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5,401
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https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11
| 6
|
Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$
|
Let the midpoint of $BC$ be $M$ . Angle-chase and observe that $\Delta AMD~\Delta ABM~\Delta MCD$ . Let $BM=CM=a$ and $AM=x$ and $DM=y$ . As a result of this similarity, we write
\[\dfrac2a=\dfrac a3,\]
which gives $a=\sqrt 6$ . Similarly, we write
\[\dfrac2x=\dfrac x7\]
and
\[\dfrac3y=\dfrac y7\]
to get $x=\sqrt{14}$ and $y=\sqrt{21}$
We now have all required side lengths; we can find the area of $\Delta AMD$ with Heron's formula. Doing so yields $\dfrac72\sqrt5$ . We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between $\Delta AMD$ and $\Delta ABM$ is $\dfrac{\sqrt{14}}7=\sqrt{\dfrac27}$ , and between $\Delta AMD$ and $\Delta MCD$ is $\dfrac{\sqrt{21}}7=\sqrt{\dfrac37}$ . Thus, the area ratios are $\dfrac27$ and $\dfrac37$ , respectively, so adding together we have $\dfrac27+\dfrac37=\dfrac57$ . Multiplying this by our $\dfrac72\sqrt5$ , we have $\dfrac52\sqrt5$ as their total area. Adding this to our original area, we have $\dfrac52\sqrt5+\dfrac72\sqrt5=\sqrt5\left(\dfrac52+\dfrac72\right)=\sqrt5\left(\dfrac{12}2\right)=6\sqrt5$
The square of this is $\boxed{180}$
| 180
|
5,402
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13
| 1
|
There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$
|
Because $0 < x < 1$ , we have \begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}
Denote by $c_{2022}$ the coefficient of $P \left( x \right)$ .
Thus, \begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}
Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy \[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]
Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$ .
Hence, we can write $b = 2 b'$ . Plugging this into (1), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy \[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]
Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$ .
Hence, we can write $c = 3 c'$ . Plugging this into (2), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy \[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]
Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$ .
Hence, we can write $d = 5 d' + 1$ . Plugging this into (3), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy \[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]
Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$ .
Hence, we can write $e = 7 e' + 3$ . Plugging this into (4), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy \[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]
The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} = \boxed{220}$
| 220
|
5,403
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13
| 2
|
There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$
|
Note that $2022 = 210\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ $70$ $42$ , or $30$ is $132 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$ , it can only be expressed as a sum in terms of only one of the numbers ( $105$ $70$ $42$ , or $30$ ). Thus, the answer is (by sticks and stones) \[\binom{12}{3} = \boxed{220}\]
| 220
|
5,404
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_14
| 1
|
For positive integers $a$ $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$
|
This should be $\lfloor \frac{1000}{c} \rfloor$ . The current function breaks when $c \mid 1000$ and $b \mid c$ . Take $c = 200$ and $b = 20$ . Then, we have $\lfloor \frac{999}{200} \rfloor = 4$ stamps of value 200, $\lfloor \frac{199}{20} \rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is $200 \cdot 4 + 20 \cdot 9 +19 \cdot 1 = 999$ , just shy of the needed 1000. As for the rest of solution, proceed similarly, except use $1000$ instead of $999$
Also, some explanation: $b-1$ one cent stamps cover all residues module $b$ . Having $\lfloor \frac{c-1}{b} \rfloor$ stamps of value b covers all residue classes modulo $c$ . Finally, we just need $\lfloor \frac{1000}{c} \rfloor$ to cover everything up to 1000.
In addition, note that this function sometimes may not always minimize the number of stamps required. This is due to the fact that the stamps of value $b$ and of value $1$ have the capacity to cover values greater than or equal to $c$ (which occurs when $c-1$ has a remainder less than $b-1$ when divided by $b$ ). Thus, in certain cases, not all $\lfloor \frac{1000}{c} \rfloor$ stamps of value c may be necessary, because the stamps of value $b$ and 1 can replace one $c$
CrazyVideoGamez
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Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$ , and $b-1$ stamps of value $1$ , all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.
So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$ . We can get the answer by solving this equation.
$c > \lfloor \frac{c-1}{b} \rfloor + b-1$
$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$
$c^2 - 97c + 999 > 0$ $c > 85.3$ or $c < 11.7$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$
$97 > \frac{999}{c}$ $c>10.3$
The $3$ least values of $c$ are $11$ $88$ $89$ $11 + 88+ 89 = \boxed{188}$
| 188
|
5,405
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15
| 1
|
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]
|
First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$ . Let points $A'$ and $B'$ be the reflections of $A$ and $B$ , respectively, about the perpendicular bisector of $\overline{O_1O_2}$ . Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid. [asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label("$A'$",Ap,dir(origin--Ap)); label("$B'$",Bp,dir(origin--Bp)); label("$O_1$",O1,dir(origin--O1)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); label("$O_2$",O2,dir(origin--O2)); draw(O2--O1,linetype("4 4")); draw(Ap--D^^Bp--C,linetype("2 2")); [/asy] Next, remark that $B'O_1 = DO_2$ , so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$ , and similarly $A'C = 15$ . Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\cdot B'C + 2\cdot 16 = 15^2$ , whence $A'D = B'C = \sqrt{193}$ . Let $\alpha = \angle A'B'D$ . The Law of Cosines on triangle $A'B'D$ yields \[\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,\] and hence $\sin\alpha = \tfrac 45$ . Thus the distance between bases $A’B’$ and $CD$ is $12$ (in fact, $\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\tfrac12\cdot 12\cdot(2+16) = 108$
Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$ ; the tangency of circles $\omega_1$ and $\omega_2$ implies $r_1 + r_2 = 15$ . Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$ , which implies the measure of angle $A'O_2D$ is $180^\circ - \alpha$ . Therefore, the Law of Cosines applied to triangle $\triangle A'O_2D$ yields \begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. \end{align*}
Thus $r_1r_2 = 40$ , and so the area of triangle $A'O_2D$ is $\tfrac12r_1r_2\sin\alpha = 16$
Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\cdot 16 = \boxed{140}$
| 140
|
5,406
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15
| 2
|
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]
|
Denote by $O$ the center of $\Omega$ .
Denote by $r$ the radius of $\Omega$
We have $O_1$ $O_2$ $A$ $B$ $C$ $D$ are all on circle $\Omega$
Denote $\angle O_1 O O_2 = 2 \theta$ .
Denote $\angle O_1 O B = \alpha$ .
Denote $\angle O_2 O A = \beta$
Because $B$ and $C$ are on circles $\omega_1$ and $\Omega$ $BC$ is a perpendicular bisector of $O_1 O$ . Hence, $\angle O_1 O C = \alpha$
Because $A$ and $D$ are on circles $\omega_2$ and $\Omega$ $AD$ is a perpendicular bisector of $O_2 O$ . Hence, $\angle O_2 O D = \beta$
In $\triangle O O_1 O_2$ \[ O_1 O_2 = 2 r \sin \theta . \]
Hence, \[ 2 r \sin \theta = 15 . \]
In $\triangle O AB$ \begin{align*} AB & = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}
Hence, \[ 15 \cos \frac{\alpha + \beta}{2} - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 2 . \hspace{1cm} (1) \]
In $\triangle O CD$ \begin{align*} CD & = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\ & = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\ & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ & = 15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . \end{align*}
Hence, \[ 15 \cos \frac{\alpha + \beta}{2} + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} = 16 . \hspace{1cm} (2) \]
Taking $\frac{(1) + (2)}{30}$ , we get $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ .
