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2021 SMT Guts Round 5 p17-20 - Stanford Math Tournament
parmenides51   5
N 4 hours ago by MATHS_ENTUSIAST
p17. Let the roots of the polynomial $f(x) = 3x^3 + 2x^2 + x + 8 = 0$ be $p, q$, and $r$. What is the sum $\frac{1}{p} +\frac{1}{q} +\frac{1}{r}$ ?


p18. Two students are playing a game. They take a deck of five cards numbered $1$ through $5$, shuffle them, and then place them in a stack facedown, turning over the top card next to the stack. They then take turns either drawing the card at the top of the stack into their hand, showing the drawn card to the other player, or drawing the card that is faceup, replacing it with the card on the top of the pile. This is repeated until all cards are drawn, and the player with the largest sum for their cards wins. What is the probability that the player who goes second wins, assuming optimal play?


p19. Compute the sum of all primes $p$ such that $2^p + p^2$ is also prime.


p20. In how many ways can one color the $8$ vertices of an octagon each red, black, and white, such that no two adjacent sides are the same color?


PS. You should use hide for answers. Collected here.
5 replies
parmenides51
Feb 11, 2022
MATHS_ENTUSIAST
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geometry parabola problem
smalkaram_3549   10
N Apr 13, 2025 by ReticulatedPython
How would you solve this without using calculus?
10 replies
smalkaram_3549
Apr 11, 2025
ReticulatedPython
Apr 13, 2025
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geometry parabola problem
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smalkaram_3549
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smalkaram_3549
#1 h  Apr 11, 2025, 9:52 PM • 1 Y
Y by Kizaruno
How would you solve this without using calculus?
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ShadowDragonRules
418 posts
ShadowDragonRules
#2 h  Apr 11, 2025, 10:12 PM • 1 Y
Y by Kizaruno
hmm... I would suggest to find a relationship of graphing this through first finding the parabola's graph to find the eventual diameter. BTW
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mathmax001
14 posts
mathmax001
#3 h  Apr 11, 2025, 11:15 PM • 1 Y
Y by Kizaruno
smalkaram_3549 wrote:
How would you solve this without using calculus?

is it $ R=\frac{\sqrt{11}}{2} $ ?
I found it solving a quadratic equation.
This post has been edited 1 time. Last edited by mathmax001, Apr 11, 2025, 11:16 PM
Reason: mistype
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rchokler
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rchokler
#4 h  Apr 11, 2025, 11:42 PM
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Solution 1:

$y'=2x$, so the normal line through $(a,a^2)$ is $y=a^2-\frac{x-a}{2a}$,

Put $(x,y)=(0,3)$ to get $a^2+\frac{1}{2}=3\implies a^2=\frac{5}{2}$.

So the points of tangency are $\left(\pm\frac{\sqrt{10}}{2},\frac{5}{2}\right)$.

$r=\sqrt{\left(\frac{\sqrt{10}}{2}-0\right)^2+\left(\frac{5}{2}-3\right)^2}=\sqrt{\frac{5}{2}+\frac{1}{4}}=\frac{\sqrt{11}}{2}$.

Solution 2:

$r(t)=\text{dist}[(0,3),(t,t^2)]=\sqrt{t^2+(t^2-3)^2}=\sqrt{t^4-5t^2+9}=\sqrt{\left(t^2-\frac{5}{2}\right)^2+\frac{11}{4}}\geq\frac{\sqrt{11}}{2}$ with equality when $t^2=\frac{5}{2}$.
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smalkaram_3549
168 posts
smalkaram_3549
#5 h  Apr 12, 2025, 12:05 AM • 1 Y
Y by Kizaruno
mathmax001 wrote:
smalkaram_3549 wrote:
How would you solve this without using calculus?

is it $ R=\frac{\sqrt{11}}{2} $ ?
I found it solving a quadratic equation.

yes it is. How did you do that? I got it using calculus but can't figure out the algebraic method
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joeym2011
493 posts
joeym2011
#6 h  Apr 12, 2025, 12:25 AM
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We let the circle equation be $x^2+(y-3)^2=r^2$. We can solve for $y$:
$$y^2-6y+9+y=r^2.$$Due to tangency, there is only one solution of $y$, and we have a double root and
$$5^2=4\left(9-r^2\right)\implies r=\boxed{\frac{\sqrt{11}}2}.$$@rchokler's solution 2 is also algebraic and works well.
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ReticulatedPython
697 posts
ReticulatedPython
#7 h  Apr 12, 2025, 12:33 AM
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Solution
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mathmax001
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mathmax001
#8 h  Apr 12, 2025, 7:30 PM
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ReticulatedPython wrote:
Solution

This is exactly the method I used .

If you don't understand the meaning of this sentence " Since we want the roots to be negations of each other, the discriminant is equal to $0.$ " , I'll explain it more :
If the discriminant $ 5^2-4(9-r^2) $ were $ \neq 0 $ then the equation would have 2 different solutions for $ x^2 $ ( because the discriminant is positive ) , that means four 4 points of tangency which is not true .
This post has been edited 2 times. Last edited by mathmax001, Apr 12, 2025, 7:41 PM
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jb2015007
1955 posts
jb2015007
#9 h  Apr 12, 2025, 7:41 PM • 1 Y
Y by Kizaruno
sol

also hi reticulated python!

also sorry for a bad sol
i didnt have time to explain everything clearly
This post has been edited 1 time. Last edited by jb2015007, Apr 12, 2025, 7:42 PM
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smbellanki
181 posts
smbellanki
#11 h  Apr 13, 2025, 3:40 AM
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The equation of the circle is \( x^2 + (y - 3)^2 = r^2 \). Subbing in \( y = x^2 \) to find the intersection gets \( x^2 + (x^2 - 3)^2 = r^2 \) which is \( x^4 - 5x^2 + 9 - r^2 = 0 \) now since, the circle only intersects the parabola twice there must be 2 double roots so we consider the determinant which is \( 5^2 - 4(9 - r^2) = 0 \) which becomes \( 4r^2 - 11 = 0 \) so \( r = \sqrt{11}/2 \) only since the radius can't be negative.
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ReticulatedPython
697 posts
ReticulatedPython
#12 h  Apr 13, 2025, 2:05 PM
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jb2015007 wrote:
sol

also hi reticulated python!

also sorry for a bad sol
i didnt have time to explain everything clearly

Looks good to me!
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