[u]Round 6[/u]


[b]p16.[/b] Given that $x$ and $y$ are positive real numbers such that $x^3+y = 20$, the maximum possible value of $x + y$ can be written as $\frac{a\sqrt{b}}{c}$ +d where $a$, $b$, $c$, and $d$ are positive integers such that $gcd(a,c) = 1$ and $b$ is square-free. Find $a +b +c +d$.


[b]p17.[/b] In $\vartriangle DRK$ , $DR = 13$, $DK = 14$, and $RK = 15$. Let $E$ be the intersection of the altitudes of $\vartriangle DRK$. Find the value of $\lfloor DE +RE +KE \rfloor$.


[b]p18.[/b] Subaru the frog lives on lily pad $1$. There is a line of lily pads, numbered $2$, $3$, $4$, $5$, $6$, and $7$. Every minute, Subaru jumps from his current lily pad to a lily pad whose number is either $1$ or $2$ greater, chosen at random from valid possibilities. There are alligators on lily pads $2$ and $5$. If Subaru lands on an alligator, he dies and time rewinds back to when he was on lily pad number $1$. Find the expected number of jumps it takes Subaru to reach pad $7$.


[u]Round 7[/u]

This set has problems whose answers depend on one another.

[b]p19.[/b] Let $B$ be the answer to Problem $20$ and let $C$ be the answer to Problem $21$. Given that $$f (x) = x^3-Bx-C = (x-r )(x-s)(x-t )$$ where $r$, $s$, and $t$ are complex numbers, find the value of $r^2+s^2+t^2$.


[b]p20.[/b] Let $A$ be the answer to Problem $19$ and let $C$ be the answer to Problem $21$. Circles $\omega_1$ and $\omega_2$ meet at points $X$ and $Y$ . Let point $P \ne Y$ be the point on $\omega_1$ such that $PY$ is tangent to $\omega_2$, and let point $Q  \ne Y$ be the point on $\omega_2$ such that $QY$ is tangent to $\omega_1$. Given that $PX = A$ and $QX =C$, find $XY$ .


[b]p21.[/b] Let $A$ be the answer to Problem $19$ and let $B$ be the answer to Problem $20$. Given that the positive difference between the number of positive integer factors of $A^B$ and the number of positive integer factors of $B^A$ is $D$, and given that the answer to this problem is an odd prime, find $\frac{D}{B}-40$.


[u]Round 8[/u]


[b]p22.[/b] Let $v_p (n)$ for a prime $p$ and positive integer $n$ output the greatest nonnegative integer $x$ such that $p^x$ divides $n$. Find $$\sum^{50}_{i=1}\sum^{i}_{p=1} { v_p (i )+1 \choose 2},$$ where the inner summation only sums over primes $p$ between $1$ and $i$ .


[b]p23.[/b] Let $a$, $b$, and $c$ be positive real solutions to the following equations. $$\frac{2b^2 +2c^2 -a^2}{4}= 25$$
$$\frac{2c^2 +2a^2 -b^2}{4}= 49$$
$$\frac{2a^2 +2b^2 -c^2}{4}= 64$$ The area of a triangle with side lengths $a$, $b$, and $c$ can be written as $\frac{x\sqrt{y}}{z}$ where $x$ and $z$ are relatively prime positive integers and $y$ is square-free. Find $x + y +z$.


[b]p24.[/b] Alan, Jiji, Ina, Ryan, and Gavin want to meet up. However, none of them know when to go, so they each pick a random $1$ hour period from $5$ AM to $11$ AM to meet up at Alan’s house. Find the probability that there exists a time when all of them are at the house at one time.