Thus, $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$
Taking these into (1), we get $2 r \cos \theta = \frac{35}{4}$ .
Hence, \begin{align*} 2 r & = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\ & = \frac{5}{4} \sqrt{193} . \end{align*}
Hence, $\cos \theta = \frac{7}{\sqrt{193}}$
In $\triangle O O_1 B$ \[ O_1 B = 2 r \sin \frac{\alpha}{2} . \]
In $\triangle O O_2 A$ , by applying the law of sines, we get \[ O_2 A = 2 r \sin \frac{\beta}{2} . \]
Because circles $\omega_1$ and $\omega_2$ are externally tangent, $B$ is on circle $\omega_1$ $A$ is on circle $\omega_2$ \begin{align*} O_1 O_2 & = O_1 B + O_2 A \\ & = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\ & = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . \end{align*}
Thus, $\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}$
Now, we compute $\sin \alpha$ and $\sin \beta$
Recall $\cos \frac{\alpha + \beta}{2} = \frac{3}{5}$ and $\sin \frac{\alpha + \beta}{2} = \frac{4}{5}$ .
Thus, $e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}$
We also have \begin{align*} \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} & = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right) \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right) . \end{align*}
Thus, \begin{align*} \sin \alpha + \sin \beta & = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha} + e^{i \beta} - e^{-i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( e^{i \alpha} + e^{i \beta} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) \left( \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2 - 2 e^{i \frac{\alpha + \beta}{2}} \right) \\ & = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right) \left( \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2} {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 } + 1 \right) \\ & = \frac{167 \cdot 8}{193 \cdot 5 } . \end{align*}
Therefore, \begin{align*} {\rm Area} \ ABO_1CDO_2 & = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1 + {\rm Area} \ \triangle O_3 O_1 C \\ & \quad + {\rm Area} \ \triangle O_3 C D + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right) + \sin \beta + \sin \beta \right) \\ & = \frac{1}{2} r^2 \left( \sin \left( 2 \theta - \alpha - \beta \right) - \sin \left( 2 \theta + \alpha + \beta \right) + 2 \sin \alpha + 2 \sin \beta \right) \\ & = r^2 \left( - \cos 2 \theta \sin \left( \alpha + \beta \right) + \sin \alpha + \sin \beta \right) \\ & = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} + \sin \alpha + \sin \beta \right) \\ & = \boxed{140}
| 140
|
5,407
|
https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15
| 3
|
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]
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Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \[B'O_2 = BO_1 = O_1 P = O_1 C,\] \[A'O_1 = AO_2 = O_2 P = O_2 D.\] We establish the equality of the arcs and conclude that the corresponding chords are equal \[\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}\] \[\implies A'D = B'C = O_1 O_2 = 15.\] Similarly $A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.$
Ptolemy's theorem on $A'CDB'$ yields \[B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies\] \[B'D^2 + 2 \cdot 16 = 15^2 \implies B'D = A'C = \sqrt{193}.\] The area of the trapezoid $A'CDB'$ is equal to the area of an isosceles triangle with sides $A'D = B'C = 15$ and $A'B' + CD = 18.$
The height of this triangle is $\sqrt{15^2-9^2} = 12.$ The area of $A'CDB'$ is $108.$
\[\sin \angle B'CD = \frac{12}{15} = \frac{4}{5},\] \[\angle B'CD + \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D = \frac{4}{5}.\]
Denote $\angle B'O_2 D = 2\alpha.$ $\angle B'O_2 D > \frac{\pi}{2},$ hence $\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.$ \[\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.\]
Semiperimeter of $\triangle B'O_2 D$ is $s = \frac {15 + \sqrt{193}}{2}.$
The distance from the vertex $O_2$ to the tangent points of the inscribed circle of the triangle $B'O_2 D$ is equal $s – B'D = \frac{15 – \sqrt{193}}{2}.$
The radius of the inscribed circle is $r = (s – B'D) \tan \alpha.$
The area of triangle $B'O_2 D$ is $[B'O_2 D] = sr = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.$
The hexagon $ABO_1 CDO_2$ has the same area as hexagon $B'A'O_1 CDO_2.$
The area of hexagon $B'A'O_1 CDO_2$ is equal to the sum of the area of the trapezoid $A'CDB'$ and the areas of two equal triangles $B'O_2 D$ and $A'O_1 C,$ so the area of the hexagon $ABO_1 CDO_2$ is \[108 + 16 + 16 = \boxed{140}.\]
| 140
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5,408
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2
| 1
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In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]
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Again, let the intersection of $AE$ and $BC$ be $G$ . By AA similarity, $\triangle AFG \sim \triangle CDG$ with a $\frac{7}{3}$ ratio. Define $x$ as $\frac{[CDG]}{9}$ . Because of similar triangles, $[AFG] = 49x$ . Using $ABCD$ , the area of the parallelogram is $33-18x$ . Using $AECF$ , the area of the parallelogram is $63-98x$ . These equations are equal, so we can solve for $x$ and obtain $x = \frac{3}{8}$ . Thus, $18x = \frac{27}{4}$ , so the area of the parallelogram is $33 - \frac{27}{4} = \frac{105}{4}$
Finally, the answer is $105+4=\boxed{109}$
| 109
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5,409
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2
| 2
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In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]
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Let $P = AD \cap FC$ , and $K = AE \cap BC$ . Also let $AP = x$
$CK$ also has to be $x$ by parallelogram properties. Then $PD$ and $BK$ must be $11-x$ because the sum of the segments has to be $11$
We can easily solve for $PC$ by the Pythagorean Theorem: \begin{align*} DC^2 + PD^2 &= PC^2\\ 9 + (11-x)^2 &= PC^2 \end{align*} It follows shortly that $PC = \sqrt{x^2-22x+30}$
Also, $FC = 9$ , and $FP + PC = 9$ . We can then say that $PC = \sqrt{x^2-22x+30}$ , so $FP = 9 - \sqrt{x^2-22x+30}$
Now we can apply the Pythagorean Theorem to $\triangle AFP$ \begin{align*} AF^2 + FP^2 = AP^2\\ 49 + \left(9 - \sqrt{x^2-22x+30}\right)^2 = x^2 \end{align*}
This simplifies (not-as-shortly) to $x = \dfrac{35}{4}$ . Now we have to solve for the area of $APCK$ . We know that the height is $3$ because the height of the parallelogram is the same as the height of the smaller rectangle.
