Oh we meet again sweet Aime
by shiningsunnyday, Jun 21, 2016, 4:19 PM
1987 AIME P15 wrote:
Squares
and
are inscribed in right triangle
, as shown in the figures below. Find
if area
and area
.
![[asy]
size(250);
real a=15, b=5;
real x=a*b/(a+b), y=a/((a^2+b^2)/(a*b)+1);
pair A=(0,b), B=(a,0), C=origin, X=(y,0), Y=(0, y*b/a), Z=foot(Y, A, B), W=foot(X, A, B);
draw(A--B--C--cycle);
draw(W--X--Y--Z);
draw(shift(-(a+b), 0)*(A--B--C--cycle^^(x,0)--(x,x)--(0,x)));
pair point=incenter(A,B,C);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$A$", (A.x-a-b,A.y), dir(point--A));
label("$B$", (B.x-a-b,B.y), dir(point--B));
label("$C$", (C.x-a-b,C.y), dir(point--C));
label("$S_1$", (x/2-a-b, x/2));
label("$S_2$", intersectionpoint(W--Y, X--Z));
dot(A^^B^^C^^(-a-b,0)^^(-b,0)^^(-a-b,b));[/asy]](//latex.artofproblemsolving.com/6/0/d/60d41d88b7334a6696438152f60a005ae8e6d4e3.png)






![[asy]
size(250);
real a=15, b=5;
real x=a*b/(a+b), y=a/((a^2+b^2)/(a*b)+1);
pair A=(0,b), B=(a,0), C=origin, X=(y,0), Y=(0, y*b/a), Z=foot(Y, A, B), W=foot(X, A, B);
draw(A--B--C--cycle);
draw(W--X--Y--Z);
draw(shift(-(a+b), 0)*(A--B--C--cycle^^(x,0)--(x,x)--(0,x)));
pair point=incenter(A,B,C);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$A$", (A.x-a-b,A.y), dir(point--A));
label("$B$", (B.x-a-b,B.y), dir(point--B));
label("$C$", (C.x-a-b,C.y), dir(point--C));
label("$S_1$", (x/2-a-b, x/2));
label("$S_2$", intersectionpoint(W--Y, X--Z));
dot(A^^B^^C^^(-a-b,0)^^(-b,0)^^(-a-b,b));[/asy]](http://latex.artofproblemsolving.com/6/0/d/60d41d88b7334a6696438152f60a005ae8e6d4e3.png)
First denote the small triangle above
as
Now observe that the two triangles bordering the hypotenuse of the
are just the two smaller triangles of
scaled down by a factor of
The two smaller triangles of
has area
which is equivalent to
after simplifying of which we obtain
implying which
as well as implying that the height of
which we'll call
is
that of the height of
allowing up to write
Thus,
Finally, we compute the answer: ![$$(AC+CB)^2=AC^2+BC^2+2AC \cdot CB = 21^2 \cdot 440 + 4[ABC]$$](//latex.artofproblemsolving.com/8/8/d/88dd2a42e24f9064b6d034cd3c3366be5c8a6e99.png)







![$[ABC]-41^2,$](http://latex.artofproblemsolving.com/f/d/0/fd0c76fd791fffae9e1348093e2f2e4fd27ac10f.png)
![$\frac{441}{440}([ABC]-[C]-440),$](http://latex.artofproblemsolving.com/0/a/7/0a7a60534721d5420c45431880d208c77c3e6ce5.png)
![$[ABC]=441[C],$](http://latex.artofproblemsolving.com/6/3/4/634e6435e17b2e7333a73e06352c147a138cd33e.png)






![$$[ABC]=441[C]=\frac{441 \cdot \sqrt{440} \cdot \sqrt{\frac{11}{10}}}{2} =\frac{441\cdot 22}{2}.$$](http://latex.artofproblemsolving.com/4/3/7/4378d93f13a3b657bee838dc24a6b6f4fc113c6f.png)
![$$(AC+CB)^2=AC^2+BC^2+2AC \cdot CB = 21^2 \cdot 440 + 4[ABC]$$](http://latex.artofproblemsolving.com/8/8/d/88dd2a42e24f9064b6d034cd3c3366be5c8a6e99.png)

Tidbit
This was a fun ratio problem. Apparently I solved this in Feb but forgot how to do it, and it took me way too long solve it today. How did I become dumber?? T_T
Also, if I didn't say so already, I'm slowly mastering every AIME in existence, at a much more leisurely pace compared to Feb/March. I just finished 1988, so the difficulty has been quite manageable so far. Hopefully, I'll finish all by the end of summer and start spamming USAMOs!
Also, if I didn't say so already, I'm slowly mastering every AIME in existence, at a much more leisurely pace compared to Feb/March. I just finished 1988, so the difficulty has been quite manageable so far. Hopefully, I'll finish all by the end of summer and start spamming USAMOs!
1998 IMO P2 wrote:
In a competition, there are
contestants and
judges, where
is an odd integer. Each judge rates each contestant as either “pass” or “fail.” Suppose that
is a number such that, for any two judges, their ratings coincide for at most
contestants. Prove that
.
![$\color[rgb]{0.35,0.35,0.35}a$](http://latex.artofproblemsolving.com/e/c/a/eca40b28aef52c6970cbfe347140847e9cd1d9c0.png)
![$\color[rgb]{0.35,0.35,0.35}b$](http://latex.artofproblemsolving.com/d/b/d/dbdb7d40e889fb758607704acfdc8a59eb362204.png)
![$\color[rgb]{0.35,0.35,0.35}b \ge 3$](http://latex.artofproblemsolving.com/7/7/d/77da16d394cee2bb309c7e59d356a81482dabc7d.png)
![$\color[rgb]{0.35,0.35,0.35}k$](http://latex.artofproblemsolving.com/0/4/d/04d93df2f804a46fe1da4931d7ec96909bfc8347.png)
![$\color[rgb]{0.35,0.35,0.35}k$](http://latex.artofproblemsolving.com/0/4/d/04d93df2f804a46fe1da4931d7ec96909bfc8347.png)
![$\color[rgb]{0.35,0.35,0.35}\dfrac{k}{a} \ge \dfrac{b-1}{2b}$](http://latex.artofproblemsolving.com/7/6/2/7624e70dcefc22a943751e7fead1bb3ec89d0c53.png)
We will prove the statement by finding an upper and lower bound of the total number of agreements between pairs of judges. By the definition of
, we can first achieve an upper bound of
On the other hand, let
be the number of judges that rate a random contestant a "pass." It follows the number of agreements for that contestant is
which is minimized when
But because
is odd, the minimum occurs at
Plugging this in, we obtain
Because this is only for one contestant, the lower bound is thus
Finally, we can write the inequality 










Tidbit
Rip I didn't completely solve this. I obtained the inequality by first getting the upper bound (easy), but for the lower bound I did obtain
but used a non-rigorous argument by saying that the expected number of passes/fails should be
and
(since they're closest to half of
), from which follows the lower bound.
...so partial, I guess?




...so partial, I guess?
This post has been edited 3 times. Last edited by shiningsunnyday, Jun 22, 2016, 3:30 PM