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.
+3 w
, draw exterior equilateral triangles on sides 
and 
to obtain 
and 
, respectively. Let 
be the intersection of the altitude through 
and the median through 
. Let 
be the intersection of the altitude through 
and line 
. Let 
be the intersection of the median through and the line 
. Prove that , , and lie on a common line.
be a square with sides length 
. Let 
be a path within which does not meet itself and which is composed of line segments 
with 
. Suppose that for every point 
on the boundary of there is a point of at a distance from no greater than 
. Prove that there are two points and of such that the distance between and is not greater than 
and the length of the part of which lies between and is not smaller than 
.
has lengths 
and 
with 
. What is the length of the shorter diagonal of ?
so that the sum of its edges' lengths is minumum.
with equilateral. Let be the area of the triangle of sides 
, 
and 
. Show that 
where 
is the circumradius and 
is the volume of the tetrahedron. be circumscribed about circle 
, and let 
be the orthocenter of 
. The circle touches line 
at 
. The tangent to the circle 
at meets at . Let 
be the midpoint of 
, and let the line 
meet again at . The tangent to parallel to meets the line 
at 
. Prove that 
is tangent to .
we consider a variable point 
which moves on line 
towards . Halfway there, it stops and starts moving in a straight line line towards . Halfway there, it stops and starts moving in a straight line towards , and halfway there it stops and starts moving in a straight line towards , and so on. Show that will get as close as we want to the vertices of a fixed triangle with area 
.
cents, where is a positive integer. He uses the so-called greedy algorithm, successively choosing the coin of greatest value that does not cause the value of his collection to exceed 
For example, to get 42 cents, Silas will choose a 25-cent coin, then a 10-cent coin, then 7 1-cent coins. However, this collection of 9 coins uses more coins than necessary to get a total of 42 cents; indeed, choosing 4 10-cent coins and 2 1-cent coins achieves the same total value with only 6 coins. In general, the greedy algorithm succeeds for a given if no other collection of 1-cent, 10-cent, and 25-cent coins gives a total value of cents using strictly fewer coins than the collection given by the greedy algorithm. Find the number of values of between and 
inclusive for which the greedy algorithm succeeds. in the plane of a square for which ![\[\max (PA, PC)=\frac1{\sqrt2}(PB+PD).\]](http://latex.artofproblemsolving.com/3/9/8/3984b1970837a85af0bac93fd711843630a1fd99.png)
be a triangle with incenter and -excenter 
, whose incircle touches 
at 
. The line 
meets at and is the midpoint of 
. Show that 
.
be a triangle, and let be an arbitrary point on line 
, where 
is the circumcenter of . Define 
and 
as the orthocenters of triangles 
and 
. Prove that 
passes through a fixed point which is independent of the choice of .
satisfy 
?
be an equilateral triangle, be a point inside it, and 
be the intersections of 
with the sides of . If 
are the midpoints of , 
, , show that there is a triangle with sides 
, 
and 
.
![[asy]usepackage("tikz");label("\begin{tikzpicture}
\coordinate[label=above:$A$] (A) at (0,4);
\coordinate[label=below:$C$] (C) at (0,0);
\coordinate[label=left:$B$] (B) at (-1.24,3.57);
\coordinate[label=below:$D$] (D) at (6,0);
\coordinate[label=left:$M$] (M) at (0,2);
\coordinate[label=below:$O$] (O) at (1.94,2);
\coordinate[label=above:$N$] (N) at (3,2);
\draw[thick] (A)--(B)--(C)--(D)--cycle;
\draw[thick] (A)--(C);
\draw[thick] (B)--(D);
\coordinate[label=below left:$E$] (E) at (0,2.95);
\draw[thick,red] (B)--(M)--(N);
\foreach \s in {A,B,C,D,M,O,N,E} \filldraw (\s) circle (2pt);
\path (C)--(D) node[midway,below]{30};
\path (A)--(E) node[right,midway]{5};
\path (E)--(M) node[right,midway]{5};
\path (C)--(M) node[right,midway]{10};
\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/7/5/3/7539933764d9bc616aecd50dadabb72b103459cc.png)
and be a midpoint of and 
, respectively. Let 
and 
. We have![\[\tan \theta = \frac{CE}{CD} = \frac{CE}{CD} = \frac{1}{2}\quad \text{and}\quad \tan (\theta + \alpha) = \frac{AC}{AD} = \frac{2}{3}\]](http://latex.artofproblemsolving.com/9/e/8/9e86f6dbf1b1d156c273e33286ea0df05a5162ab.png)