From the area of a parallelogram (we know that the base is $\dfrac{35}{4}$ and the height is $3$ ), it is clear that the area is $\dfrac{105}{4}$ , giving an answer of $\boxed{109}$
| 109
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5,410
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2
| 3
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In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]
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Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$
Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\overline{CE}$ is $(x-11)^2+y^2=49.$
We now have a system of two equations with two variables. Expanding and rearranging respectively give \begin{align*} x^2+y^2-6y&=72, &(1) \\ x^2+y^2-22x&=-72. &(2) \end{align*} Subtracting $(2)$ from $(1),$ we obtain $22x-6y=144.$ Simplifying and rearranging produce \[x=\frac{3y+72}{11}. \hspace{34.5mm} (*)\] Substituting $(*)$ into $(1)$ gives \[\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,\] which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: \begin{align*} \left(3y+72\right)^2+121y^2-726y&=8712 \\ \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ 130y^2-294y-3528&=0 \\ 2(5y+21)(13y-84)&=0 \\ y&=-\frac{21}{5},\frac{84}{13}. \end{align*} Since $E$ is in Quadrant IV, we have $E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).$ It follows that the equation of $\overleftrightarrow{AE}$ is $y=-\frac{4}{3}x+3.$
Let $G$ be the intersection of $\overline{AD}$ and $\overline{FC},$ and $H$ be the intersection of $\overline{AE}$ and $\overline{BC}.$ Since $H$ is the $x$ -intercept of $\overleftrightarrow{AE},$ we get $H=\left(\frac94,0\right).$
By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},$ from which the requested sum is $105+4=\boxed{109}.$
| 109
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5,411
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2
| 4
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In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]
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Let the intersection of $AE$ and $BC$ be $G$ . It is useful to find $\tan(\angle DAE)$ , because $\tan(\angle DAE)=\frac{3}{BG}$ and $\frac{3}{\tan(\angle DAE)}=BG$ . From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = $33-3BG=33-\frac{9}{\tan(\angle DAE)}$
let $\angle CAD = \alpha$ . Let $\angle CAE = \beta$ . Note, $\alpha+\beta=\angle DAE$
$\alpha=\tan^{-1}\left(\frac{3}{11}\right)$
$\beta=\tan^{-1}\left(\frac{7}{9}\right)$
$\tan(\angle DAE) = \tan\left(\tan^{-1}\left(\frac{3}{11}\right)+\tan^{-1}\left(\frac{7}{9}\right)\right) = \frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}} = \frac{\frac{104}{99}}{\frac{78}{99}} = \frac{4}{3}$
$\mathrm{Area}=33-\frac{9}{\frac{4}{3}} = 33-\frac{27}{4 } = \frac{105}{4}$ . The answer is $105+4=\boxed{109}$
| 109
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5,412
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4
| 1
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Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
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Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$ . Continuing this pattern until $21$ coins in the first pile, we have the sum \begin{align*} 31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\ &= 176+155 \\ &= \boxed{331}
| 331
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5,413
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4
| 2
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Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
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We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\geq b\geq c.$ By stars and bars, we have $\dbinom{65}{2}=2080$ to assign positive integer solutions to $a + b + c = 66.$ Now we need to subtract off the cases where it doesn't satisfy the condition.
We start with $a = b.$ We can write that as $2b + c = 66.$ We can find there are 32 integer solutions to this equation. There are $32$ solutions for $b=c$ and $a = c$ by symmetry. We also need to add back $2$ because we subtracted $(a,b,c)=(22,22,22)$ $3$ times.
We then have to divide by $6$ because there are $3!=6$ ways to order $a, b,$ and $c.$ Therefore, we have $\dfrac{\dbinom{65}{2}-96+2}{6} = \dfrac{1986}{6} = \boxed{331}.$
| 331
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5,414
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4
| 3
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Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
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Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$ . Then, let $b = a + k_1$ , and $c = b + k_2$ , such that each $k_i \geq 1$ . The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66$ . This is simply the number of positive solutions to the equation $3x+2y+z = 66$ . Now, we take cases on $a$
If $a = 1$ , then $2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31$ . Each value of $k_1$ corresponds to a unique value of $k_2$ , so there are $31$ solutions in this case. Similarly, if $a = 2$ , then $2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29$ , for a total of $29$ solutions in this case. If $a = 3$ , then $2k_1 + k_2 = 57 \implies 1 \leq k_1 \leq 28$ , for a total of $28$ solutions. In general, the number of solutions is just all the numbers that aren't a multiple of $3$ , that are less than or equal to $31$
We then add our cases to get \begin{align*} 1 + 2 + 4 + 5 + \cdots + 31 &= 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) \\ &= \frac{31(32)}{2} - 3(55) \\ &= 31 \cdot 16 - 165 \\ &= 496 - 165 \\ &= \boxed{331} as our answer.
| 331
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5,415
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6
| 1
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Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$
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First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations \begin{align*} (s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\ x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\ x^2+y^2+(s-z)^2&=\left(10\sqrt{2}\right)^2, \\ (s-x)^2+(s-y)^2+(s-z)^2&=\left(3\sqrt{7}\right)^2. \end{align*} These simplify into \begin{align*} s^2+x^2+y^2+z^2-2sx&=250, \\ s^2+x^2+y^2+z^2-2sy&=125, \\ s^2+x^2+y^2+z^2-2sz&=200, \\ 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. \end{align*} Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$ .
Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$ , so $x^2+y^2+z^2=256$ . This means $PA=16$ . However, we scaled down everything by $12$ so our answer is $16*12=\boxed{192}$
| 192
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5,416
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6
| 2
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Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$
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Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\] Subtracting the fourth equation gives \begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\ x^2 + y^2 + z^2 &= 256 \\ \sqrt{x^2 + y^2 + z^2} &= 16. \end{align*} Since point $A = (0,0,0), PA = 16$ , and since we scaled the answer is $16 \cdot 12 = \boxed{192}$
| 192
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5,417
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6
| 3
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Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$
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Let $E$ be the vertex of the cube such that $ABED$ is a square.
Using the British Flag Theorem , we can easily show that \[PA^2 + PE^2 = PB^2 + PD^2\] and \[PA^2 + PG^2 = PC^2 + PE^2\] Hence, by adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$
Thus, we can solve for $PA$ , which ends up being $\boxed{192}$
| 192
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5,418
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6
| 4
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Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$
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For all points $X$ in space, define the function $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$ . Then $f$ is linear; let $O=\frac{2A+G}{3}$ be the center of $\triangle BCD$ . Then since $f$ is linear, \begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\ \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\ \left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2}, \end{align*} where $x$ denotes the side length of the cube. Thus \begin{align*} 36\text000+18\text000+28\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ 82\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ 73\text728&=2PA^{2} \\ 36\text864&=PA^{2} \\ PA&=\boxed{192}
| 192
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5,419
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 1
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*S,linewidth(4)); dot("$G$",G,SE,linewidth(4)); dot("$H$",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label("$x$",midpoint(A--B),N); label("$y$",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.
Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\] Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$
Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$
By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\] Solving this system of equations ( $(1)$ and $(2)$ ), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$
| 567
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5,420
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 2
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$ , and the length of the altitude from $B$ to $AC$ is also $10$ . Let the foot of this altitude be $F$ , and let the foot of the altitude from $A$ to $BC$ be denoted as $E$ . Then, $\Delta BCF \sim \Delta ACE$ . So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$ . Now, notice that $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$ , where $[ABC]$ denotes the area of triangle $ABC$ . Letting $AB = x$ , this equality becomes $AC = \frac{9x}{5}$ . Also, from $\frac{BC}{AC} = \frac{2}{3}$ , we have $BC = \frac{6x}{5}$ . Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$ , we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Notice that $AC = AF + CF$ , so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$ , and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$ . Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$ . Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$
Now, let the foot of the perpendicular from $A$ to $CD$ be $G$ . Then let $DG = y$ . Let the foot of the perpendicular from $B$ to $CD$ be $H$ . Then, $CH$ is also equal to $y$ . Notice that $ABHG$ is a rectangle, so $GH = x$ . Now, we have $CG = GH + CH = x + y$ . By the Pythagorean theorem applied to $\Delta AGC$ , we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$ . We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$ , so we can plug this into this equation. Solving for $x+y$ , we get $x+y=\frac{63}{2\sqrt{2}}$
Finally, to find $[ABCD]$ , we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$ . The problem asks us for $K \cdot \sqrt{2}$ , which comes out to be $\boxed{567}$
| 567
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5,421
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 3
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Make $AE$ perpendicular to $BC$ $AG$ perpendicular to $BD$ $AF$ perpendicular $DC$
It's obvious that $\triangle{AEB} \sim \triangle{AFD}$ . Let $EB=5x; AB=5y; DF=6x; AD=6y$ . Then make $BQ$ perpendicular to $DC$ , it's easy to get $BQ=18$
Since $AB$ parallel to $DC$ $\angle{ABG}=\angle{BDQ}$ , so $\triangle{ABG} \sim \triangle{BDQ}$ . After drawing the altitude, it's obvious that $FQ=AB=5y$ , so $DQ=5y+6x$ . According to the property of similar triangles, $AG/BQ=BG/DQ$ . So, $\frac{5}{9}=\frac{GB}{(6x+5y)}$ , or $GB=\frac{(30x+25y)}{9}$
Now, we see the $\triangle AEB$ , pretty easy to find that $15^2+(5x)^2=(5y)^2$ , then we get $x^2+9=y^2$ , then express $y$ into $x$ form that $y=\sqrt{x^2+9}$ we put the length of $BG$ back to $\triangle AGB$ $BG^2+100=AB^2$ . So, \[\frac{[30x+25\sqrt{(x^2+9)}]^2}{81}+100=(5\sqrt{x^2+9})^2.\] After calculating, we can have a final equation of $x^2+9=\sqrt{x^2+9}\cdot3x$ . It's easy to find $x=\frac{3\sqrt{2}}{4}$ then $y=\frac{9\sqrt{2}}{4}$ . So, \[\sqrt{2}\cdot K = \sqrt{2}\cdot(5y+5y+6x+6x)\cdot9=\boxed{567}.\] ~bluesoul
| 567
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5,422
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 4
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Let the foot of the altitude from $A$ to $BC$ be $P$ , to $CD$ be $Q$ , and to $BD$ be $R$
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$ . As $\angle QAB = 90^\circ$ , we have that $\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ$ , with the last equality coming from cyclic quadrilateral $APBR$ . Thus, $\triangle QAR \sim \triangle QPA$ and we have that $\frac{AQ}{AR} = \frac{PQ}{PA}$ or that $\frac{18}{10} = \frac{QP}{15}$ , which we can see gives us that $QP = 27$ . Further ratios using the same similar triangles gives that $QR = \frac{25}{3}$ and $RP = \frac{56}{3}$
We also see that quadrilaterals $APBR$ and $ARQD$ are both cyclic, with diameters of the circumcircles being $AB$ and $AD$ respectively. The intersection of the circumcircles are the points $A$ and $R$ , and we know $DRB$ and $QRP$ are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center $A$ taking $\triangle APQ$ to $\triangle APD$ . Because we know a lot about $\triangle APQ$ but very little about $\triangle APD$ and we would like to know more, we wish to find the ratio of similitude between the two triangles.
To do this, we use the one number we have for $\triangle APD$ : we know that the altitude from $A$ to $BD$ has length $10$ . As the two triangles are similar, if we can find the height from $A$ to $PQ$ , we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$ . Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$ . Then, let $QH = x$ and thus $HP=27-x$ . We then have by the Pythagorean Theorem on $\triangle AQH$ and $\triangle APH$ \begin{align*} 15^2 - x^2 &= 18^2 - (27-x)^2 \\ 225 - x^2 &= 324 - (x^2-54x+729) \\ 54x &= 630 \\ x &= \frac{35}{3}. \end{align*} Then, $RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}$ . This gives us then from right triangle $\triangle ARH$ that $AH = \frac{20\sqrt{2}}{3}$ and thus the ratio of $\triangle APQ$ to $\triangle ABD$ is $\frac{3\sqrt{2}}{4}$ . From this, we see then that \[AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}\] and \[AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.\] The Pythagorean Theorem on $\triangle AQD$ then gives that \[QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.\] Then, we have the height of trapezoid $ABCD$ is $AQ = 18$ , the top base is $AB = \frac{45\sqrt{2}}{4}$ , and the bottom base is $CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}$ . From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}$ , so the answer is $K\sqrt{2} = \boxed{567}$
| 567
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5,423
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 5
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$ , respectively.
Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$
Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: \begin{align*} \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square \end{align*} Let $AD=a$ . We obtain from the similarities $AB = \frac{5a}{6}$ and $AC=BD=\frac{3a}{2}$
By Ptolemy, $\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD$ , so $\frac{5a^2}{4} = \frac{5a}{6} \cdot CD$
We obtain $CD=\frac{3a}{2}$ , so $DF=\frac{CD-AB}{2}=\frac{a}{3}$
Applying the Pythagorean theorem on $\triangle ADF$ , we get $324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}$
Thus, $a=\frac{27}{\sqrt{2}}$ , and $[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}$ , yielding $\sqrt2\cdot[ABCD]=\boxed{567}$
| 567
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5,424
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 6
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Let $AD=BC=a$ . Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$ $F$ be the foot of the perpendicular from $A$ to line $BC$ , and $H$ be the foot of the perpendicular from $A$ to $DC$
Note that $\triangle CBG\sim\triangle CAF$ , and we get that $\frac{10}{15}=\frac{a}{AC}$ . Therefore, $AC=\frac32 a$ . It then follows that $\triangle ABF\sim\triangle ADH$ . Using similar triangles, we can then find that $AB=\frac{5}{6}a$ . Using the Law of Cosines on $\triangle ABC$ , We can find that the $\cos\angle ABC=-\frac{1}{3}$ . Since $\angle ABF=\angle ADH$ , and each is supplementary to $\angle ABC$ , we know that the $\cos\angle ADH=\frac{1}{3}$ . It then follows that $a=\frac{27\sqrt{2}}{2}$ . Then it can be found that the area $K$ is $\frac{567\sqrt{2}}{2}$ . Multiplying this by $\sqrt{2}$ , the answer is $\boxed{567}$
| 567
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5,425
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 7
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Draw the distances in terms of $B$ , as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$ . As a result, let $AB=u$ , then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$ . The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$ . By angle subtraction, $\cos(180-\theta)=-\cos\theta$ . Therefore, $AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}$ and $AD=BC=\frac{27}{\sqrt{2}}$ . By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}$ which $\sqrt{2}\cdot k=\boxed{567}$
| 567
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5,426
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 8
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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[asy] size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); label("$E$",E,NE); label("$F$",F, S); label("$G$",G,SE); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); label("$5x$",midpoint(A--B),S); label("$6x$",midpoint(A--D),1.5*(-1,0)); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] Let the points formed by dropping altitudes from $A$ to the lines $BC$ $CD$ , and $BD$ be $E$ $F$ , and $G$ , respectively.
