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User:Ddk001

Revision as of 18:37, 4 February 2024 by Cxsmi (talk | contribs) (→‎Problems Sharing Contest)

Introduction

I am a 5th grader who likes making and doing problems, doing math, and redirecting pages (see Principle of Insufficient Reasons). I like geometry and don't like counting and probability. My number theory skill are also not bad.

User Counts

If this is your first time visiting this page, please change the number below by one. (Add 1, do NOT subtract 1)

$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{13}}}}}}}}}}}}}}}}$

For those of you who want more boxes, me too. However, this is the max number of boxes. Also, I check the pages history so I know if someone edited something.

(Please don't mess with the user count)

Doesn't that look like a number on a pyramid

Cool asyptote graphs

Asymptote is fun! [asy]draw((0,0)----(0,6));draw((0,-3)----(-3,3));draw((3,0)----(-3,6));draw((6,-6)----(-6,3));draw((6,0)----(-6,0));[/asy]

[asy]draw(circle((0,0),1));draw((1,0)----(0,1));draw((1,0)----(0,2));draw((0,-1)----(0,2));draw(circle((0,3),2));draw(circle((0,4),3));draw(circle((0,5),4));draw(circle((0,2),1));draw((0,9)----(0,18));[/asy]


Problems Sharing Contest

Here, you can post all the math problems that you have. Everyone will try to come up with a appropriate solution. The person with the first solution will post the next problem. I'll start:

1. There is one and only one perfect square in the form

\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]

where $p$ and $q$ are prime. Find that perfect square.

$\textbf{Solution by cxsmi}$

1. We can expand the product in the expression. $(p^2+1)(q^2+1)-((pq)^2-pq+1) = p^2q^2+p^2+q^2+1-((pq)^2-pq+1) = p^2 + q^2 + pq$. Suppose this equals $m^2$ for some positive integer $m$. We rewrite using the square of a binomial pattern to find that $m^2 = (p + q)^2 - pq$. Through trial and error on small values of $p$ and $q$, we find that $p$ and $q$ must equal $3$ and $5$ in some order. The perfect square formed using these numbers is $\boxed{49}$.

Note: I will be the first to admit that this solution is somewhat lucky.


2. A diamond is created by connecting the points at which a square circumscribed around the incircle of an isosceles right triangle $\triangle ABC$ intersects $\triangle ABC$ itself. $\triangle ABC$ has leg length $2024$. The perimeter of this diamond is expressible as $a\sqrt{b}-c$, where $a$, $b$, and $c$ are integers, and $c$ is not divisible by the square of any prime. What is the remainder when $a + b + c$ is divided by $1000$?

[asy] unitsize(1inch); draw((0,0)--(0,2)); draw((0,2)--(2,0)); draw((2,0)--(0,0)); draw(circle((0.586,0.586),0.586)); draw((0,0)--(0,1.172),red); draw((0,1.172)--(1.172,1.172)); draw((1.172,1.172)--(1.172,0)); draw((1.172,0)--(0,0),red); draw((0,1.172)--(0.828,1.172),red); draw((0.828,1.172)--(1.172,0.828),red); draw((1.172,0.828)--(1.172,0),red); draw((0,0.1)--(0.1,0.1)); draw((0.1,0.1)--(0.1,0)); label("$A$",(0,2.1)); label("$B$",(0,-0.1)); label("$C$",(2,-0.1)); label("$2024$",(-0.2,1)); label("$2024$",(1,-0.2)); [/asy]

Contributions

2005 AMC 8 Problems/Problem 21 Solution 2

2022 AMC 12B Problems/Problem 25 Solution 5 (Now it's solution 6)

2023 AMC 12B Problems/Problem 20 Solution 3

2016 AIME I Problems/Problem 10 Solution 3

2017 AIME I Problems/Problem 14 Solution 2

2019 AIME I Problems/Problem 15 Solution 6

2022 AIME II Problems/Problem 3 Solution 3

Restored diagram for 1994 AIME Problems/Problem 7

Divergence Theorem

Stokes' Theorem

Principle of Insufficient Reasons

Problems I made

Aime styled

Introductory

1. There is one and only one perfect square in the form

\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]

where $p$ and $q$ are prime. Find that perfect square.


2. $m$ and $n$ are positive integers. If $m^2=2^8+2^{11}+2^n$, find $m+n$.

Intermediate

3.The fraction,

\[\frac{ab+bc+ac}{(a+b+c)^2}\]

where $a,b$ and $c$ are side lengths of a triangle, lies in the interval $(p,q]$, where $p$ and $q$ are rational numbers. Then, $p+q$ can be expressed as $\frac{r}{s}$, where $r$ and $s$ are relatively prime positive integers. Find $r+s$.


4. Suppose there is complex values $x_1, x_2,$ and $x_3$ that satisfy

\[(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}\]

Find $x_{1}^3+x_{2}^3+x_{2}^3$.


5. Suppose

\[x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}\]

Find the remainder when $\min{x}$ is divided by $1000$.


6. Suppose that there is $192$ rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are $2$ other pegs positioned sufficiently apart. A $move$ is made if

(i) $1$ ring changed position (i.e., that ring is transferred from one peg to another)

(ii) No rings are on top of smaller rings.

Then, let $x$ be the minimum possible number $moves$ that can transfer all $192$ rings onto the second peg. Find the remainder when $x$ is divided by $1000$.


7. Suppose $f(x)$ is a $10000000010$-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \dots, r_{10000000010}$. Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$. Also, suppose that

\[(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!\]

for an integer $m$. If $p$ is the minimum possible positive integral value of

\[(1+r_1)(1+r_2) \dots (1+r_{10000000010})\].

