
\\ &= \frac12 \cdot |CD| \cdot R \cdot \sin{(90^{\circ}-\alpha)}
\\ &= [DCO] \end{align*}](http://latex.artofproblemsolving.com/2/0/7/2077507efb4bb170f9521cc3605737a404640b23.png)
Thus, we're done.
![$[OAE] = [OBD]$](http://latex.artofproblemsolving.com/9/7/4/974fa13f7d17d4337a05b16d65aea467a2478bc6.png)
.
be the intersection points of 
with 
and 
with 
respectively.
and 
.
.
.
since 
.
.
;
is cyclic, by extended law of sines we have 
.
and similarly 
by isogonality of 
is cyclic we have 
,![$$\frac{\sin(\angle OCE)}{\sin(\angle OBF)} = \frac{HF}{HE} = \frac{BF}{CE} \Leftrightarrow [BFO] = \tfrac{1}{2} \cdot R \cdot BF \cdot \sin( \angle OBF) = \tfrac{1}{2} \cdot R \cdot CE \cdot \sin(\angle OCE) = [CEO]$$](http://latex.artofproblemsolving.com/0/a/a/0aa9ade123bc0de3beb3b9094633086610df35b9.png)
where 
is the circumradius of ![$[BDO] = [AEO]$](http://latex.artofproblemsolving.com/9/1/c/91c25627831c3ac7124fcbfff9b2b1c47268b77b.png)
and ![$[CDO] = [AFO]$](http://latex.artofproblemsolving.com/f/f/b/ffbee31f28ff32733d1106b3c7b9eeb44ab648a5.png)
,
![$$\frac {[AOE]}{[BOD]} = \frac{\frac{1}{2} \cdot OA \cdot AE \cdot \sin y}{\frac{1}{2} \cdot OB \cdot BD \cdot \sin x} = \frac{AE \cdot \cos B}{BD \cdot \cos A} = \frac{AE \cdot \frac{BD}{AB}}{BD \cdot \frac{AE}{AB}} = 1$$](http://latex.artofproblemsolving.com/c/7/c/c7ce86e4f10088bd1458aa1743ab60abd2914afc.png)
![$[AOE] = [BOD]$](http://latex.artofproblemsolving.com/3/0/a/30a64748ee342e359bf10e89bd245ff2dfe3c59c.png)
and similarly , 
and 
equal by showing that the ratio of the areas of 
and 
.
Now![\begin{align*}
\frac{[AOE]}{[ABC]}&=\frac{1}{(\sum {a^2S_A})\cdot (S_{BC}+S_{AB})}\begin{vmatrix}1 & 0 & 0\\ a^2S_A & b^2S_B & c^2S_C\\ S_{BC} & 0 & S_{AB}\end{vmatrix}\\
&=\frac{1}{(\sum {a^2S_A})\cdot (S_{BC}+S_{AB})}(b^2S_B^2S_A)\\
&=\frac{b^2S_{AB}}{(\sum {a^2S_A})\cdot (S_{C}+S_{A})}
\end{align*}](http://latex.artofproblemsolving.com/f/7/5/f75977cc9101ba6780d2679f4c6e8ea2179ac8d5.png)
We can now remove the sum and 
because both are symmetric in 
symmetric but this is trivial since this is equal to 
,
.
be the midpoint of side 
and 
are similar, triangles 
and 
are similar, so 
and 
and 
are similar, triangles 
and 
are similar, so 
and 
and 
are similar, triangles 
and 
and 
.
.
and 
.![$$x = \frac{[AOF]}{[OCD]} = \frac{AO \cdot AF \cdot \frac{1}{2} \cdot \sin (a)}{OC \cdot DC \cdot \frac{1}{2} \cdot \sin (b)} = \frac{AF}{DC} \cdot \frac{\sin (a)}{\sin (b)}$$](http://latex.artofproblemsolving.com/5/e/a/5eae8dd3dc88b86959b76ec1f5d9f2b5f28dda3c.png)
Note that since 
,
.
,
and similarly, 
.
