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km long. From any point on her path she can see exactly 
km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?
be a convex quadrilateral with 
and 
such that the bisectors of acute angles 
and 
intersect at the midpoint of 
Find the square of the area of 
and 
to meet at some point 
. Let 
, with 
, 
, 
. Then 
. Let the midpoint of 
be 
so that 
.
.
and
We also have
, so ![$[ABCD]=[APD]-[BPC]=\frac{1}{2}\cdot\frac{\sqrt{5}}{3}(8\cdot 9-6\cdot 6)=6\sqrt{5}$](http://latex.artofproblemsolving.com/2/7/b/27baf770d369080b53abd3885562eb5feae4c521.png)
, so the answer is 
.
be the midpoint of 
Extend 
to meet at 
So is the incenter of 
The fact that is the midpoint of gives 
and 
is isosceles. Then, angle chase gives 
to 
and 
are 
respectively, 
but also 
so they are both 
So by similar triangle relationships 
and the area of can be computed to be 
The area of 
is then 
Subtracting yields 
be the midpoint of Extend to meet at So is the incenter of The fact that is the midpoint of gives and is isosceles. Then, angle chase gives to and are respectively, but also so they are both So by similar triangle relationships and the area of can be computed to be The area of is then Subtracting yields must be incenter of 
what if is outside of 
. Since 
and 
and 
, calculating in two ways gives 
.
. This means 
so 
. Now, we use my favorite technique of Area=Area:![\begin{align*}
[ABC]&=sr=(\ell^2+6)\ell\cos(A/2)=\frac{1}{2}(AB)(AC)\sin A=\frac{1}{2}(\ell^2+2)(\ell^2+3)\sin A=(\ell^2+2)(\ell^2+3)\sin(A/2)\cos(A/2).
\end{align*}](http://latex.artofproblemsolving.com/d/3/a/d3a2ddebf3d68cf20360e3102f868d039c6e8402.png)
cancels out so we have
This immediately gives 
. Thus, we have 
![\begin{align*}
[BGHC]=[BGI]+[BIC]+[CIH]=\frac{1}{2}\cdot 2\cdot\ell\cos(A/2)+\frac{1}{2}\cdot 7\cdot\ell\cos(A/2)+\frac{1}{2}\cdot 3\cdot\ell\cos(A/2)=6\ell\cos(A/2)=6\sqrt{5},
\end{align*}](http://latex.artofproblemsolving.com/a/d/4/ad40dd5f45964da73870e7e880e2e9ef5afbbcc4.png)
which means ans is 180 and we are done.
, let be the midpoint of , and employ barycentrics on 
with 
. Let 
and 
and 
. Then we can crank out 
and 
. Normalizing, we have 
and 
. Since is the midpoint of , it has coordinates 
However, is also the incenter, so it has coordinates 
Hence, we have 
and we can compare components to get the vector-component equation 
where we used unhomogenized coordinates. We can solved this to get 
. So 
and 
and we can normalize these to 
and 
.
is 
. By Heron's, we have ![$[ABC]=\sqrt{12(12-7)(12-8)(12-9)}=12\sqrt{5}$](http://latex.artofproblemsolving.com/b/e/8/be8952dbc6039459f5effc011c2388ad93ff948d.png)
. To finish, note that ![$[ABCD]=[BAD]+[DCB]$](http://latex.artofproblemsolving.com/d/c/8/dc844306804da9be2cb0babd44efe190da592816.png)
. We have ![$$[BAD]=\begin{vmatrix}1/4&3/4&0\\ 0&1&0\\ 0&0&1\end{vmatrix}\cdot[ABC]=\frac14\cdot 12\sqrt{5}=3\sqrt5 $$](http://latex.artofproblemsolving.com/3/7/e/37e6d0966aece7c2b41aa400917e780599f72a87.png)
and ![$$[DCB]=\begin{vmatrix}0&0&1\\ 1/3&0&2/3\\ 1/4&3/4&0\end{vmatrix}\cdot[ABC]=\frac14\cdot 12\sqrt{5}=3\sqrt5$$](http://latex.artofproblemsolving.com/7/2/f/72f665fceb1ff607c2a35de047dae7a2f57bcb9b.png)
so ![$[ABCD]=6\sqrt5$](http://latex.artofproblemsolving.com/f/6/b/f6bc204626dca26de02abeea92d634b67bed5e10.png)
, so the answer is .