[b]Round 9 [/b] 

[b]p25.[/b] Let $n$ be the number of registered participantsin this $LMT$. Estimate the number of digits of $\left[ {n \choose 2} \right]$ in base $10$. If your answer is $A$ and the correct answer is $C$, then your score will be
$$\left \lfloor \max \left( 0,20 - \left| \ln \left( \frac{A}{C}\right) \cdot 5 \right|\right| \right \rfloor.$$

[b]p26.[/b] Let $\gamma$ be theminimum value of $x^x$ over all real numbers $x$. Estimate $\lfloor 10000\gamma \rfloor$. If your answer is $A$ and the correct answer is $C$, then your score will be
$$\left \lfloor \max \left( 0,20 - \left| \ln \left( \frac{A}{C}\right) \cdot 5 \right|\right| \right \rfloor.$$

[b]p27.[/b]  Let $$E = \log_{13} 1+log_{13}2+log_{13}3+...+log_{13}513513.$$ Estimate $\lfloor E \rfloor$. If your answer is $A$ and the correct answer is $C$, your score will be $$\left \lfloor \max \left( 0,20 - \left| \ln \left( \frac{A}{C}\right) \cdot 5 \right|\right| \right \rfloor.$$


PS. You should use hide for answers. Rounds 1-5 have been posted [url=https://artofproblemsolving.com/community/c3h3167127p28823220]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve this problem, we need to find the orthocenter \( E \) of triangle \( \triangle DRK \) and then calculate the sum of the distances from \( E \) to the vertices \( D \), \( R \), and \( K \). Finally, we will take the floor of this sum.

1. **Determine the coordinates of the vertices:**
   - Let \( D = (0, 0) \)
   - Let \( K = (14, 0) \)
   - Let \( R = (9, 12) \)

2. **Calculate the semi-perimeter \( s \) and the area \( \Delta \) of the triangle:**
   - The semi-perimeter \( s \) is given by:
     \[
     s = \frac{DR + DK + RK}{2} = \frac{13 + 14 + 15}{2} = 21
     \]
   - Using Heron's formula, the area \( \Delta \) is:
     \[
     \Delta = \sqrt{s(s - DR)(s - DK)(s - RK)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84
     \]

3. **Find the feet of the perpendiculars from each vertex to the opposite side:**
   - The altitude from \( D \) to \( RK \) is:
     \[
     \text{Altitude} = \frac{2 \Delta}{RK} = \frac{2 \cdot 84}{15} = 11.2
     \]
   - The altitude from \( R \) to \( DK \) is:
     \[
     \text{Altitude} = \frac{2 \Delta}{DK} = \frac{2 \cdot 84}{14} = 12
     \]
   - The altitude from \( K \) to \( DR \) is:
     \[
     \text{Altitude} = \frac{2 \Delta}{DR} = \frac{2 \cdot 84}{13} \approx 12.923
     \]

4. **Find the equations of the altitudes:**
   - The equation of the altitude from \( D \) to \( RK \) (line \( RK \)) is:
     \[
     y = \frac{4}{3} x
     \]
   - The equation of the altitude from \( R \) to \( DK \) (line \( DK \)) is:
     \[
     y = 0
     \]
   - The equation of the altitude from \( K \) to \( DR \) (line \( DR \)) is:
     \[
     y = \frac{-12}{5} x + \frac{168}{5}
     \]

5. **Find the orthocenter \( E \):**
   - The orthocenter \( E \) is the intersection of the altitudes. Solving the equations of the altitudes, we find:
     \[
     E = \left(9, \frac{15}{4}\right)
     \]

6. **Calculate the distances \( DE \), \( RE \), and \( KE \):**
   - Using the distance formula:
     \[
     DE = \sqrt{(9 - 0)^2 + \left(\frac{15}{4} - 0\right)^2} = \sqrt{81 + \left(\frac{225}{16}\right)} = \frac{25}{4}
     \]
     \[
     RE = \sqrt{(9 - 9)^2 + \left(\frac{15}{4} - 12\right)^2} = \sqrt{0 + \left(\frac{-33}{4}\right)^2} = \frac{33}{4}
     \]
     \[
     KE = \sqrt{(14 - 9)^2 + \left(0 - \frac{15}{4}\right)^2} = \sqrt{25 + \left(\frac{-15}{4}\right)^2} = \frac{39}{4}
     \]

7. **Sum the distances and take the floor:**
   \[
   DE + RE + KE = \frac{25}{4} + \frac{33}{4} + \frac{39}{4} = \frac{97}{4} = 24.25
   \]
   \[
   \lfloor 24.25 \rfloor = 24
   \]

The final answer is \( \boxed{ 24 } \)