We have \[\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB\] and \[BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.\] For convenience, let $AB = 5x$ . By Heron's formula on $\triangle ABD$ , we have sides $5x,6x,9x$ and semiperimeter $10x$ , so \[\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},\] so $AB = 5x = \frac{45}{2\sqrt{2}}$
Then, \[BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}\] and \[\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.\] Finally, recalling that $ABCD$ is isosceles, \[K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},\] so $\sqrt{2}\cdot K = \boxed{567}$
| 567
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5,427
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 9
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$ .
Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$ . Denote $\theta = \angle{CBH}.$ It is clear that \[BG' = AG, CH = AF, \triangle CBH \ =\triangle ADF,\] the area of $ABCD$ is equal to the area of the rectangle $AFCH.$
The problem is reduced to finding $AH$
In triangle $ABC$ all altitudes are known: \[AB : BC : AC = \frac{1}{CH}\ : \frac{1}{AE}\ : \frac{1}{BG'}\ =\] \[= \frac{1}{AF}\ : \frac{1}{AE}\ : \frac{1}{AG}\ = 5 : 6 : 9.\] We apply the Law of Cosines to $\triangle ABC$ and get $:$ \begin{align*} 2\cdot AB\cdot BC \cdot \cos\theta = AC^2 – AB^2 – BC^2, \end{align*} \begin{align*} 2\cdot 5\cdot 6\cdot \cos\theta = 60 \cos\theta = 9^2 – 5^2 – 6^2 = 20, \cos\theta =\frac{1}{3}. \end{align*} \begin{align*} BH = BC \cos\theta = \frac{BC}{3}.\end{align*} We apply the Pythagorean Law to $\triangle HBC$ and get $:$ \begin{align*} HC^2 = 18^2 = BC^2 – BH^2 = 9\cdot BH^2 – BH^2 = 8 BH^2.\end{align*} \begin{align*} BH = \frac{9}{\sqrt2}, AH = (\frac{5}{2} + 1)\cdot BH = \frac{63}{2\cdot \sqrt2}. \end{align*} Required area is \begin{align*} K = \frac{63}{2\cdot \sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}} \implies \sqrt{2} K=\boxed{567}
| 567
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5,428
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9
| 10
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
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Let $F$ be on $DC$ such that $AF \| DC$ . Let $G$ be on $BD$ such that $AG \| BD$
Let $m$ be the length of $AB$ . Let $n$ be the length of $AD$
The area of $\triangle ABD$ can be expressed in three ways: $\frac{1}{2}(15)(BC) = \frac{1}{2}(15)(n)$ $\frac{1}{2}(18)(m)$ , and $\frac{1}{2}(10)(BD)$
\[\frac{1}{2}(15)(n) = \frac{1}{2}(18)(m)\] \[15n = 18m\] \[5n = 6m\] \[n = \frac{6}{5}m\]
Now, $BD = BG + GD = \sqrt{m^2-100} + \sqrt{n^2-100}$ . We can substitute in $n = \frac{6}{5}m$ to get
$BD = \sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}$
We have \[\frac{1}{2}(10)\left(\sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}\right) = \frac{1}{2}(18)(m)\] After a fairly straightforward algebraic bash, we get $m = \frac{45\sqrt{2}}{4}$ , and $n = (\frac{6}{5})(\frac{45\sqrt{2}}{4}) = \frac{27\sqrt{2}}{2}$ . By the Pythagorean Theorem on $\triangle ADF$ $DF^2 = n^2 - 18^2 = \frac{729}{2} - 324 = \frac{81}{2}$ , and $DF = \frac{9\sqrt{2}}{2}$
Thus, $DC = 2DF + AB = 9\sqrt{2}+\frac{45\sqrt{2}}{4} = \frac{81\sqrt{2}}{4}$ . Therefore, $K = \frac{1}{2}(\frac{45\sqrt{2}}{4}+\frac{81\sqrt{2}}{4}) \cdot 18 = \frac{63\sqrt{2}}{2} \cdot 18 = \frac{567\sqrt{2}}{2}$ . The requested answer is $K \cdot \sqrt{2} = \boxed{567}$
| 567
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5,429
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11
| 1
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Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively: \begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos\angle BEC&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\angle CED&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos\angle DEA&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ \end{alignat*} We subtract $(1\star)+(3\star)$ from $(2\star)+(4\star):$ \begin{align*} 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ 2\cdot\cos\theta\cdot59&=22 \\ \cos\theta&=\frac{11}{59}. \end{align*} Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$
| 301
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5,430
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11
| 2
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Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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Let the brackets denote areas.
