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)
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]](http://latex.artofproblemsolving.com/6/0/d/60d41d88b7334a6696438152f60a005ae8e6d4e3.png)
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 ![$[ABC]-41^2,$](http://latex.artofproblemsolving.com/f/d/0/fd0c76fd791fffae9e1348093e2f2e4fd27ac10f.png)
which is equivalent to ![$\frac{441}{440}([ABC]-[C]-440),$](http://latex.artofproblemsolving.com/0/a/7/0a7a60534721d5420c45431880d208c77c3e6ce5.png)
after simplifying of which we obtain ![$[ABC]=441[C],$](http://latex.artofproblemsolving.com/6/3/4/634e6435e17b2e7333a73e06352c147a138cd33e.png)
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, ![$$[ABC]=441[C]=\frac{441 \cdot \sqrt{440} \cdot \sqrt{\frac{11}{10}}}{2} =\frac{441\cdot 22}{2}.$$](http://latex.artofproblemsolving.com/4/3/7/4378d93f13a3b657bee838dc24a6b6f4fc113c6f.png)
Finally, we compute the answer: ![$$(AC+CB)^2=AC^2+BC^2+2AC \cdot CB = 21^2 \cdot 440 + 4[ABC]$$](http://latex.artofproblemsolving.com/8/8/d/88dd2a42e24f9064b6d034cd3c3366be5c8a6e99.png)
1998 IMO P2 wrote:
In a competition, there are ![$\color[rgb]{0.35,0.35,0.35}a$](//latex.artofproblemsolving.com/e/c/a/eca40b28aef52c6970cbfe347140847e9cd1d9c0.png)
contestants and ![$\color[rgb]{0.35,0.35,0.35}b$](//latex.artofproblemsolving.com/d/b/d/dbdb7d40e889fb758607704acfdc8a59eb362204.png)
judges, where ![$\color[rgb]{0.35,0.35,0.35}b \ge 3$](//latex.artofproblemsolving.com/7/7/d/77da16d394cee2bb309c7e59d356a81482dabc7d.png)
is an odd integer. Each judge rates each contestant as either “pass” or “fail.” Suppose that ![$\color[rgb]{0.35,0.35,0.35}k$](//latex.artofproblemsolving.com/0/4/d/04d93df2f804a46fe1da4931d7ec96909bfc8347.png)
is a number such that, for any two judges, their ratings coincide for at most contestants. Prove that ![$\color[rgb]{0.35,0.35,0.35}\dfrac{k}{a} \ge \dfrac{b-1}{2b}$](//latex.artofproblemsolving.com/7/6/2/7624e70dcefc22a943751e7fead1bb3ec89d0c53.png)
.
![$\color[rgb]{0.35,0.35,0.35}a$](http://latex.artofproblemsolving.com/e/c/a/eca40b28aef52c6970cbfe347140847e9cd1d9c0.png)
contestants and ![$\color[rgb]{0.35,0.35,0.35}b$](http://latex.artofproblemsolving.com/d/b/d/dbdb7d40e889fb758607704acfdc8a59eb362204.png)
judges, where ![$\color[rgb]{0.35,0.35,0.35}b \ge 3$](http://latex.artofproblemsolving.com/7/7/d/77da16d394cee2bb309c7e59d356a81482dabc7d.png)
is an odd integer. Each judge rates each contestant as either “pass” or “fail.” Suppose that ![$\color[rgb]{0.35,0.35,0.35}k$](http://latex.artofproblemsolving.com/0/4/d/04d93df2f804a46fe1da4931d7ec96909bfc8347.png)
is a number such that, for any two judges, their ratings coincide for at most contestants. Prove that ![$\color[rgb]{0.35,0.35,0.35}\dfrac{k}{a} \ge \dfrac{b-1}{2b}$](http://latex.artofproblemsolving.com/7/6/2/7624e70dcefc22a943751e7fead1bb3ec89d0c53.png)
.
,
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 
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 This post has been edited 3 times. Last edited by shiningsunnyday, Jun 22, 2016, 3:30 PM
May 2017
April 2017