,
.
as desired. By symmetry, the segments 
and circumradius be ![\[[AOF] = AF \cdot \delta(O, AB) = b\cos{\angle A} \cdot R\cos{\angle C}\]](http://latex.artofproblemsolving.com/f/b/4/fb4b80e887ad221b34ffd869d1c0b522ec68147c.png)
![\[[BOF] = BF \cdot \delta(O, AB) = a\cos{\angle B} \cdot R\cos{\angle C}\]](http://latex.artofproblemsolving.com/c/e/0/ce0c1a280383b4bca7d2ebbe5e692a81b1f15924.png)
![\[[BOD] = BD \cdot \delta(O, BC) = c\cos{\angle B} \cdot R\cos{\angle A}\]](http://latex.artofproblemsolving.com/4/4/a/44a21952fa11f1147c1c1e2bae5dc3cb0b60bbfe.png)
![\[[COD] = CD \cdot \delta(O, BC) = b\cos{\angle C} \cdot R\cos{\angle A}\]](http://latex.artofproblemsolving.com/d/7/b/d7b8d4f0ce214256bd409d6854320a3b6bbd0379.png)
![\[[COE] = CE \cdot \delta(O, AC) = a\cos{\angle C} \cdot R\cos{\angle B}\]](http://latex.artofproblemsolving.com/1/d/e/1de718025360f1eb0292e5db511acf09c50edffd.png)
![\[[AOE] = AE \cdot \delta(O, AC) = c\cos{\angle A} \cdot R\cos{\angle B}\]](http://latex.artofproblemsolving.com/0/9/b/09b4912e30ed898d235d567bc33bc2413b586bdb.png)
hence ![$[AOF] = [COD]$](http://latex.artofproblemsolving.com/1/5/e/15ec5dae74b8c7b448c8fea2d1d82788e6874226.png)
and ![$[BOF] = [COE]$](http://latex.artofproblemsolving.com/9/c/4/9c4e4032e10dbd3d79e52617eb4443eba77e27de.png)
and ![$[BOD] = [AOE]$](http://latex.artofproblemsolving.com/a/9/a/a9a82ee23e2960d25c48aeb55f8e1a1e9f371e64.png)
as desired. ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
represents two times the area of triangle 
be the reflection of 
is parallel to 
is parallel to ![\begin{align*}
[BOF] &= \dfrac{1}{2} [BFA'] \\
&= \dfrac{1}{2} [BCA'] \\
&= \dfrac{1}{2} [ECA'] \\
&= [ECO]
\end{align*}](http://latex.artofproblemsolving.com/2/5/b/25b05eb0f4599b95b5f2a0392a02f61894d9c9e2.png)
where the first and last equalities are due to the base-length triangle area formula.
be the midpoints of 
.
,
.
.
and 
.
and the result easily follows. 
?
respectively![$$[AOF]= \frac12\cdot AF \cdot OX = \frac12 \cdot b\cos{\angle A}\cdot R\cos{\angle C} = \frac12\cdot R\cos{\angle A}\cdot b\cos{\angle C} = \frac12 \cdot CD \cdot OY = [COD] $$](http://latex.artofproblemsolving.com/4/0/0/400989ad6192390da813ae6373bba10bb826f87d.png)
and so we are done 
and 
Hence,![\begin{align*}
[AEO]=
\frac{1}{2}\cdot AO\cdot AE\cdot\sin\angle OAE &= \\
\frac{1}{2}\cdot BO\cdot AE\sin\angle BAD &= \\
\frac{1}{2}\cdot BO\cdot AE\cdot\frac{BD}{AB} &= \\
\frac{1}{2}\cdot BO\cdot BD\cdot\frac{AE}{AB} &= \\
\frac{1}{2}\cdot BO\cdot BD\cdot\sin\angle ABE &= \\
\frac{1}{2}\cdot BO\cdot BD\cdot\sin\angle OBD &= \\
[BDO],
\end{align*}](http://latex.artofproblemsolving.com/e/a/4/ea4dfa8bfc6121497d8ff739da28a7682d6af7b6.png)
the other triangle pairs can be handled similarly. 
![$[AOE]=[BOD]$](http://latex.artofproblemsolving.com/0/8/5/0859dcd268ef7177eec2c32e400833739f9bef37.png)
.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/e/e/9/ee94b5163a5f96eb9e66f62984228cbbbada6f0c.png)
.
,
and 
.
as desired. ![$[BOF]=[COE], [AOF]=[COD]$](http://latex.artofproblemsolving.com/a/e/a/aeacd56596cf32b7b047e6a9e1d1ff694103413c.png)
.