Then, we have 
, and![\[\sin \alpha = \frac{1}{\sqrt{65}} \quad \text{and}\quad \sin \theta = \frac{1}{\sqrt{5}}\quad \text{and}\quad \cos \theta = \frac{2}{\sqrt{5}}\]](http://latex.artofproblemsolving.com/1/c/a/1caa344f2b303ad9413de07be93d228e66d537da.png)
Because 
, then![\[\angle MOB =\angle CDO = \theta\]](http://latex.artofproblemsolving.com/7/4/6/74682669a09e847d244f29d5068fcb66dfccb214.png)
We have 
which give us 
.Therefore, 
. From LoS:![\[\frac{BM}{\sin \theta} = \frac{BO}{\sin (180^\circ- 2\theta)} = \frac{BO}{2\sin \theta \cos \theta}\Longrightarrow BO = 8\sqrt{5}\]](http://latex.artofproblemsolving.com/9/8/7/98794056865ca8143ec1bbf417383ab8e47a9333.png)
From 
we get 
. Then 
. Note that 
.![\begin{align*}
[ABCD] &= [BCD] + [ADB] \\
&= \frac{1}{2} \cdot 30 \cdot 18\sqrt{5} \cdot \frac{1}{\sqrt{5}} + \frac{1}{2} \cdot 18\sqrt{5} \cdot 10\sqrt{13} \cdot \frac{1}{\sqrt{65}}\\
&= 270 + 90\\
&= \boxed{360}
\end{align*}](http://latex.artofproblemsolving.com/2/2/b/22b635a3f8859aebd926d81038c086203c6173aa.png)
as the origin, 
, and 
. Notice that the circle with diameter , which we will call 
has equation 
. Furthermore, letting 
, we see that is one of the intersections of 
with . We now find the intersections: has equation 
, and plugging this into our equation for gives![\[ (x -10)^2 + (2x - 10)^2 = 100\implies 5x^2 - 60x + 200 = 100\implies x^2 - 12x + 20 = 0\implies x = 10, 2.\]](http://latex.artofproblemsolving.com/0/3/3/0334b2cae12f1c717bf5b1990aaf00a5284faba6.png)
Hence, the intersections are 
and 
. The former solution cannot be since 
. Thus, 
. This gives![\[ [ABC] =\frac{1}{2} \text{dist}(B, AC)\cdot AC = \frac{1}{2}\cdot 6\cdot 20 = 60.\]](http://latex.artofproblemsolving.com/b/7/1/b7190060d1718456c187c0f85daf4fd363eb99d5.png)
FInally, our answer is ![$[ABC] + [ACD] = 60 + 300 = \boxed{360}$](http://latex.artofproblemsolving.com/c/7/7/c779844c1559c4a5af1865fc3f9bcf2532b5dfbc.png)
.
be the foot from to . Since 
and 
(and 
), we have 
. If 
, then obtain 
, 
, 
. It's well known that the altitude to a hypotenuse is the geometric mean of the lengths of the two pieces it cuts the hypotenuse into, so![\[ (5-a)(15+a)=4a^2.\]](http://latex.artofproblemsolving.com/b/c/e/bcefa2a213bdc82a48f3b14c06c546024021a9f3.png)
Solving yields 
. The rest is straightforward as![\[ [ABCD]=[ABC]+[ACD]=60+300=\boxed{360}.\]](http://latex.artofproblemsolving.com/0/8/7/0870854ad8aabc3d6e49f3a5d13ceb7e9f523c8c.png)
,
.
and 
intersect at point 
,
.
and 
for simplicity. Note that 
,
.
and 
.
and 
,
since 
.
,
,
.![$$[ABCD] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} + \frac{1}{2} \cdot 20 \cdot 30 = \boxed{360}.$$](http://latex.artofproblemsolving.com/b/b/8/bb82e735bb9e98c59633e94281668baf28535f3a.png)
.
and 
which gives 
and the answer is $360#
is 300, and when u draw the circumcircle of 
and so the length of the altitude from 
by symmetry with 
so area of 
) w/ graph paper and compass, just draw lines with slopes 
and 
for 
and then draw the circle w/ diameter 
![[asy]
size(300);
draw(circle((0,0), 10));
draw((0,10)--(30,10));
draw((30,10)--(-6,-8));
draw((-6,-8)--(0,10));
draw((0,10)--(0,-10));
draw((0,-10)--(-6,-8));
draw((0,0)--(10,0));
dot((0,0));
label("$O$", (0.29,0.6725), NE * 0.5);
dot((0,10));
label("$A$", (0.29,10.6625), N);
dot((30,10));
label("$D$", (30.26,10.6625), N);
dot((0,-10));
label("$C$", (0.29,-9.3175), S * 2);
dot((10,0));
label("$P$", (10.28,0.6725), NE);
dot((0,-5));
label("$E$", (0.29,-4.3225), S + E * 0.5);
dot((-6,-8));
label("$B$", (-7.0675,-8.9125), SW * 0.25);
[/asy]](http://latex.artofproblemsolving.com/9/c/9/9c9950c9b71a4545f5cdbdddc9414d20dffbd27c.png)
,
such that 
.
,
and the Pythagorean Theorem gives 
,
and 
.![$[ABCD]=\frac{20\cdot 30}{2}+\frac{2\sqrt{10}\cdot 6\sqrt{10}}{2}=360$](http://latex.artofproblemsolving.com/0/8/2/082fba762ebfbcc3c57039d57aac09a000ee5c76.png)
.
,
is similar to triangle 
,
,
as 
by our ratio. Then we also know that triangle 
is similar to triangle 
by 
.![$[ABCD]=[ABC]+[ACD]=\frac{1}{2}(6)(20)+\frac{1}{2}(20)(30)=360 \rightarrow \fbox{D}$](http://latex.artofproblemsolving.com/d/f/3/df39bae95562104428de8cf6e94faa744ddeb644.png)