Find the number of factors of the prime $999999937$ in $p$.


Olympiad

8. (Much harder) $\Delta ABC$ is an isosceles triangle where $CB=CA$. Let the circumcircle of $\Delta ABC$ be $\Omega$. Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$, and point $D$ lies on minor arc $BC$. Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$. Ray $CF$ intersects $\Omega$ at point $G$ (not $C$) and is extended past $G$ to point $I$, and $IG=AC$. Point $H$ is also on $\Omega$ and $AH=GI<HB$. Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$. Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$. Let $O’$ be the reflection of point $O$ over line $IJ$. There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$. $IO’$ intersect $\Omega_1$ at $L$. Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$, and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$, then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$, where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$.

Someone mind making a diagram for this?


9. Suppose \[\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}\] where $p$ and $q$ are relatively prime positive integers. Find $p+q$.


Proofs

10. In $\Delta ABC$ with $AB=AC$, $D$ is the foot of the perpendicular from $A$ to $BC$. $E$ is the foot of the perpendicular from $D$ to $AC$. $F$ is the midpoint of $DE$. Prove that $AF$ is perpendicular to $BE$.

I will leave a big gap below this sentence so you won't see the answers accidentally.






















































Answer key

1. 049

2. 019

3. 092

4. 170

5. 736

6. 895

7. 011

8. 054

9. 077

Solutions

  • Note: All the solutions so far have been made by me :)

$\textbf{I wrote a couple of solutions here. Hope it's okay :) - cxsmi (please feel free to delete this note and/or the solutions)}$

Problem 1

There is one and only one perfect square in the form

\[(p^2+1)(q^2+1)-((pq)^2-pq+1)\]

where $p$ and $q$ is prime. Find that perfect square.

Solution 1

$(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2 \cdot q^2 +p^2+q^2+1-p^2 \cdot q^2 +pq-1=p^2+q^2+pq$. Suppose $n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)$. Then,

\[n^2=(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=(p+q)^2-pq \implies pq=(p+q)^2-n^2=(p+q-n)(p+q+n)\]

, so since $n=\sqrt{p^2+q^2+pq}>\sqrt{p^2+q^2}$, $n>p,n>q$ so $p+q-n$ is less than both $p$ and $q$ and thus we have $p+q-n=1$ and $p+q+n=pq$. Adding them gives $2p+2q=pq+1$ so by Simon's Favorite Factoring Trick, $(p-2)(q-2)=3 \implies (p,q)=(3,5)$ in some order. Hence, $(p^2+1)(q^2+1)-((pq)^2-pq+1)=p^2+q^2+pq=\boxed{049}$.$\square$

Problem 2

$m$ and $n$ are positive integers. If $m^2=2^8+2^{11}+2^n$, find $m+n$.

Solution 1

\[m^2=2^8+2^{11}+2^n\]

\[\implies 2^n=m^2-2^8-2^{11}\]

\[\implies 2^n=(m+48)(m-48)\]

Let $m+48=2^t$ and $m-48=2^s$. Then,

\[2^t-2^s=96 \implies 2^s(2^{t-s}-1)=2^5 \cdot 3 \implies 2^{t-s}-1=3,2^s=2^5 \implies (t,s)=(7,5) \implies m+n=80+12=\boxed{092}\] $\square$

Solution 2

Recall that a perfect square $(a + b)^2$ can be written as $a^2 + 2ab + b^2$. Since $m^2$ is a perfect square, the $RHS$ must be in this form. We substitute $2^4$ for $a$ to get that $2^8 + 2^5 \cdot 2^b + 2^{2b} = m^2$. To make the middle term have an exponent of $11$, we must have $b = 6$. Then $n = 12$ and $m = (2^4 + 2^6)^2 = (16 + 64)^2 = 80^2$, so $m + n = \boxed{092}$.

~ cxsmi

Solution 3

Calculating the terms on the RHS, we find that $256 + 2048 + 2^n = 2304 + 2^n = m^2$. We use trial-and-error to find a power of two that makes the RHS a perfect square. We find that $4096 = 2^12$ works, and it produces $6400 = 80^2$. Then $m + n = \boxed{092}$.

~ (also) cxsmi

Problem 3

The fraction,

\[\frac{ab+bc+ac}{(a+b+c)^2}\]

where $a,b$ and $c$ are side lengths of a triangle, lies in the interval $(p,q]$, where $p$ and $q$ are rational numbers. Then, $p+q$ can be expressed as $\frac{r}{s}$, where $r$ and $s$ are relatively prime positive integers. Find $r+s$.

Solution 1(Probably official MAA, lots of proofs)

Lemma 1: $\text{max} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{3}$

Proof: Since the sides of triangles have positive length, $a,b,c>0$. Hence,

\[\frac{ab+bc+ac}{(a+b+c)^2}>0 \implies \text{max} (\frac{ab+bc+ac}{(a+b+c)^2})= \frac{1}{\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})}\]

, so now we just need to find $\text{min} (\frac{(a+b+c)^2}{ab+bc+ac})$.

Since $(a-c)^2+(b-c)^2+(a-b)^2 \ge 0$ by the Trivial Inequality, we have

\[a^2-2ac+c^2+b^2-2bc+c^2+a^2-2ab+b^2 \ge 0\]

\[\implies a^2+b^2+c^2 \ge ac+bc+ab\]

\[\implies a^2+b^2+c^2+2(ac+bc+ab) \ge 3(ac+bc+ab)\]

\[\implies (a+b+c)^2 \ge 3(ac+bc+ab)\]

\[\implies \frac{(a+b+c)^2}{ab+bc+ac} \ge 3\]

\[\implies \frac{ab+bc+ac}{(a+b+c)^2} \le \frac{1}{3}\]

as desired. $\square$

To show that the minimum value is achievable, we see that if $a=b=c$, $\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{3}$, so the minimum is thus achievable.

