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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?
, and the area of the trapezoid is 
. Let the parallel sides of the trapezoid have lengths 
and 
, with 
. Find 
![[asy]
unitsize(0.5 cm);
real r_base = 12 + 6*sqrt(3);
real s_base = 12 - 6*sqrt(3);
real h = 6;
pair A = (-r_base/2, 0);
pair B = ( r_base/2, 0);
pair C = ( s_base/2, h);
pair D = (-s_base/2, h);
draw(A--B--C--D--cycle);
pair O = (0, h/2);
draw(circle(O, 3));
dot(A); label("$A$", A, SW);
dot(B); label("$B$", B, SE);
dot(C); label("$C$", C, NE);
dot(D); label("$D$", D, NW);
dot(O);
draw((0,0)--(0,h), dashed+linewidth(1));
label("$O$", (0,h/2), E);
label("$r$", midpoint(A--B), S);
label("$s$", midpoint(C--D), N);
[/asy]](http://latex.artofproblemsolving.com/7/2/9/7297f663419276a9de5d37a1c530c7277b9c61ab.png)
, another for 
. We can easily find from the formula for an area of a trapezoid:![\[\frac{r + s}{2} \cdot 6 = 72\]](http://latex.artofproblemsolving.com/5/7/b/57bf5824eac6d6a1096693677bdc468038b0d860.png)
![\[r + s = 24\]](http://latex.artofproblemsolving.com/8/1/c/81c5bffb646a79ca75a4c1c7f868f2d4c55e440d.png)
, we utilize Pitot's theorem, which states that 
. Since 
(where x is just an auxiliary variable), we have 
. Thus, 
from the relation we found earlier. using the pythagorean theorem:![\[\left(\frac{r - s}{2}\right)^2 + 6^2 = 12^2\]](http://latex.artofproblemsolving.com/b/f/c/bfcfc460585bde205421a90e1645c595cd14321c.png)
![\[(r - s)^2 = 4 \cdot 108\]](http://latex.artofproblemsolving.com/8/c/a/8cabd1b12d27ad78adaf9c7a926ac2e04e5c0d0c.png)
, we calculate the final answer using the identity![\[\frac{(r + s)^2 + (r - s)^2}{2} = r^2 + s^2\]](http://latex.artofproblemsolving.com/7/a/b/7ab7b04e4b070a4a9d31a1460932ef619f4c9bb4.png)
![\[r^2 + s^2 = \frac{4 \cdot 108 + 24^2}{2} = \boxed{504}\]](http://latex.artofproblemsolving.com/2/2/8/22802e0b7192690cd2653c78bf684d05c40a7361.png)
and 
. We have 
so 
. Furthermore, the legs both have length 
(equal tangents) Drop an altitude from one of the smaller base's vertices, giving a right triangle with hypotenuse 
and legs 
and 
. Thus, 
and it is trivial to find 
.
, so r and s are 
and 
for an answer of 504.
be the bottom base and be the top base,
, 
. Pretty easy for a problem 6, ngl.
We have 
so 
Thus the sum of the other two sides is 
by Pitot, so each of the equal sides is equal to 
Then 
so 
and 
Extract 
![[asy]
unitsize(0.5 cm);
real r_base = 12 + 6*sqrt(3);
real s_base = 12 - 6*sqrt(3);
real h = 6;
pair A = (-r_base/2, 0);
pair B = ( r_base/2, 0);
pair C = ( s_base/2, h);
pair D = (-s_base/2, h);
draw(A--B--C--D--cycle);
pair O = (0, h/2);
draw(circle(O, 3));
dot(A); label("$A$", A, SW);
dot(B); label("$B$", B, SE);
dot(C); label("$C$", C, NE);
dot(D); label("$D$", D, NW);
dot(O);
draw((0,0)--(0,h), dashed+linewidth(1));
label("$O$", (0,h/2), E);
label("$r$", midpoint(A--B), S);
label("$s$", midpoint(C--D), N);
[/asy]](http://latex.artofproblemsolving.com/b/e/5/be52de3ec3b40bf6fff01766dbe16b23d2f8ebbb.png)
Note that 
and let 
be the foot of the altitude from 
to 
We know by equal tangents that 
and 
and by similar 
and 
we have that 
so 
Then, 
, and the area of the trapezoid is . Let the parallel sides of the trapezoid have lengths and , with . Find , so the area is 
which means 
. We can also calculate the area as 
, the product of the semiperimeter and inradius where 
is the length of either leg of the trapezoid, since it has an inscribed circle. Solving, we get 
.
where 
, and let 
be the foot of the perpendicular from 
to 
. We have 
and 
, so by the Pythagorean theorem 
. Then WLOG 
so 
. Recalling that , we can solve for 
to get 
and 
, then:
with 
and 
. Let 
. Let 
be the foot of perpendicular line from 
to . WLOG 
., we have 
. By the area condition, we have 
. By Pythagoras on 
, we should have 
.
, 
, 
. In particular, .
for 
so the bases are 
.
.
,
,
by Pitot.
.
,
implying 
.
.
from the area and radius. Using Pivots theorem, 
equals the sum of the non-parallel sides, so each of them is 
.
which has a length of 
.
,
,
,
.
,
and 
.
.
Then, since the tangents from a point to a circle have equal lengths, it follows that the length of both of the legs of the isosceles trapezoid is 
it then follows that the difference between the lengths of the bases is
Hence, we have 
from which it follows that
meant that it would've had to be a really wide trapezoid, but I decided that it was probably fine.
.
to get 