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.
be a nonzero polynomial such that 
for every real 
, and 
. Then 
, where 
and 
are relatively prime positive integers. Find 
.
coins of the same appearance. It is known that one of them weighs 
g, two coins weigh 
g each, three coins weigh 
g each, four coins weigh 
g each, and five coins weigh 
g each. There are inscriptions on the coins, indicating their weight. It is allowed to perform two weighings on a balance without additional weights. Find a way to check that there are no wrong inscriptions.
is a rational number.
and other permutations.
, 
, 
be non-negative real numbers, no two of which are zero. Such that 
Prove that :
. Prove that
be positive real numbers. Prove that 
.
, 
and 
are the zeros of a polynomial 
, and 
. The points corresponding to , , and in the complex plane are the vertices of a right triangle with hypotenuse 
. Find 
.
I would have been trolled Butch and Sundance....
with 
. Let 
be the reflection of 
over 
, and we are given 
and 
. Find 
.
and 
, the diagonals. By Parallelogram Law,![\[a^2 + b^2 = 2(r^2 + s^2) \implies a^2 + b^2 + c^2 = 6r^2 + 2s^2 = 250.\]](http://latex.artofproblemsolving.com/6/d/5/6d56ca065945cdefb44019f93b807d5d01095e09.png)
Lastly, by perpendicularity, 
, thus![\[Y'Z^2 = Y'X^2 + 4s^2 = 9r^2 + 3s^2 = 250 \cdot \frac32 = 375\]](http://latex.artofproblemsolving.com/7/d/a/7dadd6e514f8c0dcd058fcdf3c7453a5c2fc76da.png)
as desired.
is the complex number for the centroid of a triangle with vertices as the complex numbers 
and . In this problem, it is given that the sum of the roots is 
, so the centroid is at ., it is actually just the sum of the squares of the distances from the centroid to each vertex of the triangle.
, we see that 
, where 
and 
are the lengths of the medians of the triangle. and be the legs of the right triangles. We can find the squares of the lengths of each median in terms of these legs. Then substituting, we get that
so the square of the hypotenuse is 
.
, 
, and 
for reals . These three roots form an isosceles triangle because ![\[|c - (a+bi)|=|c-(a-bi)|\]](http://latex.artofproblemsolving.com/1/5/7/157da8c9bac0ee2fd3a7ed26fa1fb86f4e04d202.png)
..
since they must also sum to 0.![\[250 = |a|^2+|b|^2+|c|^2 = 4x^2+10x^2+10x^2=24x^2\]](http://latex.artofproblemsolving.com/c/c/7/cc7bf9282e09089e0b9505081060da03928a2518.png)
![\[h^2 = (3x-(-3x))^2 = 36x^2 = 250 \cdot \frac{3}{2} = 375\]](http://latex.artofproblemsolving.com/6/3/f/63f5cf83bb255ae1c3d63b24e20c26c3acf76b0c.png)
be the vertex that has a right angle on it.![\[a+b=c = 0\]](http://latex.artofproblemsolving.com/4/c/0/4c023dd0c6b5b319e6a4541465d12d0cb6f31a38.png)
![\[|a|^2+|b|^2+|c|^2 = 250 \implies |b+c|^2+|b|^2+|c^2| = 250\]](http://latex.artofproblemsolving.com/d/c/8/dc89b049a10ec992ece76e212fc0842e2aa74b5f.png)
and 
. Then, we write the equation as![\[b_{1}^2+2b_{1}c_{1}+c_{1}^2+b_{2}^2+2b_{2}c_{2}+c_{2}^2+b_{1}^2+b_{2}^2+c_{1}^2+c_{2}^2 = 250\]](http://latex.artofproblemsolving.com/8/a/0/8a0a5c01dbb5f165ac1754c9d83d008878236965.png)
![\[b_{1}^2+b_{1}c_{1}+c_{1}^2+b_{2}^2+b_{2}c_{2}+c_{2}^2 = 125\]](http://latex.artofproblemsolving.com/8/0/3/80318860660b7d84c646d34aa576803daca3293b.png)
. Note that 
. Thus, our answer is 
. Because the three roots sum to zero, the real part of the two other roots, which are conjugates of each other, is 
, while their imaginary parts are 
as the three points form an isosceles right triangle in the complex plane. This means that 
Thus 
and 
.