Thus, $q=\frac{1}{3}$.

Lemma 2: $\frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}$

Proof: By the Triangle Inequality, we have

\[a+b>c\]

\[b+c>a\]

\[a+c>b\].

Since $a,b,c>0$, we have

\[c(a+b)>c^2\]

\[a(b+c)>a^2\]

\[b(a+c)>b^2\].

Add them together gives

\[a^2+b^2+c^2<c(a+b)+a(b+c)+b(a+c)=2(ab+bc+ac)\]

\[\implies a^2+b^2+c^2+2(ab+bc+ac)<4(ab+bc+ac)\]

\[\implies (a+b+c)^2<4(ab+bc+ac)\]

\[\implies \frac{(a+b+c)^2}{ab+bc+ac}<4\]

\[\implies \frac{ab+bc+ac}{(a+b+c)^2}>\frac{1}{4}\] $\square$

Even though unallowed, if $a=0,b=c$, then $\frac{ab+bc+ac}{(a+b+c)^2}=\frac{1}{4}$, so

\[\lim_{b=c,a \to 0} (\frac{ab+bc+ac}{(a+b+c)^2})=\frac{1}{4}\].

Hence, $p=\frac{1}{4}$, since by taking $b=c$ and $a$ close $0$, we can get $\frac{ab+bc+ac}{(a+b+c)^2}$ to be as close to $\frac{1}{4}$ as we wish.

$p+q=\frac{1}{3}+\frac{1}{4}=\frac{7}{12} \implies r+s=7+12=\boxed{019}$ $\blacksquare$

Solution 2 (Fast, risky, no proofs)

By the Principle of Insufficient Reason, taking $a=b=c$ we get either the max or the min. Testing other values yields that we got the max, so $q=\frac{1}{3}$. Another extrema must occur when one of $a,b,c$ (WLOG, $a$) is $0$. Again, using the logic of solution 1 we see $p=\frac{1}{4}$ so $p+q=\frac{7}{12}$ so our answer is $\boxed{019}$. $\square$

Problem 4

Suppose there are complex values $x_1, x_2,$ and $x_3$ that satisfy

\[(x_i-\sqrt[3]{13})((x_i-\sqrt[3]{53})(x_i-\sqrt[3]{103})=\frac{1}{3}\]

Find $x_{1}^3+x_{2}^3+x_{2}^3$.

Solution 1

To make things easier, instead of saying $x_i$, we say $x$.

Now, we have \[(x-\sqrt[3]{13})(x-\sqrt[3]{53})(x-\sqrt[3]{103})=\frac{1}{3}\]. Expanding gives

\[x^3-(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}) \cdot x^2+(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})x-(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})=0\].

To make things even simpler, let

\[a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}\]

, so that $x^3-ax^2+bx-c=0$.

Then, if $P_n=x_{1}^n+x_{2}^n+x_{3}^n$, Newton's Sums gives

\[P_1+(-a)=0\] $(1)$

\[P_2+(-a) \cdot P_1+2 \cdot b=0\] $(2)$

\[P_3+(-a) \cdot P_1+b \cdot P_1+3 \cdot (-c)=0\] $(3)$

Therefore,

\[P_3=0-((-a) \cdot P_1+b \cdot P_1+3 \cdot (-c))\]

\[=a \cdot P_2-b \cdot P_1+3 \cdot c\]

\[=a(a \cdot P_1-2b)-b \cdot P_1 +3 \cdot c\]

\[=a(a^2-2b)-ab+3c\]

\[=a^3-3ab+3c\]

Now, we plug in $a=\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103}, b=\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103}, c=\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3}:$

\[P_3=(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})^3-3(\sqrt[3]{13}+\sqrt[3]{53}+\sqrt[3]{103})(\sqrt[3]{13 \cdot 53}+\sqrt[3]{13 \cdot 103}+\sqrt[3]{53 \cdot 103})+3(\sqrt[3]{13 \cdot 53 \cdot 103}+\frac{1}{3})\].

We substitute $x=\sqrt[3]{13},y=\sqrt[3]{53},z=\sqrt[3]{103}$ to get

\[P_3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3(abc+\frac{1}{3})\]

\[=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3(x^2y+y^2x+x^2z+z^2x+z^2y+y^2z+3xyz)+3xyz+1\]

\[=x^3+y^3+z^3+3x^2y+3y^2x+3x^2z+3z^2x+3z^2y+3y^2z+6xyz-3x^2y-3y^2x-3x^2z-3z^2x-3z^2y-3y^2z-9xyz+3xyz+1\]

\[=x^3+y^3+z^3+1\]

\[=13+53+103+1\]

\[=\boxed{170}\]. $\square$

Note: If you don't know Newton's Sums, you can also use Vieta's Formulas to bash.

Problem 5

Suppose

\[x \equiv 2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6 \pmod{7!}\]

Find the remainder when $\min{x}$ is divided by 1000.

