2006 iTest Problems/Ultimate Question

The following problems are from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem 41

Problem U1

Find the real number $x$ such that \[\sqrt{x-9} + \sqrt{x-6} = \sqrt{x-1}.\]


Problem U2

Points $A$ and $B$ lie on a circle centered at $O$ such that $\angle AOB$ is right. Points $C$ and $D$ lie on radii $OA$ and $OB$ respectively such that $AC = 7$, $CD = 5$, and $BD = 6$. Determine the area of quadrilateral $ACDB$.

[asy] draw(circle((0,0),10)); draw((0,10)--(0,0)--(10,0)--(0,10)); draw((0,3)--(4,0)); label("O",(0,0),SW); label("C",(0,3),W); label("A",(0,10),N); label("D",(4,0),S); label("B",(10,0),E); [/asy]


Problem U3

When properly sorted, $9$ math books on a shelf are arranged in alphabetical order from left to right. An eager student checked out and read all of them. Unfortunately, the student did not realize how the books were sorted, and so after finishing the student put the books back on the shelf in a random order. If all arrangements are equally likely, the probability that exactly $6$ of the books were returned to their correct (original) position can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m + n$.


Problem 42

Problem U4

As $n$ ranges over the integers, the expression $n^4 - 898n^2 + 1$ evaluates to just one prime number. Find this prime.


Problem U5

In triangle $ABC$, points $D$, $E$, and $F$ are the feet of the angle bisectors of $\angle A$, $\angle B$, $\angle C$ respectively. Let point $P$ be the intersection of segments $AD$ and $BE$, and let $p$ denote the perimeter of $ABC$. If $AP = 3PD$, $BE = 9$, and $CF = 9$, then the value of $\frac{AD}{p}$ can be expressed uniquely as $\frac{\sqrt{m}}{n}$ where $m$ and $n$ are positive integers such that $m$ is not divisible by the square of any prime. Find $m + n$.


Problem U6

$x$ and $y$ are nonzero real numbers such that

\[18x - 4x^2 + 2x^3 - 9y - 10xy - x^2y + 6y^2 + 2xy^2 - y^3 = 0\]

The smallest possible value of $\frac{y}{x}$ is equal to $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.