. Then is incenter of 
. Now 
bisects 
and moreover 
. So 
, implying 
. Now easy angle chase yields
Now we simple compute 
by cosine rule in 
:
Hence,![$$ [ABCD] = [AMD] \left( 1 + \left( \frac{\sqrt{14}}{7} \right)^2 + \left( \frac{\sqrt{21}}{7} \right)^2 \right) = \frac{1}{2} \cdot \sqrt{14} \cdot \sqrt{21} \cdot \frac{12}{7} = 6 \sqrt{5} $$](http://latex.artofproblemsolving.com/5/0/4/50402c5c07740d5c0e0e1a34be1b432374e7d056.png)
and 
meet at 
. It is clear that is the incenter of 
. This means 
which implies that 
is isosceles and that 
is the perpendicular bisector of the triangle.
and 
. Here 
which means 
, giving 
suffices to show that 
, so similar triangles again on and 
gives 
Heron's formula on 
gives 
![$$[ABCD]=[AFD]-[BFC] \implies 12 \sqrt{5} - [BFC]$$](http://latex.artofproblemsolving.com/2/d/8/2d8dee69754982e92974c043ecd7ef8a822e7159.png)
By Pythagorean theorem 
, so ![$[BFC]=6 \sqrt{5}$](http://latex.artofproblemsolving.com/e/5/b/e5bf30b9e5a207d3a704aa53bbf319b05da9c1f7.png)
. The answer is 
and let be the midpoint of . Notice that is the incenter of 
. Then drawing in the angle bisector of 
we find that 
from the angle-bisector theorem. This implies 
from SSS.
and let 
. Then notice that 
and that 
. The latter implies 
so 
. This means that 
and 
are similar, and doing a similar angle chase reveals that these two triangles are also similar to 
.
. Using this we find 
and 
. Using Heron's on gives us ![$[ABM]=\sqrt{5}$](http://latex.artofproblemsolving.com/2/5/f/25f72d87dbce20d872caef05e4a92a2b6dcdb30b.png)
. Then from similar triangles, ![$[MCD]=\frac{3\sqrt{5}}{2}$](http://latex.artofproblemsolving.com/8/2/3/8234bb076e3785cf6d133d33a9fc866e3b67da7f.png)
and ![$[AMD]=\frac{7\sqrt{5}}{2}$](http://latex.artofproblemsolving.com/5/f/1/5f1213f42a80824e7d7ed98922031c59b6bc902e.png)
. Thus ![$[ABCD]=6\sqrt{5}\implies 180$](http://latex.artofproblemsolving.com/8/3/4/834f915aa60f62b0e7c6647a4077cf486aa07ba1.png)
.
be the midpoint of . As lies on the bisector of 
, is equidistant from and 
, and as it also lies on the bisector of 
, it is also equidistant from and . As a result, there exists a circle with radius 
centered at that is tangent to , , and . Let , 
, and 
be the respective points of tangency, which are equivalently the projections of onto those sides.
, 
, and 
, triangles 
and 
are congruent by HL. Set 
. We also have that triangles 
and 
are congruent by HL, as are 
and 
, so 
and 
. Then,
It follows that 
and 
.
, 
, 
, and 
are 
by looking along , so
By the angle addition formulas,
and doubling gives
Setting this equal to and taking 
of both sides,
whose only positive real solution is 
. Finally,![$$[ABCD]=[ABM]+[CDM]+[ADM]=\frac{2r}{2}+\frac{3r}{2}+\frac{7r}{2}=6r=6\sqrt{5},$$](http://latex.artofproblemsolving.com/8/3/e/83e21966e28f7a761801bb9359a76350275e0a02.png)
and squaring yields .
be an acute triangle with circumcircle 
. Let 
be the 
-mixtillinear incircle of -- that is, the unique circle internally tangent to that is also tangent to 
and 
at 
and 
, respectively. Suppose 
, 
, and 
. Compute the square of the area of .