Solution 1 (Euler's Totient Theorem)

We first simplify $2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6:$

\[2^4 \cdot 3^4 \cdot 7^4+2^7 \cdot 3^7 \cdot 5^6=42^4+6 \cdot 30^6=(\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)}\]

so

\[x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)} \equiv 1 \pmod{5}\]

\[x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 0 \pmod{6}\]

\[x \equiv (\frac{5 \cdot 6 \cdot 7}{5})^{\phi (5)}+6\cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)}+0 \cdot (\frac{5 \cdot 6 \cdot 7}{6})^{\phi (6)} \equiv 6 \cdot (\frac{5 \cdot 6 \cdot 7}{7})^{\phi (7)} \equiv 6 \pmod{7}\].

where the last step of all 3 congruences hold by the Euler's Totient Theorem. Hence,

\[x \equiv 1 \pmod{5}\]

\[x \equiv 0 \pmod{6}\]

\[x \equiv 6 \pmod{7}\]

Now, you can bash through solving linear congruences, but there is a smarter way. Notice that $5|x-6,6|x-6$, and $7|x-6$. Hence, $210|x-6$, so $x \equiv 6 \pmod{210}$. With this in mind, we proceed with finding $x \pmod{7!}$.

Notice that $7!=5040= \text{lcm}(144,210)$ and that $x \equiv 0 \pmod{144}$. Therefore, we obtain the system of congruences :

\[x \equiv 6 \pmod{210}\]

\[x \equiv 0 \pmod{144}\].

Solving yields $x \equiv 2\boxed{736} \pmod{7!}$, and we're done. $\square$

Problem 6

Suppose that there is $192$ rings, each of different size. All of them are placed on a peg, smallest on the top and biggest on the bottom. There are $2$ other pegs positioned sufficiently apart. A $move$ is made if

(i) $1$ ring changed position (i.e., that ring is transferred from one peg to another)

(ii) No bigger rings are on top of smaller rings.

Then, let $x$ be the minimum possible number $moves$ that can transfer all $192$ rings onto the second peg. Find the remainder when $x$ is divided by $1000$.

Solution 1 (Recursion)

Let $M_n$ be the minimum possible number $moves$ that can transfer $n$ rings onto the second peg. To build the recursion, we consider what is the minimum possible number $moves$ that can transfer $n+1$ rings onto the second peg. If we use only legal $moves$, then $n+1$ will be smaller on the top, bigger on the bottom. Hence, the largest ring have to be at the bottom of the second peg, or the largest peg will have nowhere to go. In order for the largest ring to be at the bottom, we must first move the top $n$ rings to the third peg using $M_n$ $moves$, then place the largest ring onto the bottom of the second peg using $1$ $move$, and then get all the rings from the third peg on top of the largest ring using another $M_n$ $moves$. This gives a total of $2M_n+1$, hence we have $M_{n+1}=2M_{n}+1$. Obviously, $M_1=1$. We claim that $M_n=2^n-1$. This is definitely the case for $n=1$. If this is true for $n$, then

\[M_{n+1}=2M_{n}+1=2(2^n-1)+1=2^{n+1}-1\]

so this is true for $n+1$. Therefore, by induction, $M_n=2^n-1$ is true for all $n$. Now, $x=M_{192}=2^{192}-1$. Therefore, we see that

\[x+1 \equiv 0 \pmod{8}\].

But the $\text{mod 125}$ part is trickier. Notice that by the Euler's Totient Theorem,

\[2^{\phi (125)}=2^{100} \equiv 1 \pmod{125} \implies 2^{200} \equiv 1 \pmod{125}\]

so $x+1=\frac{2^{200}}{256}$ is equivalent to the inverse of $256$ in $\text{mod 125}$, which is equivalent to the inverse of $6$ in $\text{mod 125}$, which, by inspection, is simply $21$. Hence,

\[x+1 \equiv 0 \pmod{8}\]

\[x+1 \equiv 21 \pmod{125}\]

, so by the Chinese Remainder Theorem, $x+1 \equiv 896 \pmod{1000} \implies x \equiv \boxed{895} \pmod{1000}$. $\square$

Problem 7

Suppose $f(x)$ is a $10000000010$-degrees polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \dots, r_{10000000010}$. Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$. Also, suppose that

\[(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!\]

for an integer $m$. If $p$ is the minimum possible positive integral value of

\[(1+r_1)(1+r_2) \dots (1+r_{10000000010})\].

Find the number of factors of the prime $999999937$ in $p$.

Solution 1

Since all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$, we have that all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)-n=0$, so by the Factor Theorem,

\[n+1|f(n)-n, n|f(n)-n, \dots, n-10000000008|f(n)-n\]

\[\implies (n+1)n \dots (n-10000000008)|f(n)-n\].

\[\implies f(n)=a(n+1)n \dots (n-10000000008)+n\]

since $f(n)$ is a $10000000010$-degrees polynomial, and we let $a$ to be the leading coefficient of $f(n)$.

Also note that since $r_1, r_2, \dots, r_{10000000010}$ is the roots of $f(n)$, $f(n)=a(n-r_1)(n-r_2) \dots (n-r_{10000000010})$

Now, notice that

\[m!=(2+r_1)(2+r_2) \dots (2+r_{10000000010})\]

\[=(-2-r_1)(-2-r_2) \dots (-2-r_{10000000010})\]

\[=\frac{f(-2)}{a}\]

\[=\frac{a(-1) \cdot (-2) \dots (-10000000010)-2}{a}\]

\[=\frac{10000000010! \cdot a-2}{a}\]

\[=10000000010!-\frac{2}{a}\]

Similarly, we have

\[(1+r_1)(1+r_2) \dots (1+r_{10000000010})=\frac{f(-1)}{a}=-\frac{1}{a}\]

To minimize this, we minimize $m$. The minimum $m$ can get is when $m=10000000011$, in which case

\[-\frac{2}{a}=10000000011!-10000000010!\]

\[=10000000011 \cdot 10000000010!-10000000010!\]

\[=10000000010 \cdot 10000000010!\]

\[\implies p=(1+r_1)(1+r_2) \dots (1+r_{10000000010})\]

\[=-\frac{1}{a}\]

\[=\frac{10000000010 \cdot 10000000010!}{2}\]

\[=5000000005 \cdot 10000000010!\]