We find $[ABCD]$ in two different ways:
Equating the expressions for $[ABCD],$ we have \[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.\] Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$
| 301
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5,431
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11
| 3
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Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09]* WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \sqrt{59} - x.$ By Pythagorean theorem, we have \[49 - x^2 = 16 - \left(\sqrt{59} - x\right)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.\] Now let $y$ be $C_1D.$ (thus making $C_1B = \sqrt{59} - y$ ). Similarly, we have \[36 - y^2 = 25 - \left(\sqrt{59} - y\right)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.\] We see that $A_1C_1$ , the scaled down diagonal is just $x - y = \frac{11\sqrt{59}}{59},$ which is $\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\frac{11}{59} \cdot 22 = \frac{242}{59},$ making our answer $242+59 = \boxed{301}.$
| 301
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5,432
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11
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Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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Solution
In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to $\cos \theta,$ where $\theta$ is the acute angle between the diagonals. \begin{align*} s &= A'B' + B'C' + C'D' + D'A' \\ &= (AB + BC + CD + DA)\cos \theta \\ &= (a + b + c + d)\cos \theta \\ &= 22\cos \theta. \end{align*} In accordance with Claim 2, \begin{align*} 2(ac + bd)\cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*} \[2 \cdot 59 \cos \theta = |13 + 9|.\] \[s = 22\cos \theta = \frac{22 \cdot 11}{59} = \frac{242}{59}.\] Therefore, the answer is $242+59=\boxed{301}.$
| 301
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5,433
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 1
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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Let $O_i$ and $r_i$ be the center and radius of $\omega_i$ , and let $O$ and $r$ be the center and radius of $\omega$
Since $\overline{AB}$ extends to an arc with arc $120^\circ$ , the distance from $O$ to $\overline{AB}$ is $r/2$ . Let $X=\overline{AB}\cap \overline{O_1O_2}$ . Consider $\triangle OO_1O_2$ . The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$ . Let $H$ be the foot from $O$ to $\overline{O_1O_2}$ ; so $HX=r/2$ . We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$ . Let $O_1O_2=d$ [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy] Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$ , it has equal power with respect to both circles, so \[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\] since $O_1X+O_2X=d$ . Now we can solve for $O_1X$ and $O_2X$ , and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{align*} We want to solve for $d$ . By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$
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5,434
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 2
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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Let $O_{1}$ $O_{2}$ , and $O$ be the centers of $\omega_{1}$ $\omega_{2}$ , and $\omega$ with $r_{1}$ $r_{2}$ , and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$ . Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$ . Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).\] Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity \[\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$ , thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$ . Since $r_{1}=961$ and $r_{2}=625$ , the answer is $x=2\cdot(961-625)=\boxed{672}$
| 672
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5,435
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 4
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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Suppose we label the points as shown below [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("$A$",A,dir(A-O2/2)); dot("$B$",B,dir(B-O2/2)); dot("$O_2$",O2,right+up); dot("$O_1$",O1,left+up); dot("$O$",O,dir(O-O2)); dot("$D$",D,dir(170)); dot("$E$",E,dir(E-O1)); dot("$X$",X,dir(X-D)); dot("$Y$",Y,dir(Y-D)); label("$R$",O--E,right+up,p); label("$R$",O--D,left+down,p); label("$2R$",(X+O)/2-(150,0),down,p); label("$961$",O1--D,2*(left+down),p); label("$625$",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy] By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$ . Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$ . Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$ , and $OX = 2OP$ because $\angle POQ = 120^{\circ}$ . Thus \[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\] by similar triangles.
| 672
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5,436
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 5
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$ , and let $u=TO_{2}$ $v=TO_{1}$ $x=PT$ , where the length $O_{1}O_{2}$ splits as $u+v$ . Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.
Applying the Law of Cosines to $\triangle O_{2}PT$ $\triangle O_{1}PT$ $\triangle O_{2}OT$ , and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}
Adding the first and fourth equations, then subtracting the second and third equations gives us \[\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)\]
Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$ , the power of point $P$ with respect to either circle is \[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\]
Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \[u+v=2(R_{1}-R_{2}).\]
The requested distance \[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\] is therefore equal to $2\cdot(961-625)=\boxed{672}$
| 672
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5,437
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 6
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$ . Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$
Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$
In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median
$\hspace{10mm} \implies O,T,$ and $M$ are collinear.
\[\angle OCT = \angle ODT = \angle OET = 90^\circ \implies\]
$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ \[OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies\] \[\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies\] \[\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.\]
Since $r_{1}=625$ and $r_{2}=961$ , the answer is $2\cdot|961-625|=\boxed{672}$
| 672
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5,438
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 7
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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We are not given the radius of circle $w$ , but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$ , call it $W$ , and $w_2$ , call it $W_2$ , is perpendicular to the line containing the center of $w_1$ , call it $W_1$ and $w_2$ . Let $AB = 2h$ and $W_1W_2 = x$ . Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$ , respectively.
By Pythagorean Theorem on $\triangle WW_1W_2$ , we get \[x^2+(625+2x)^2=(961+2r)^2 \;(1)\] Note that since $\angle PWQ = 120$ $\angle PWX = 60$ . So, $WX = 2r/2 = r = W_1Y$ . We now get two more equations from Pythag: \[h^2+r^2 = 625^2 \; (2)\] \[h^2+(x-r)^2 = 961^2 \; (3)\] From subtracting $(2)$ and $(3)$ $x^2-2rx=961^2-625^2 \; (4)$ . Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$ . Plugging in our result from $(4)$ $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$
| 672
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5,439
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13
| 8
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$
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Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$ $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$
Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$
Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$
| 672
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5,440
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14
| 1
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problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object
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We first claim that $n$ must be divisible by $42$ . Since $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \neq 0,1 \pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.
Then $\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)$ . In order for this expression to be divisible by $2021=43\cdot 47$ , a necessary condition is $a^n - 1 \equiv 0 \pmod{43}$ . By Fermat's Little Theorem $a^{42} \equiv 1 \pmod{43}$ . Moreover, if $a$ is a primitive root modulo $43$ , then $\text{ord}_{43}(a) = 42$ , so $n$ must be divisible by $42$
By similar reasoning, $n$ must be divisible by $46$ , by considering $a \not\equiv 0,1 \pmod{47}$
We next claim that $n$ must be divisible by $43$ . By Dirichlet, let $a$ be a prime that is congruent to $1 \pmod{43}$ . Then $\sigma(a^n) \equiv n+1 \pmod{43}$ , so since $\sigma(a^n)-1$ is divisible by $43$ $n$ is a multiple of $43$
Alternatively, since $\left(\frac{a(a^n - 1^n)}{a-1}\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,$ which simplifies to $v_{43}(n) \geq 1,$ which implies the desired result.
Similarly, $n$ is a multiple of $47$
Lastly, we claim that if $n = \text{lcm}(42, 46, 43, 47)$ , then $\sigma(a^n) - 1$ is divisible by $2021$ for all positive integers $a$ . The claim is trivially true for $a=1$ so suppose $a>1$ . Let $a = p_1^{e_1}\ldots p_k^{e_k}$ be the prime factorization of $a$ . Since $\sigma(n)$ is multiplicative , we have \[\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.\] We can show that $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for all primes $p_i$ and integers $e_i \ge 1$ , so \[\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in}),\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for this choice of $n$ , so suppose $p_i \not\equiv 0 \pmod{43}$ and $p_i \not\equiv 0 \pmod{47}$ . Each expression in parentheses equals $\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$ . If $p_i \not\equiv 1 \pmod{43}$ , then FLT implies $p_i^n - 1 \equiv 0 \pmod{43}$ , and if $p_i \equiv 1 \pmod{43}$ , then $p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}$ (since $n$ is also a multiple of $43$ , by definition). Similarly, we can show $\sigma(p_i^{e_in}) \equiv 1 \pmod{47}$ , and a simple CRT argument shows $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ . Then $\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}$
Then the prime factors of $n$ are $2,3,7,23,43,$ and $47,$ and the answer is $2+3+7+23+43+47 = \boxed{125}$
| 125
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5,441
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14
| 2
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problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object
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$n$ only needs to satisfy $\sigma(a^n)\equiv 1 \pmod{43}$ and $\sigma(a^n)\equiv 1 \pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . The following three statements are the same:
We can show this by casework on $p$
Similar arguments for modulo $47$ lead to $46|n$ and $47|n$ . Therefore, we get $n=\operatorname{lcm}[42,43,46,47]$ . Following the last paragraph of Solution 1 gives the answer $\boxed{125}$
| 125
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5,442
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14
| 3
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problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object
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We perform casework on $a:$
Finally, the least such positive integer $n$ for all cases is \begin{align*} n&=\operatorname{lcm}(42,43,46,47) \\ &=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\ &=2\cdot3\cdot7\cdot23\cdot43\cdot47, \end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\boxed{125}.$
| 125
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5,443
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14
| 4
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problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object
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Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem , we get that $n \equiv 0 \pmod{42}$ and $n \equiv 0 \pmod{46}.$ Then, we can look at $a$ being a $1\pmod{43}$ prime and a $1\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$ . There don't seem to be any other odd "numbers" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $\boxed{125}$
| 125
|
5,444
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15
| 1
|
Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$
|
Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately:
Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\boxed{285}.$
| 285
|
5,445
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15
| 2
|
Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$
|
Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$ . Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$ . Completing the square gives us $\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}$ ; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \leq 21^2=441 \rightarrow 24k \leq 6731$ , so $k \leq 280$
For the lower bound, we need to ensure there are $4$ intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of $y=x^2-k$ . As we increase the value of $k$ , two more intersections appear on the "left branch":
$k=4$ does not work because the "leftmost" point of $x=2(y-20)^2-4$ is $(-4,20)$ which lies to the right of $\left(-\sqrt{24}, 20\right)$ , which is on the graph $y=x^2-4$ . While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, $k<4$ does not work.