. Note that by Vieta, 
, so the centroid of the triangle lies at . Then, let the distance from to be and the distance from to be 
. By properties of the centroid and the Pythagorean Theorem, we see that ![\[|a|^2=\tfrac49((2x)^2+y^2)\]](http://latex.artofproblemsolving.com/0/f/c/0fc41efa3612e8c69b43dcf65bff9f8005877022.png)
![\[|c|^2=\tfrac49(x^2+(2y)^2).\]](http://latex.artofproblemsolving.com/6/7/3/67345523b7deceb1a564bbe192b76cc89d1ecac4.png)
But from the fact that the length of median from the right angle in a right triangle is equal to half the length of the hypotenuse, we also have ![\[|b|^2=\tfrac19((2x)^2+(2y)^2).\]](http://latex.artofproblemsolving.com/1/4/d/14dc2eef10ef0a6ae8c1103895e30aeedf8f7abc.png)
Plugging these into the given equation and simplifying, we get ![\[\tfrac83(x^2+y^2)=250 \Rightarrow x^2+y^2=\tfrac{375}4.\]](http://latex.artofproblemsolving.com/e/a/6/ea64a1f16effb8fec7436e95b5ef42e9483b5d4e.png)
The length of our hypotenuse is given by 
, so the final answer is 
.
. Since the sum of the roots is zero, the real parts of the other roots is 
. Then, the imaginary part is 
because they form a right triangle. Hence,![\[|a|^2+|b|^2+|c|^2=x^2+ \frac{5x^2}{2}+\frac{5x^2}{2} = 250\]](http://latex.artofproblemsolving.com/d/8/e/d8e32cfb725d82f6173a16f70917246cb57985e2.png)
![\[\implies x = \sqrt{\frac{125}{3}}.\]](http://latex.artofproblemsolving.com/9/7/2/972b47730012d4d199b0ec85b189ad8a29a464aa.png)
.
, so 
. WLOG, let the right angle of the triangle be at . We wish to compute 
. From the given condition, we have 
. The right angle condition tells us that 
. In a similar fashion as before, we can extract the equation 
. Now, we solve 
and 
, so 
.
, and the sum of the squares of the distances from the centroid are 
. From median length formula we have in a right triangle with sidelengths , , ,
.
are positive numbers, then prove that
,
,
,
.
,
,
.
represent the hypotenuse. By the Pythagorean Theorem, 
.
,
.
.
,
.
.
and 
.
,
.
.
,
,
,
.
or 
.
so 
,
,
,
for some reals 
.
,
and 
.
and 
,![\[ |a|^2 + |b|^2 + |c|^2 = (2x)^2 + y^2 + x^2 + (2y)^2 + x^2 + y^2 = 250 \implies 6(x^2+y^2) = 250. \]](http://latex.artofproblemsolving.com/c/9/5/c95c8bf1b029775adbd87b8c027d53ce37a340a4.png)
,![\[ h^2 = 9(x^2 + y^2) = \frac{3}{2} \cdot 6(x^2+y^2) = \frac{3}{2} \cdot 250 = \boxed{375}. \]](http://latex.artofproblemsolving.com/0/7/5/075e201b017086701ac48e2122279f4fbbf09f84.png)
is the centroid of said triangle. Then the segment from the origin to a root is the distance from the centroid to a vertex. The sum of the squares of these segments is 
of the median, this is just 
the sum of the squares of the medians. But the sum of the squares of the medians is 
the sum of the squares of the sides, which is just twice the square of the hypotenuse.
.
by Vieta. Assume 
is the right angle vertex. We have 
as well.
by the Pythagorean theorem.
.
.
.
is the distance from vertex 
to the centroid. Let 
,
similarly. Using Stewart's theorem, we get 
.
.
(or some permutation of 
)
.
and 
which form a degenerate right triangle in the complex plane. We know that
,
,
,
.
,
.
,
.![\[(a-c)^2+(b-d)^2+(a-e)^2+(b-f)^2=(c-e)^2+(d-f)^2\]](http://latex.artofproblemsolving.com/2/f/6/2f609b53a6b5b481044acaf1d6d0efee28d3f243.png)
.
