AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
with two altitudes 
and 
.Let 
be the point on the line such that 
.The lines 
and 
intersect at point 
, and 
is the point on segment 
such that 
.Suppose that bisects .Prove that triangle 
is isosceles.
s such that:
such that 
. Prove that
of positive integers with 
satisfying 
is
be a positive integer and let 
and 
be sets of integers satisfying the following conditions:
, 
and is a subset of 
, 
iff 
.
be a positive integer. Find the number of permutations 
, 
, 
of the
, 
, 
, satisfying
.
positive integer the equation: 
has positive integer solutions.
be an equilateral triangle. Let 
be a point inside of it such that the square root of the distance of to one of the sides is equal to the sum of the square roots of the distances of to the other two sides. Find the geometric place of .
such that if any 
elements are removed from the set 
, one can still find distinct numbers among the remaining elements with sum .
be a triangle with incenter 
, incircle 
and circumcircle 
. Let 
be the midpoints of sides 
, 
, 
and let 
be the tangency points of with and , respectively. Let 
be the intersections of line with line 
and line 
, respectively, and let be the midpoint of arc 
of . lies on ray 
.
bisects 
.
, 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, .
such that:
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 
.
.
,
,
,
.
,
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 