, and note that 
and 
are the -mixtilinear touch-points of . If we let 
, then we have, by the above lemma,
Now, the area of ![$[PAD] = 12\sqrt{5}$](http://latex.artofproblemsolving.com/3/b/a/3ba6b43e03ac17bf2c4325cc208465b12d479925.png)
by Heron's formula, so![$$[ABCD] = [PAD] - [PBC] = 12\sqrt{5} - 12\sqrt{5}\cdot\tfrac{6}{8}\cdot\tfrac{6}{9} = 6\sqrt{5},$$](http://latex.artofproblemsolving.com/6/5/d/65d36f532e5fae170fb2375538ee312f57badcd1.png)
and therefore our desired answer is 
.
the midpoint of 
. The key claim is as follows.
and meet at . Since two angle bisectors of 
intersect at , the third one also passes through . Hence, 
is the angle bisector of 
. Now, looking at 
, we see that 
is both an angle bisector and a median, so must be isosceles.
and 
. We compute 
and consequently 
. Hence, 
. Also, 
and 
, so 
and 
. Hence, 
by AA similarity. Furthermore, 
and 
, so these two triangles are also similar to 
by AA, as desired 
. From similarity, we have 
. Now, let 
and 
. Similarity gives 
and 
. Hence, 
and 
. Now, Heron's gives ![$[ABP] = \sqrt{5}$](http://latex.artofproblemsolving.com/7/8/0/78035e69f1b0ccf33d4c2cf36eda60582220b6bf.png)
. By similarity, we have ![$[ADP] = (\sqrt{14}/2)^2[ABP]$](http://latex.artofproblemsolving.com/6/c/9/6c9cc37a0b2759ab81dfc425d678c05cfe7b11b7.png)
and ![$[PCD] = (\sqrt{3/2})^2[ABP]$](http://latex.artofproblemsolving.com/5/3/6/5365732ad4f4c195728a1102c75f06fa70ea6ce6.png)
. Thus, summing the three areas, we obtain![\[ [ABCD] = \left(1 + (\sqrt{14}/2)^2 + (\sqrt{3/2})^2\right)\cdot \sqrt{5} = \sqrt{180},\]](http://latex.artofproblemsolving.com/a/d/3/ad300ccaca951c31afe971cbf67e13a5f5a3fc14.png)
and squaring gives .
it is not too hard to finish with Heron's (if you are used to this type of radical Heron's and even if you aren't it's not too difficult).
be the midpoint of , and extend and . Let the intersection point be . is the incenter of 
. Thus, 
. Now, angle chasing on 
shows that they are similar triangles, which yields . Construct 
and since 
and , 
. Now, we can construct the law of cosines equations and set them equal:
From that, we can find the cosine value of 
via law of cosines, and we need the sine value to compute the area. Trig gives 
. Now finish with 
and we need the square so .
, 
and 
where is the mid point of BC. Then doing sim triangles we get 
and 
. Then notice if the height is dropped for from it slipts the base into 3,4. So area is 7/2 sqrt(5). And rest is just similar areas and add to get 6sqrt5. Hence 180.
and Heron's formula (3 times, or use similar areas) does the rest.
.
, let be the midpoint of , and employ barycentrics on with . Let and and . Then we can crank out and . Normalizing, we have and . Since is the midpoint of , it has coordinates However, is also the incenter, so it has coordinates Hence, we have and we can compare components to get the vector-component equation where we used unhomogenized coordinates. We can solved this to get . So and and we can normalize these to and . is . By Heron's, we have . To finish, note that . We have and so , so the answer is .
be a convex quadrilateral with and such that the bisectors of acute angles and intersect at the midpoint of Find the square of the area of and 
to meet at . Note that is the incenter of . Since 
bisects 
and , we have 
and 
.
and 
. Then 
, so 
, 
, 
, so 
. Similarly, you can get 
.
, so![\[\frac{AB}{BM}=\frac{CM}{CD}\qquad\text{and}\qquad\frac{CD}{MD}=\frac{MD}{AD}.\]](http://latex.artofproblemsolving.com/b/2/6/b2651cecf32ec34d593221e6c6bb8b73d42f8369.png)
From the first equation, 
. From the second, 
. Then via similar triangles again, . is 
, so the area of is 
, and that of 
is 
. Adding, , so the desired answer is .
, and let 
be the midpoint of . Then it is also the incenter of triangle 
. Let 
.
are the tangency points of the -mixtilinear incircle. The 
inverse of this circle is the -excircle. The lengths of the tangents from to these circles are 
and 
(half of perimeter), respectively. Thus 
, so 
.![$[XAD]=12\sqrt5$](http://latex.artofproblemsolving.com/2/8/3/28323c963937ff7c11ed9c4af5aec6ceee697a96.png)
and 
, so the answer is 
.