.
we have![\[a(c+e-a)+b(d+f-b)=ce+df\]](http://latex.artofproblemsolving.com/1/8/a/18aa2907b0c504db959e225d7368134df9228da0.png)
,
and 
.
,
.
and 
.
.
,
,
.
,
.
.
.![\[f(z)=z^3+qz+r,\]](http://latex.artofproblemsolving.com/5/4/6/546102a70b1a56d29541adeeedd7d8dd44db5a33.png)
and 
term is ![\[a+b+c=0.\]](http://latex.artofproblemsolving.com/2/f/8/2f8daee84241f7a27d7ee26c9887302f17e5cb26.png)
we find that:![\[\bar{a}+\bar{b}+\bar{c}=0.\]](http://latex.artofproblemsolving.com/0/2/d/02d0ab593a09b53c8c82471d99f986c0d5c6ae13.png)
for all complex numbers 
,![\[\mid a-b \mid ^2 +\mid b-c \mid^2 = \mid c-a \mid ^2=h^2,\]](http://latex.artofproblemsolving.com/7/0/b/70b67d48e76e9122ebe3d454efd2ffe9cc31833b.png)
![\[\mid b-c \mid ^2 +\mid c-a \mid^2 = \mid a-b \mid ^2=h^2,\]](http://latex.artofproblemsolving.com/4/4/1/441247dc3ce0b24e7135c31ad2f2241defe718db.png)
![\[\mid c-a \mid ^2 +\mid a-b \mid^2 = \mid b-c \mid ^2=h^2,\]](http://latex.artofproblemsolving.com/3/f/9/3f903157f82b86f274e605d07bc43c32fd94038d.png)
![\[h^2=\frac{\mid a-b \mid ^2 +\mid b-c \mid^2 + \mid c-a \mid ^2}{2}.\]](http://latex.artofproblemsolving.com/b/0/2/b0215c54272c361bf54fd7ec30614dbaaa7e208d.png)
![\[\mid a \mid ^2 + \mid b \mid ^2 + \mid c \mid^2= 250.\]](http://latex.artofproblemsolving.com/b/f/3/bf3a8b301da900f14176ae632ad0a2032ca10d83.png)
![\[\frac{\mid a-b \mid ^2 +\mid b-c \mid^2 + \mid c-a \mid ^2}{2}.\]](http://latex.artofproblemsolving.com/8/2/a/82a6d532aba71a496103de1fbf2cea2790033797.png)
![\[(a+b+c) \cdot (\bar{a}+\bar{b}+\bar{c}) = 0,\]](http://latex.artofproblemsolving.com/4/d/9/4d97a34cfd4d67f783f5c5223c22409a014a58e1.png)
![\[\mid a \mid ^2 +\mid b \mid ^2 +\mid c \mid ^2+\bar{a}b+\bar{a}c+\bar{b}a+\bar{b}c+\bar{c}a+\bar{c}b=0.\]](http://latex.artofproblemsolving.com/1/f/f/1ff94c4cc14c59a00cdd5ac578e7dccde098a85d.png)
![\[\bar{a}b+\bar{a}c+\bar{b}a+\bar{b}c+\bar{c}a+\bar{c}b=-250.\]](http://latex.artofproblemsolving.com/9/d/2/9d232adaa0d800d6bde9fa78c1be12a1f72bb3eb.png)
![\[\frac{\mid a-b \mid ^2 +\mid b-c \mid^2 + \mid c-a \mid ^2}{2},\]](http://latex.artofproblemsolving.com/8/4/d/84d065376d1afc0ead12d3fd6bc74015947d70d3.png)
![\[\frac{(a-b)(\bar{a}-\bar{b} + (b-c)(\bar{b}-\bar{c})+(c-a)(\bar{c}-\bar{a})}{2},\]](http://latex.artofproblemsolving.com/6/0/9/6099af10b5ed9c86d61770b0f8668bf6d0d52afd.png)
![\[\frac{\mid a \mid ^2 -\bar{a}b-a\bar{b}+\mid b \mid ^2 +\mid b \mid ^2 -\bar{b}c-b\bar{c}+\mid c \mid^2 + \mid c \mid ^2-\bar{c}a-c\bar{a}+\mid a \mid ^2}{2},\]](http://latex.artofproblemsolving.com/6/1/f/61f12f3b4054562e738a541cf6aa19c5e6d1eb5e.png)
![\[\frac{2(\mid a \mid ^2 + \mid b \mid ^2 \mid c \mid^2)-\bar{a}b+\bar{a}c+\bar{b}a+\bar{b}c+\bar{c}a+\bar{c}b}{2},\]](http://latex.artofproblemsolving.com/2/f/2/2f27df37c4b6de918fc2f5c8ae098ee79da39d0f.png)
![\[\boxed{375}\]](http://latex.artofproblemsolving.com/d/8/f/d8f80dcd27c43811ac991bf026009a989be1c162.png)
![\[\oplus\]](http://latex.artofproblemsolving.com/4/5/5/455b2cc4263f11fc5571cded66b42b3d188e6b76.png)
.
