2000 USAMO Problems/Problem 2

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Let $S$ be the set of all triangles $ABC$ for which

\[5 \left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r},\]

where $r$ is the inradius and $P, Q, R$ are the points of tangency of the incircle with sides $AB, BC, CA,$ respectively. Prove that all triangles in $S$ are isosceles and similar to one another.


We let $x = AP = s - a, y = BQ = s-b, z = CR = s-c$, and without loss of generality let $x \le y \le z$. Then $x + y + z = 3s - (a+b+c) = s$, so $r = \frac {A}{s} = \frac{\sqrt{(x+y+z)xyz}}{x+y+z} = \sqrt{\frac{xyz}{x+y+z}}$. Thus,

\[6\sqrt{\frac{x+y+z}{xyz}} = \frac{2yz + 5xy + 5xz}{xyz}\]

Squaring and simplifying yields (after much grueling work)

\[0 = 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz\]

We claim that the inequality

\begin{align}0 \le 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz \tag{*} \end{align}

holds true, with equality iff $4x = y = z$.

Note that $(*)$ is homogeneous in $x,y,z$, so without loss of generality, scale so that $x=1$. Then

\[0 \le y^2 (4z^2 - 16z + 25) + y(14z - 16z^2) + 25z^2\]

which is a quadratic in $y$. As $4z^2 - 16z + 25 = 4(z-4)^2 + 9 \ge 0$, it suffices to show that the quadratic cannot have more than one root, or the discriminant $\Delta \le 0$. Then,

\[\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \le 0\]

as desired. Equality comes when $z = 4$; since $(*)$ is symmetric in $y$ and $z$, it follows that $y = 4$ is also necessary for equality. Reversing our scaling, it follows that $x:y:z = 1:4:4$.

Then $s = x+y+z = 9x$, and $s-a = x, s-b = 4x, s-c = 4x$ yields $a:b:c = 8:5:5$. Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle. $\blacksquare$

Solution 2

Let $A$, $B$, $C$ be the three angles of the triangle, then $r/AP = \tan{(A/2)}$, $r/BQ=\tan{(B/2)}$, and $r/CR=\tan{(C/2)}$. Without loss of generality, assume $A \leq B \leq C$, then $CR=min{(AP,BQ,CR)}$. Denote $x=\tan{(A/2)}$, $x=\tan{(B/2)}$, $z=\tan{(C/2)}$, then we have, \[5(x+y+z)- 3z=6 \qquad \qquad (1)\] Meanwhile, \[z = \tan{(C/2)} = \tan{(90- (A+B)/2))} = \frac{1-xy}{x+y}\] Substitute $z$ in $(1)$, we get \[5x^{2}+8xy+5y^{2} - 6x - 6y+2=0 \qquad \qquad (2)\] Treating (2) as a quadratic equation for $x$, the discriminant is: \[\Delta = (8y-6)^2 - 4*5(5y^2-6y+2)=-36y^2+24y-4 = -(6y-2)^2 \leq 0\] For $x$, $y$ to be both real numbers, $\Delta$ must not be negative, so $\Delta = 0$, which yields $y=1/3$. Solving for x, we get $x=\frac{-(8y-6)}{2*5}=1/3$. Using the formula $\tan(A)=\frac{2\tan(A/2)}{1-\tan^2(A/2)}$, we have $\tan(A)=\tan(B)=3/4$. Therefore $A=B=\arctan{(3/4)}$. This proves that all possible triangle $ABC$ are isosceles with the same set of angles, i.e., they are similar to each other.

It's inconsequential to the proof, but it can be easily found that, since $tan(A) = 3/4$, the altitude to the base of the isosceles triangle divides the triangle into two 3-4-5 right triangles with $4$ on the base; therefore, the entire triangle is 5-5-8. $\square$.

Solution by $Mathdummy$.

See Also

2000 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions

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