, so there is $\left\lfloor \frac{10000000010}{999999937} \right\rfloor=\boxed{011}$ factors of $999999937$. $\square$

Problem 8

$\Delta ABC$ is an isosceles triangle where $CB=CA$. Let the circumcircle of $\Delta ABC$ be $\Omega$. Then, there is a point $E$ and a point $D$ on circle $\Omega$ such that $AD$ and $AB$ trisects $\angle CAE$ and $BE<AE$, and point $D$ lies on minor arc $BC$. Point $F$ is chosen on segment $AD$ such that $CF$ is one of the altitudes of $\Delta ACD$. Ray $CF$ intersects $\Omega$ at point $G$ (not $C$) and is extended past $G$ to point $I$, and $IG=AC$. Point $H$ is also on $\Omega$ and $AH=GI<HB$. Let the perpendicular bisector of $BC$ and $AC$ intersect at $O$. Let $J$ be a point such that $OJ$ is both equal to $OA$ (in length) and is perpendicular to $IJ$ and $J$ is on the same side of $CI$ as $A$. Let $O’$ be the reflection of point $O$ over line $IJ$. There exist a circle $\Omega_1$ centered at $I$ and tangent to $\Omega$ at point $K$. $IO’$ intersect $\Omega_1$ at $L$. Now suppose $O’G$ intersects $\Omega$ at one distinct point, and $O’, G$, and $K$ are collinear. If $IG^2+IG \cdot GC=\frac{3}{4} IK^2 + \frac{3}{2} IK \cdot O’L + \frac{3}{4} O’L^2$, then $\frac{EH}{BH}$ can be expressed in the form $\frac{\sqrt{b}}{a} (\sqrt{c} + d)$, where $b$ and $c$ are not divisible by the squares of any prime. Find $a^2+b^2+c^2+d^2+abcd$.

Someone mind making a diagram for this?

Solution 1

Line $IJ$ is tangent to $\Omega$ with point of tangency point $J$ because $OJ=OA \implies \text{J is on } \Omega$ and $IJ$ is perpendicular to $OJ$ so this is true by the definition of tangent lines. Both $G$ and $K$ are on $\Omega$ and line $O’G$, so $O’G$ intersects $\Omega$ at both $G$ and $K$, and since we’re given $O’G$ intersects $\Omega$ at one distinct point, $G$ and $K$ are not distinct, hence they are the same point.

Now, if the center of $2$ tangent circles are connected, the line segment will pass through the point of tangency. In this case, if we connect the center of $2$ tangent circles, $\Omega$ and $\Omega_1$ ($O$ and $I$ respectively), it is going to pass through the point of tangency, namely, $K$, which is the same point as $G$, so $O$, $I$, and $G$ are collinear. Hence, $G$ and $I$ are on both lines $OI$ and $CI$, so $CI$ passes through point $O$, making $CG$ a diameter of $\Omega$.

Now we state a few claims :

Claim 1: $\Delta O’IO$ is equilateral.

Proof:

\[\frac{3}{4} (IK+O’L)^2\]

\[=\frac{3}{4} IK^2+\frac{3}{2} IK \cdot O’L+\frac{3}{4} O’L^2\]

\[=IG^2+IG \cdot GC\]

\[=IG \cdot (IG+GC)\]

\[=IG \cdot IC\]

\[=IJ^2\]

where the last equality holds by the Power of a Point Theorem.

Taking the square root of each side yields $IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2$.

Since, by the definition of point $L$, $L$ is on $\Omega_1$. Hence, $IK=IL$, so

$IJ= \frac{\sqrt{3}}{2} (IK+O’L)^2=\frac{\sqrt{3}}{2} (IL+O’L)^2=\frac{\sqrt{3}}{2} IO’^2$, and since $O’$ is the reflection of point $O$ over line $IJ$, $OJ=O’J=\frac{OO’}{2}$, and since $IJ=\frac{\sqrt{3}}{2} IO’^2$, by the Pythagorean Theorem we have

$JO’=\frac{IO’}{2} \implies \frac{OO’}{2}=\frac{IO’}{2} \implies OO’=IO’$

Since $IJ$ is the perpendicular bisector of $OO’$, $IO’=IO$ and we have $IO=IO’=OO’$ hence $\Delta O’IO$ is equilateral. $\square$

With this in mind, we see that

\[2OJ=OO’=OI=OK+KI=OJ+GI=OJ+AC \implies OA=OJ=AC\]

Here, we state another claim :

Claim 2 : $BH$ is a diameter of $\Omega$

Proof: Since $OA=OC=AC$, we have

\[\angle AOC =60^\circ \implies \angle ABC=\frac{1}{2} \angle AOC=30^\circ \implies AB=\sqrt{3} AC\]

and the same reasoning with $\Delta CAH$ gives $CH=\sqrt{3} AC$ since $AH=IG=AC$.