, and the midpoint of 
. Then is the incenter of 
, from which is follows that 
by Angle Bisector. Then, let 
and 
. Applying the Law of Cosines on wrt to 
and 
gives us 
Drop the altitude from and to , let them be and 
respectively. Let the tangency point of the incircle with be . The semiperimeter of the triangle is 
from which it follows from the 
formulas that 
. Then, 
and 
. Now, because is the midpoint similar triangles gives us that 
so we solve the equation 
Now we have a 
triangle which by Herons is 
, We have an isosceles triangle cut out from it inscribed inside which by ratios has area 
. The desired answer is 
and to a point 
Then, if is the midpoint of 
we have that is the incenter of 
Let 
Now, drawing a height from to 
and 
and calling them and 
we get that 
so 
Let 
By Stewart's on 
and angle bisector theorem on 
we get 
so 
Thus, by simple computations, the area of the quadrilateral is 
and 
,
.
.
and 
,
.
by the definition of reflection, so![\[
\angle ABM = \angle AEM = \angle MFD = \angle MCD
\]](http://latex.artofproblemsolving.com/6/d/c/6dcd2d888e8cc5c539d25b0892d3fb7a1c9831af.png)
and, if 
is the foot of the altitude from 
.
,
,
.
,
.![\[
\angle BMA = 180^\circ - \gamma - \alpha = \beta,
\]](http://latex.artofproblemsolving.com/0/a/c/0ac896acbf765d795989ae037ebaeae75c3d0bdb.png)
and likewise 
.
,
.
yields 
,
The requested answer is 
.
.
.
;
and 
and since 
we have 
.
,![$[ABC]=12\sqrt{5}$](http://latex.artofproblemsolving.com/c/b/4/cb4da9395f77f086bf71de41dc2994e12880f2d8.png)
while ![$[AGH]=IF\cdot AG=6\sqrt{5}$](http://latex.artofproblemsolving.com/3/4/5/34516464550e62736b5205275c0ed4d7dd75254a.png)
and the answer is 
.
and 
to obtain 
and 
,
.
is isosceles, and
so 
.
and 
,
.
.
.
,
,
This means 
,
.
fails the triangle inequality, so 
.![$[ABCD]=\sqrt{5}+2\sqrt{5}+3\sqrt{5}=6\sqrt5\implies\boxed{180}$](http://latex.artofproblemsolving.com/c/9/0/c90521677a79d513fcce2cf413ac33c0260cd943.png)
.
and 
:
.
.![$K = [APD]$](http://latex.artofproblemsolving.com/e/d/f/edf88b4d9ab3a5785b652786d21c892f6b869767.png)
.![\[x = \frac{PI}{\sin \frac P2} = \frac{r}{\sin \frac P2 \cos \frac P2} = \frac{2r}{\sin P} = \frac{\frac{2K}{s}}{\frac{2K}{(x+3)(x+2)}} = \frac{(x+3)(x+2)}{x+6}\]](http://latex.artofproblemsolving.com/b/1/0/b1021ce4962be6047e17e0fed165856313c2f1f1.png)
Solving, we get 
.![$K = 12\sqrt{5} \implies [PBC] = 6 \sqrt{5} \implies [ABCD] = 6\sqrt{5}$](http://latex.artofproblemsolving.com/e/3/3/e33c453081f870777dd8b256a671e87462043ca9.png)
.
.
.
.
.![\[B = (x:2:0); I = (x+3:7:x+2); C = (0:3:x)\]](http://latex.artofproblemsolving.com/a/d/4/ad4b0524c91ee038ed99f755b890b6879e056298.png)
lies on 
.
are collinear, we may solve to get 
.![$[ADE] = 12 \sqrt5$](http://latex.artofproblemsolving.com/8/5/0/85014c0096831875490da9b5519fd0c97c06e975.png)
.![$[BCE] = [ADE] \cdot \frac{2}{2 + 6} \cdot \frac{3}{3 + 6} = [ADE] \cdot \frac12 = 6 \sqrt5$](http://latex.artofproblemsolving.com/3/5/8/3581b444088db2e2da3b2b283013a9b3204563c1.png)
,![$[ABCD] = 12 \sqrt 5 - 6 \sqrt 5 = 6 \sqrt5$](http://latex.artofproblemsolving.com/0/8/9/08959ee54f48456441a6c7170b0671adfa05ab77.png)
.