![$[OBF]=[OCE]$](http://latex.artofproblemsolving.com/e/6/d/e6d196b9012e376fd7c0573b89d90a7146d3d935.png)
because they both have area 
.
= 
and use same approach to prove other pairs.![$[AOF]=[COD],$](http://latex.artofproblemsolving.com/8/5/1/851d8ab1842f4e8b87876dae4e7958d70463dbf3.png)
after which the rest of the problem follows by symmetry. Let 
.
Since 
is cyclic, 
and 
have the same circumradius, so 
![$$2[AOF]=AF\cdot AO\cdot \sin\angle OAF=AF\cdot AO\cdot \sin(90-\gamma),$$](http://latex.artofproblemsolving.com/6/1/f/61fa8b3f78c870857f83fa14e5525f8236437eb0.png)
![$$2[COD]=CD\cdot CO\cdot \sin\angle OCD=CD\cdot CO\cdot \sin(90-\alpha)$$](http://latex.artofproblemsolving.com/d/d/8/dd8fdc542bb705e982f73ce526218607594d430d.png)
By our claim and the fact that 
,![$[AOF]=[COD]$](http://latex.artofproblemsolving.com/c/d/c/cdcf06f2060d797bff2c64b57d870b1b4d714cbd.png)
,
and 
,![\begin{align*}[OAF] & =\frac{i}{4}\begin{vmatrix}0&0&1\\a&\frac{1}{a}&1\\\frac{a+b+c-\frac{ab}{c}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{c}{ab}}{2}&1\end{vmatrix} \\
& =\frac{i}{4}\begin{vmatrix}a&\frac{1}{a}\\\frac{a+b+c-\frac{ab}{c}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{c}{ab}}{2}\end{vmatrix} \\
& =\frac{i}{8}\left (\frac{a}{b}+\frac{a}{c}-\frac{c}{b}-\frac{b}{a}-\frac{c}{a}+\frac{b}{c}\right ) \\
& =\frac{i}{4}\begin{vmatrix}\frac{a+b+c-\frac{bc}{a}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{2}\\c&\frac{1}{c}\end{vmatrix} \\
& =\frac{i}{4}\begin{vmatrix}0&0&1\\\frac{a+b+c-\frac{bc}{a}}{2}&\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}}{2}&1\\c&\frac{1}{c}&1\end{vmatrix} \\
& =[OCD].
\end{align*}](http://latex.artofproblemsolving.com/3/c/b/3cb81683e540b9861a24c5c57c9b77ce909f15c6.png)
Done.
.
.
.
![\begin{align*}
[AOF] &= [DOC] \\
[BOF] &= [EOC] \\
[BOD] &=[AOE].
\end{align*}](http://latex.artofproblemsolving.com/2/8/b/28b0a78d240dd24c4f3f5c45d1585f08534ac167.png)
We show each of the equalities above.![$[AOF] = [DOC]$](http://latex.artofproblemsolving.com/a/1/3/a13839eb9424d3b0c4abaeffb52f738b3641cb4c.png)
which is obviously true, so the result follows as all steps are reversible.![$[BOF] = [OEC]$](http://latex.artofproblemsolving.com/4/4/1/4416e5a529e2039385f88501798aba0c41d6e8b2.png)
giving the desired equality.
and we are done. 
![$$[BOD] = \frac{1}{2} \cdot BO \cdot BD \cdot \sin(\angle{OBD}),$$](http://latex.artofproblemsolving.com/0/2/d/02ddd22600366b3bbe5c477a7ea4d7307b3af448.png)
and ![$$[AOE] = \frac{1}{2} \cdot AO \cdot AE \cdot \sin(\angle{OAE}).$$](http://latex.artofproblemsolving.com/4/2/d/42d3c06deb17fc1e0c6ea7f0f649b66f4f4f3902.png)
Since 
it follows that 
so it suffices to prove that 
First, it is easy to see that 
so we have
Now, it is easy to get, through angle-chasing, that 
so it follows that 
Therefore,
as desired. 
and 
.
,
so:
,
and 
respectively.![$[ODC]=[OAF]$](http://latex.artofproblemsolving.com/f/3/4/f34e8bd1ef1010b08e92864567cb1043e1168b1e.png)
and ![$[BFO]=[CEO]$](http://latex.artofproblemsolving.com/e/a/c/eac05ee1677a82716f35af6e3b8d12fce1fa3560.png)
are we're done 