Now, apply Ptolemy’s Theorem gives

\[BH \cdot AC+BC \cdot AH=CH \cdot AB \implies BH \cdot AC+AC^2=3AC^2 \implies BH=2AC=2OA\]

so $BH$ is a diameter. $\square$

From that, we see that $\angle BEH=90^\circ$, so $\frac{EH}{BH}=\cos{BHE}$. Now,

\[\angle BHE=\angle BAE=\frac{1}{2} \angle CAB=15^\circ\]

, so

\[\frac{EH}{BH}=\cos{15}=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\sqrt{2}}{4} (\sqrt{3}+1)\]

, so

\[a=4, b=2, c=3, d=1 \implies a^2+b^2+c^2+d^2+abcd=1+4+9+16+24=\boxed{054}\]

, and we’re done. $\blacksquare$

Problem 9

Suppose \[\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]=\frac{p}{q}\] where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution 1(Wordless endless bash)

\[\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}\]

\[=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n+2)}\]

\[=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{m(m+n+2)}\]

\[=\sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=1}^{\infty} \frac{1}{n+2} (\frac{1}{m}-\frac{1}{m+n+2})\]

\[=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \sum_{m=1}^{\infty} (\frac{1}{m}-\frac{1}{m+n+2})\]

\[=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot [(1-\frac{1}{n+3})+(\frac{1}{2}-\frac{1}{n+4})+ \dots]\]

\[=\sum_{n=1}^{\infty} \frac{1}{n(n+2)} \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})\]

\[=\sum_{n=1}^{\infty} (\frac{\frac{1}{2}}{n}-\frac{\frac{1}{2}}{n+2}) \cdot (1+\frac{1}{2}+\frac{1}{3}+ \dots \frac{1}{n+2})\]

\[=\frac{1}{2} [(1-\frac{1}{3})(1+\frac{1}{2}+\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+ \dots\]

\[=\frac{1}{2} [[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\dots](1+\frac{1}{2}+\frac{1}{3})+[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+\dots](1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+\dots](\frac{1}{4}+\frac{1}{5})+\dots]\]

\[=\frac{1}{2} [(1+\frac{1}{2}+\frac{1}{3})+\frac{1}{2} (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{3} (\frac{1}{4}+\frac{1}{5})+\dots]\]

\[=\frac{1}{2} [\frac{11}{6}+\frac{1}{2} \cdot \frac{25}{12}+(\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\dots)+(\frac{1}{3 \cdot 5}+\frac{1}{4 \cdot 6}+\dots)]\]

\[=\frac{1}{2} [\frac{69}{24}+[(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots ]+\frac{1}{2} [(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+\dots ]\]

\[=\frac{1}{2} [\frac{69}{24}+\frac{1}{3}+\frac{1}{6}+\frac{1}{8}]\]

\[=\frac{1}{2} \cdot \frac{84}{24}\]

\[=\frac{7}{4}\]

\[(1+\frac{1}{x})^x=e^{x \cdot \ln (1+\frac{1}{x})}\]

\[=e^{x \cdot [(\frac{1}{x})-\frac{(\frac{1}{x})^2}{2}+\frac{(\frac{1}{x})^3}{3}+\dots]}\]

\[=e^{1-\frac{1}{2} (\frac{1}{x})+\frac{1}{3} (\frac{1}{x})^2+\dots}\]

\[=e \cdot e^{-\frac{1}{2} (\frac{1}{x})} \cdot e^{\frac{1}{3} (\frac{1}{x})^2} \dots\]

\[=e \cdot [1-\frac{1}{2x}+\frac{1}{2!} (\frac{1}{2x})^2- \dots] \cdot [1+\frac{1}{3x^2}+\frac{1}{2!} (\frac{1}{3x^2})^2+ \dots]\]

\[=e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]\]

\[\implies \lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]\]

\[=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{e[1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots]}{e}-1]]\]

\[=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (1-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots-1)]\]

\[=\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 (-\frac{1}{2x}+\frac{11}{24} (\frac{1}{x})^2+\dots)]\]

\[=\lim_{x\rightarrow \infty} (\frac{x}{2}-\frac{x}{2}+\frac{11}{24}+\dots)\]

\[=\frac{11}{24}\]

\[\implies \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n \cdot m^2+m \cdot n^2+2mn}+\lim_{x\rightarrow \infty} [\frac{x}{2}+x^2 [\frac{(1+\frac{1}{x})^{x}}{e}-1]]\]

\[=\frac{7}{4}+\frac{11}{24}\]

\[=\frac{53}{24}\]

\[\implies p=53,q=24\]

\[\implies p+q=\boxed{077}\] $\square$

Problem 10

In $\Delta ABC$ with $AB=AC$, $D$ is the foot of the perpendicular from $A$ to $BC$. $E$ is the foot of the perpendicular from $D$ to $AC$. $F$ is the midpoint of $DE$. Prove that $AF \perp BE$.

Solution 1 (Analytic geo)

Let

\[A=(0,0)\]

\[B=(4a,4b)\]

\[C=(4 \sqrt{a^2+b^2},0)\]

We set it this way to simplify calculation when we calculate the coordinates of $E$ and $F$ (Notice to find $E$, you just need to take the x coordinate of $D$ and let the y coordinate be $0$).

Obviously,

\[D=(\frac{4a+4 \sqrt{a^2+b^2}}{2},\frac{4b+0}{2})=(2a+2 \sqrt{a^2+b^2},2b)\]

\[\implies E=(2a+2 \sqrt{a^2+b^2},0)\]

\[\implies F=(\frac{2a+2 \sqrt{a^2+b^2}+2a+2 \sqrt{a^2+b^2}}{2},\frac{2b+0}{2})=(2a+2 \sqrt{a^2+b^2},b)\]

Now, we see that

\[\text{Slope} _ {AF}=\frac{b}{2a+2 \sqrt{a^2+b^2}}\]

\[\text{Slope} _ {BE}=\frac{0-4b}{2a+2 \sqrt{a^2+b^2}-4a}=\frac{-2b}{\sqrt{a^2+b^2}-a}\]

\[\implies \text{Slope} _ {AF} \cdot \text{Slope} _ {BE}=\frac{b}{2a+2 \sqrt{a^2+b^2}} \cdot \frac{-2b}{a+ \sqrt{a^2+b^2}-2a}=\frac{-2b^2}{2(a+\sqrt{a^2+b^2})(\sqrt{a^2+b^2}-a)}=\frac{-2b^2}{2b^2}=-1\]