be the hypotenuse. The magnitudes of 
because 
is two thirds of the median from 
.
.
.
.
,
.
is 
.
yields 
,
.
form a right triangle, so we have another equation 
,
,
.
,
.
so 
;
to get 
in this model triangle, and we note that this ratio is scale-invariant, hence the answer is 375.
.
.
to get
This gives 
.
,
.
.
.
Now since 
is right angle we know 
after a bit of simplification.Motivated by so many common terms we make the substitutions 
and 
.
and solve this system we get 
and 
.
,
,
,
.
using the provided condition. We want to find 
.
![[asy]
pair A = (-1, -.75);
pair B = (-1, 1.5);
pair C = (2, -.75);
pair G = (0,0);
pair A1 = extension(A, G, B, C);
pair B1 = extension(B, G, C, A);
pair C1 = extension(C, G, A, B);
draw(A--B--C--cycle);
draw(A--A1);
draw(B--B1);
draw(C--C1);
label("$A$", A, dir(A));
label("$B$", B, dir(B));
label("$C$", C, dir(C));
label("$A_1$", A1, dir(B)/2+dir(C)/2);
label("$B_1$", B1, dir(C)/2+dir(A)/2);
label("$C_1$", C1, dir(A)/2+dir(B)/2);
label("$G$", G, SSW*2);
[/asy]](http://latex.artofproblemsolving.com/3/f/e/3fe5bf649d11c58d80d624bfa0b4918ae8977d7f.png)
be the centroid, and let 
be the midpoint of 
,![\[AG^2+BG^2+CG^2=250.\]](http://latex.artofproblemsolving.com/2/b/8/2b8e5ef42bbc8c3a7696dcd8f5144c39ddb1bf62.png)
From the fact that ![\[AA_1^2+BB_1^2+CC_1^2=\frac{1125}2.\]](http://latex.artofproblemsolving.com/3/f/c/3fc1e408d1d131998d7bce9a3cf7b01f45ad1845.png)
From the median length theorem, we have that![\[\frac{2CA^2+2AB^2-BC^2}4
+\frac{2AB^2+2BC^2-CA^2}4
+\frac{2BC^2+2CA^2-AB^2}4
=\frac{1125}2.\]](http://latex.artofproblemsolving.com/f/7/7/f774571132d9286a932a2914807c5220b614f450.png)
This simplifies to![\[AB^2+BC^2+CA^2=750.\]](http://latex.artofproblemsolving.com/8/2/5/825c53531a1903c01ad0205599ca3a5146b99241.png)
From the Pythagorean Theorem, we have that 
.![\[BC^2=\boxed{375}.\]](http://latex.artofproblemsolving.com/a/c/3/ac325e6bd970d810302d3a9190f5ab431c1b073d.png)
and 
and 
respectively. By symmetry, the triangle described by 
Now since 
and thus 
Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, 
and thus 
We can then solve for 
Now 
so 
and thus 
it follows that 
Therefore, we have 
as well which means that the centroid of the right triangle must be at the origin. Therefore, it follows that 
and 
are 
of the lengths of the medians of the triangle, which implies that the sum of the squares of the medians is 
On the other hand, if we let the legs of the triangle have lengths 
then the sum of the squares of the medians is also
Hence, we have
and 
.
we know that 
However, since this forms a right triangle, if we shift our points by 
then 
Plugging this into our equation, we have 