, so $AF \perp BE$, as desired. $\square$

Solution 2 (Hard vector bash)

Solution 2a (Hard)

\[\overrightarrow{AF} \cdot \overrightarrow{BE}\]

\[=(\overrightarrow{AE}+\overrightarrow{EF}) \cdot (\overrightarrow{BD}+\overrightarrow{DE})\]

\[=\overrightarrow{AE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{DE}+\overrightarrow{AE} \cdot \overrightarrow{DE}\]

\[=\overrightarrow{AE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{DE}\]

\[=(\overrightarrow{AD}+\overrightarrow{DE}) \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD} + \overrightarrow{EF} \cdot \overrightarrow{DE}\]

\[=\overrightarrow{DE} \cdot \overrightarrow{BD}+\overrightarrow{EF} \cdot \overrightarrow{BD} + \overrightarrow{EF} \cdot \overrightarrow{DE}\]

\[=\overrightarrow{DE} \cdot \overrightarrow{DC}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{BD}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DE}\]

\[=\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DC}-\frac{\overrightarrow{DE}}{2} \cdot \overrightarrow{DE}\]

\[=\overrightarrow{DE} \cdot (\frac{\overrightarrow{DC}-\overrightarrow{DE}}{2})\]

\[=\frac{\overrightarrow{DE} \cdot \overrightarrow{EC}}{2}\]

\[=0\]

Hence, $AF \perp BE$. $\square$

Solution 2b (Harder)

\[\angle ACD=\angle ECD\]

\[\angle ADC=\angle DEC\]

\[\implies \Delta ADC \sim \Delta DEC\]

\[\implies \frac{EC}{DC}=\frac{DC}{AC}\]

\[\implies EC=\frac{DC^2}{AC}\]

\[\implies \overrightarrow{E}=\overrightarrow{C}+\overrightarrow{CE}\]

\[=\overrightarrow{C}+\frac{CE}{AC} \cdot \overrightarrow{CA}\]

\[=\overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{CA}\]

\[=\overrightarrow{C}+\frac{DC^2}{AC^2} (\overrightarrow{A}-\overrightarrow{C})\]

\[=\frac{AC^2-DC^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}\]

\[=\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}\]

\[\overrightarrow{D}=\frac{\overrightarrow{B}+\overrightarrow{C}}{2}\]

Since $F$ is the midpoint of $DE$,

\[\overrightarrow{F}=\frac{\overrightarrow{D}+\overrightarrow{E}}{2}\]

\[=\frac{\frac{\overrightarrow{B}+\overrightarrow{C}}{2}+\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}}{2}\]

\[=\frac{\overrightarrow{B}}{4}+\frac{AC^2+2AD^2}{4AC^2} \overrightarrow{C}+\frac{DC^2}{2AC^2} \overrightarrow{A}\]

\[\overrightarrow{AF}=\overrightarrow{F}-\overrightarrow{A}=\frac{\overrightarrow{B}}{4}+\frac{AC^2+2AD^2}{4AC^2} \overrightarrow{C}+\frac{DC^2-2AC^2}{2AC^2} \overrightarrow{A}\]

\[\overrightarrow{BE}=\overrightarrow{E}-\overrightarrow{B}=\frac{AD^2}{AC^2} \overrightarrow{C}+\frac{DC^2}{AC^2} \overrightarrow{A}-\overrightarrow{B}\]

Now come the coordinates. Let

\[A=(0,0)\]

\[B=(-a,-b)\]

\[C=(a,-b)\]

so that

\[\overrightarrow{A}=\langle 0,0 \rangle\]

\[\overrightarrow{B}=\langle -a,-b \rangle\]

\[\overrightarrow{C}=\langle a,-b \rangle\].

Therefore,

\[\overrightarrow{AF}=\langle \frac{-a}{4},\frac{-b}{4} \rangle + \frac{(a^2+b^2)+2b^2}{4(a^2+b^2)} \langle a,-b \rangle=\langle \frac{ab^2}{2(a^2+b^2)},\frac{-a^2b-2b^3}{2(a^2+b^2)} \rangle\]

\[\overrightarrow{BE}=\frac{b^2}{a^2+b^2} \langle a,-b \rangle-\langle -a,-b \rangle=\langle \frac{2ab^2+a^3}{a^2+b^2},\frac{a^2b}{a^2+b^2} \rangle\]

\[\implies \overrightarrow{AF} \cdot \overrightarrow{BE}=\langle \frac{ab^2}{2(a^2+b^2)},\frac{-a^2b-2b^3}{2(a^2+b^2)} \rangle \cdot \langle \frac{2ab^2+a^3}{a^2+b^2},\frac{a^2b}{a^2+b^2} \rangle\]

\[=\frac{1}{2(a^2+b^2)^2}[ab^2(a^3+2ab^2)-a^2b(a^2b+2b^3)]\]

\[=\frac{ab}{2(a^2+b^2)^2} (a^3b+2ab^3-a^3b-2ab^3)\]

\[=0\]

Hence, we have that $AF$ is perpendicular to $BE$. $\square$

Vandalism area

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EX:

2021 Fall AMC 10A (Answer Key)
Printable versions: Wiki • Fall AoPS Resources • Fall PDF

Instructions

  1. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
  2. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
  3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator).
  4. Figures are not necessarily drawn to scale.
  5. You will have 75 minutes working time to complete the test.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Problem 1

What is the value of $\frac{(2112-2021)^2}{169}$?

$\textbf{(A) } 7 \qquad\textbf{(B) } 21 \qquad\textbf{(C) } 49 \qquad\textbf{(D) } 64 \qquad\textbf{(E) } 91$

Solution

Problem 2

Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?

$\textbf{(A) }16\qquad\textbf{(B) }17\qquad\textbf{(C) }18\qquad\textbf{(D) }19\qquad\textbf{(E) }20$

Solution

Problem 3

What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution

Problem 4

Mr. Lopez has a choice of two routes to get to work. Route A is $6$ miles long, and his average speed along this route is $30$ miles per hour. Route B is $5$ miles long, and his average speed along this route is $40$ miles per hour, except for a $\frac{1}{2}$-mile stretch in a school zone where his average speed is $20$ miles per hour. By how many minutes is Route B quicker than Route A?

$\textbf{(A)}\ 2 \frac{3}{4} \qquad\textbf{(B)}\ 3 \frac{3}{4} \qquad\textbf{(C)}\ 4 \frac{1}{2} \qquad\textbf{(D)}\ 5 \frac{1}{2} \qquad\textbf{(E)}\ 6 \frac{3}{4}$

Solution

Problem 5

The six-digit number $\underline{2}\,\underline{0}\,\underline{2}\,\underline{1}\,\underline{0}\,\underline{A}$ is prime for only one digit $A.$ What is $A?$

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 9$

Solution

Problem 6

Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$st pole along this road is exactly one mile ($5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?

$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$

Solution

Problem 7

As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE$?

[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]

$\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

Solution

Problem 8

A two-digit positive integer is said to be $\emph{cuddly}$ if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$

Solution

Problem 9

When a certain unfair die is rolled, an even number is $3$ times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?

$\textbf{(A) }\dfrac38\qquad\textbf{(B) }\dfrac49\qquad\textbf{(C) }\dfrac59\qquad\textbf{(D) }\dfrac9{16}\qquad\textbf{(E) }\dfrac58$

Solution

Problem 10

A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$?

$\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$

Solution

Problem 11

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?

$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }98\qquad\textbf{(D) }105\qquad\textbf{(E) }126$

Solution

Problem 12

The base-nine representation of the number $N$ is $27{,}006{,}000{,}052_{\text{nine}}.$ What is the remainder when $N$ is divided by $5?$

$\textbf{(A) } 0\qquad\textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) }4$

Solution

Problem 13

Each of $6$ balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other $5$ balls?

$\textbf{(A) } \frac{1}{64}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{5}{16}\qquad\textbf{(E) }\frac{1}{2}$

Solution

Problem 14

How many ordered pairs $(x,y)$ of real numbers satisfy the following system of equations? \begin{align*} x^2+3y&=9 \\ (|x|+|y|-4)^2 &= 1 \end{align*} $\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 7$

Solution

Problem 15

Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$, and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$. What is the area of the circle that passes through vertices $A$, $B$, and $C?$

$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

Solution

Problem 16

The graph of \[f(x) = |\lfloor x \rfloor| - |\lfloor 1 - x \rfloor|\] is symmetric about which of the following? (Here $\lfloor x \rfloor$ is the greatest integer not exceeding $x$.)

$\textbf{(A) }\text{the }y\text{-axis}\qquad \textbf{(B) }\text{the line }x = 1\qquad \textbf{(C) }\text{the origin}\qquad \textbf{(D) }\text{ the point }\left(\dfrac12, 0\right)\qquad \textbf{(E) }\text{the point }(1,0)$

Solution

Problem 17

An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$, $B$, and $C$ are $12$, $9$, and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$?

$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution

Problem 18

A farmer's rectangular field is partitioned into $2$ by $2$ grid of $4$ rectangular sections as shown in the figure. In each section the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field? [asy] draw((0,0)--(100,0)--(100,50)--(0,50)--cycle); draw((50,0)--(50,50)); draw((0,25)--(100,25)); [/asy] $\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 64 \qquad \textbf{(C)}\ 84 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 144$

Solution

Problem 19

A disk of radius $1$ rolls all the way around the inside of a square of side length $s>4$ and sweeps out a region of area $A$. A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$. The value of $s$ can be written as $a+\frac{b\pi}{c}$, where $a,b$, and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$?

$\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14$

Solution

Problem 20

For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$

Solution

Problem 21

Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$

Solution

Problem 22

Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?

$\textbf{(A)}\ \frac{3}{2} \qquad\textbf{(B)}\ \frac{90-40\sqrt{3}}{11} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{144-25\sqrt{3}}{44} \qquad\textbf{(E)}\ \frac{5}{2}$

Solution

Problem 23

For each positive integer $n$, let $f_1(n)$ be twice the number of positive integer divisors of $n$, and for $j \ge 2$, let $f_j(n) = f_1(f_{j-1}(n))$. For how many values of $n \le 50$ is $f_{50}(n) = 12?$

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution

Problem 24

Each of the $12$ edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$?

$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$

Solution

Problem 25

A quadratic polynomial with real coefficients and leading coefficient $1$ is called $\emph{disrespectful}$ if the equation $p(p(x))=0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maximized. What is $\tilde{p}(1)$?

$\textbf{(A) } \frac{5}{16} \qquad\textbf{(B) } \frac{1}{2} \qquad\textbf{(C) } \frac{5}{8} \qquad\textbf{(D) } 1 \qquad\textbf{(E) } \frac{9}{8}$

Solution

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
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2021 AMC 10B 		Followed by
2021 Fall AMC 10B
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All AMC 10 Problems and Solutions

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See also

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