$k=5$ does work because the two graphs intersect at $(-5,20)$ , and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is $5+280=\boxed{285}$
| 285
|
5,446
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15
| 3
|
Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$
|
Claim
Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$ . Suppose that these parabolas intersect at four points.
Then these points lie on the circle centered at point $(p_2, p_1)$ with radius $r = \sqrt{2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2)}.$
Proof
Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first (red) parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $A(0, –y_1), y_1 > 0.$ Let second (blue) parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $B(–x_2,0), x_2 > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\theta,$ its length $\rho_1(\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\phi,$ its length $\rho_2(\phi).$ Then \[\rho_1(\theta) = \frac{p_1}{1 - \cos(2\theta)}, \rho_2(\phi) = \frac{p_2}{1 - \cos(2\phi)}.\]
Abscissa of the point of intersection is \begin{align*} x =\rho_1 \sin(2\theta) = p_1\cot\theta = \rho_2 \cos (2\phi) - x_2 = \frac{p_2}{2} (\cot^2\phi - 1)- x_2,\end{align*} \begin{align*} x^2 = p_1^2 \cot ^2 \theta , 2 p_1\cot\theta = p_2 \cos^2 \phi - p_2 - 2x_2 .\end{align*} Ordinate of the point of intersection is \begin{align*} y =\rho_2 \sin 2\phi = p_2\cot\phi = \rho_1 \cos 2\theta - y_1 = \frac{p_1}{2} (\cot^2\theta - 1)- y_1,\end{align*} \begin{align*} y^2 = p_2^2 \cot ^2 \phi , 2 p_2\cot\phi = p_1 \cos^2 \theta - p_1 - 2y_1 .\end{align*} The square of the distance from point of intersection to the point $(p_2, p_1)$ is \begin{align*} r^2 = (x-p_2)^2 + (y-p_1)^2 = x^2 + y^2 - 2 p_1 y - 2 p_2 x + p_1^2 + p_2^2 .\end{align*} After simple transformations, we get $r^2 = 2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2).$
Hence, any intersection point has the same distance $r$ from the point $(p_2, p_1).$
Solution
Parameters of the parabola $y = x^2 – k$ are $p_1 = \frac{1}{2}, y_1 = 20 + k – \frac{1}{2}.$
Parameters of the parabola $\frac{x}{2} = (y – 20)^2 – \frac{k}{2}$ are $p_2 = \frac{1}{4}, x_2 = k – \frac{1}{4} \implies r^2 = 20 + \frac{3k}{2}.$
If $r \le 21, k \le \frac{842}{3},$ then integer $k \le 280.$
The vertex of the second parabola is point $(– k,20)$ can be on the parabola $y = x^2 – k$ or below the point of the parabola with the same abscissa. So \[20 \ge (– k)^2 – k \implies 5 \le k \le 280.\] Therefore, the answer is $5+280=\boxed{285}$
| 285
|
5,447
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1
| 1
|
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)
|
Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=\boxed{550},$ which is the final answer.
| 550
|
5,448
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1
| 2
|
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)
|
For any palindrome $\underline{ABA},$ note that $\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ The average for $A$ is $5$ since $A$ can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ The average for $B$ is $4.5$ since $B$ is either $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ Therefore, the answer is $505 + 45 = \boxed{550}.$
| 550
|
5,449
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1
| 3
|
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)
|
For every three-digit palindrome $\underline{ABA}$ with $A\in\{1,2,3,4,5,6,7,8,9\}$ and $B\in\{0,1,2,3,4,5,6,7,8,9\},$ note that $\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \begin{align*} \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\ &=1100. \end{align*} For instances: \begin{align*} 171+929&=1100, \\ 262+838&=1100, \\ 303+797&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*} and so on.
From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1100}{2}=\boxed{550}.$
| 550
|
5,450
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1
| 4
|
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)
|
We notice that a three-digit palindrome looks like this: $\underline{aba}.$
And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\times{10}=90$ three-digit palindromes.
We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean.
How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$
Thus, all of these $90$ palindromes can be broken into this form.
Thus, the sum of these $90$ palindromes will be $101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$ ). The same logic explains why we multiply by $9$ when computing the total sum of $b.$
We get a sum of $45\times{1100},$ but don't compute this! Divide this by $90$ and you will get $\boxed{550}.$
| 550
|
5,451
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1
| 5
|
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)
|
The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$
| 550
|
5,452
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2
| 1
|
Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
|
By angle chasing, we conclude that $\triangle AGF$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle, and $\triangle BED$ is a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle.
Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\triangle BED,$ we have $DE=\frac{840-x}{2}$ and $DB=\frac{840-x}{2}\cdot\sqrt3.$
Let the brackets denote areas. We have \[[AFG]=\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2}\] and \[[BED]=\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right).\]
We set up and solve an equation for $x:$ \begin{align*} \frac{[AFG]}{[BED]}&=\frac89 \\ \frac{\frac12\cdot x^2\cdot\frac{\sqrt3}{2}}{\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right)}&=\frac89 \\ \frac{2x^2}{(840-x)^2}&=\frac89 \\ \frac{x^2}{(840-x)^2}&=\frac49. \end{align*} Since $0<x<840,$ it is clear that $\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \begin{align*} \frac{x}{840-x}&=\frac23 \\ 3x&=1680-2x \\ 5x&=1680 \\ x&=\boxed{336}
| 336
|
5,453
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2
| 2
|
Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
|
We express the areas of $\triangle BED$ and $\triangle AFG$ in terms of $AF$ in order to solve for $AF.$
We let $x = AF.$ Because $\triangle AFG$ is isosceles and $\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$
Let the height of $\triangle ABC$ be $h$ and the height of $\triangle AEF$ be $h'.$ Then we have that $h = \frac{\sqrt{3}}{2}(840) = 420\sqrt{3}$ and $h' = \frac{\sqrt{3}}{2}(EF) = \frac{\sqrt{3}}{2}x.$
Now we can find $DB$ and $BE$ in terms of $x.$ $DB = h - h' = 420\sqrt{3} - \frac{\sqrt{3}}{2}x,$ $BE = AB - AE = 840 - x.$ Because we are given that $\angle DBC = 90,$ $\angle DBE = 30.$ This allows us to use the sin formula for triangle area: the area of $\triangle BED$ is $\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x).$ Similarly, because $\angle AFG = 120,$ the area of $\triangle AFG$ is $\frac{1}{2}(\sin 120)(x^2).$
Now we can make an equation: \begin{align*} \frac{\triangle AFG}{\triangle BED} &= \frac{8}{9} \\ \frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x)} &= \frac{8}{9} \\ \frac{x^2}{\left(420 - \frac{x}{2}\right)(840-x)} &= \frac{8}{9}. \end{align*} To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind). \begin{align*} \frac{x^2}{\left(1-\frac{x}{2}\right)(2-x)} &= \frac{8}{9} \\ 8\left(1-\frac{x}{2}\right)(2-x) &= 9x^2 \\ 16-16x + 4x^2 &= 9x^2 \\ 5x^2 + 16x -16 &= 0 \\ (5x-4)(x+4) &= 0. \end{align*} Thus $x = \frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $\frac{4}{5}(420) = \boxed{336}.$
| 336
|
5,454
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2
| 3
|
Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
|
$\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}$ So, If $\Delta AFG$ is isosceles, it means that $AF = FG$
Let $AF = FG = AE = EF = x$
So, $[\Delta AFG] = \frac{1}{2} \cdot x^{2} \textup{sin} 120^{0} = \frac{\sqrt{3}}{4}x^{2}$
In $\Delta BED$ $BE = 840 - x$ , Hence $DE = \frac{840 - x}{2}$ (because $\textup{sin} 30^{0} = \frac{1}{2}$
Therefore, $[\Delta BED] = \frac{1}{2} (840 - x) \left (\frac{840-x}{2} \right) \textup{sin} 60^{0}$
So, $[\Delta BED] = \frac{\sqrt{3}}{4} (840 - x) \left (\frac{840-x}{2} \right) = \frac{\sqrt{3}}{8} (840 - x)^{2}$
Now, as we know that the ratio of the areas of $\Delta AFG$ and $\Delta BED$ is $8:9$
Substituting the values, we get
$\frac{\frac{\sqrt{3}}{4}x^{2}}{\frac{\sqrt{3}}{8} (840 - x)^{2}} = \frac{8}{9} \Rightarrow \left (\frac{x}{840 - x} \right)^{2} = \frac{4}{9}$ Hence, $\frac{x}{840 - x} = \frac{2}{3}$ . Solving this, we easily get $x = 336$
We have taken $AF = x$ , Hence, $AF = \boxed{336}$
| 336
|
5,455
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2
| 4
|
Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
|
Since $\triangle AFG$ is isosceles, $AF = FG$ , and since $\triangle AEF$ is equilateral, $AF = EF$ . Thus, $EF = FG$ , and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from $E$ and $F$ to line $BC$ ; call the meeting points $P$ and $Q$ , respectively. $\triangle BEP$ is clearly congruent to both $\triangle BED$ and $\triangle FQC$ , and thus each of these new triangles has the same area as $\triangle BED$ . But we can "slide" $\triangle BEP$ over to make it adjacent to $\triangle FQC$ , thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$ , the ratio of their sides must be $3:2$ . So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$ $AF$ must equal two-fifths of $840$ , or $\boxed{336}$
| 336
|
5,456
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4
| 1
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
|
By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\sqrt{n}\cdot i$ and $\overline{z}=m-\sqrt{n}\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\overline{z}.$
We know that \begin{align*} z+\overline{z}&=2m, & (1) \\ z\overline{z}&=m^2+n. & (2) \end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$
Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\overline{z}+z\overline{z}=0,$ or $-21(z+\overline{z})+z\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$
Finally, the answer is $m+n=\boxed{330}.$
| 330
|
5,457
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4
| 2
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
|
$(-20)^{3} + (-20)a + b = 0$ , hence $-20a + b = 8000$
Also, $(-21)^{3} + c(-21)^{2} + d = 0$ , hence $441c + d = 9261$
$m + i \sqrt{n}$ satisfies both $\Rightarrow$ we can put it in both equations and equate to 0.
In the first equation, we get $(m + i \sqrt{n})^{3} + a(m + i \sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0$
Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 \rightarrow (1)$
In the second equation, we get $(m + i \sqrt{n})^{3} + c(m + i \sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0$
Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)$
Comparing (1) and (2),
$a = 2mc$ and $am + b = m^{2}c - nc + d \rightarrow (3)$
$b = 8000 + 20a \Rightarrow b = 40mc + 8000$ $d = 9261 - 441c$
Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$
This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261$
Hence, $c|1261 \Rightarrow c \in {1,13,97,1261}$
Consider case of $c = 1$
$c = 1 \Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$
$am + b = m^{2} - n + 8820$ (because c = 1)
Also, $m^{2} + n + 40m = 820 \rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \rightarrow (5)$
Solving (4) and (5) simultaneously gives $m = 10, n = 320$
[AIME can not have more than one answer, so we can stop here ... Not suitable for Subjective exam]
Hence, $m + n = 10 + 320 = \boxed{330}$
| 330
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5,458
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4
| 3
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
|
start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$ . After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$ , respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore $(-20)^3-20(a)+b=0$ and $(-21)^3+c*(-21)^2+d=0$ and you can set these two equations equal to each other while also substituting the values of $a$ $b$ $c$ , and $d$ above to give you $21m^2+21n-1682m+8000=0$ , then you can rearrange the equation into $21n = -21m^2+1682m-8000$ . With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$ therefore that means $1682m-8000=0(mod 21)$ which results in $2m-20=0(mod 21)$ which finally gives us m=10 mod 21. We can test the first obvious value of $m$ which is $10$ and we see that this works as we get $m=10$ and $n=320$ . That means your answer will be $m + n = 10 + 320 = \boxed{330}$
| 330
|
5,459
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4
| 4
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
|
We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$
Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$ , and $Q(x) = x^2 + (c-21)x + (441 - 21c)$
By the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$
Therefore, $c-21 = -20$ and $441-21c = 400 + a$ . Solving the system of equations, we get $a = 20$ and $c = 1$ , so $P(x) = Q(x) = x^2 - 20x + 420$
Finally, by the quadratic formula, we have roots of $\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i$ , so our final answer is $10 + 320 = \boxed{330}$
| 330
|
5,460
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4
| 5
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
|
We plug -20 into the equation obtaining $(-20)^3-20a+b$ , likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$
Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$
By Vieta's, the sum of the roots in the first equation is $0$ , so $m$ must be $10$
Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$ . However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$
Given that $m$ is $10$ , then $n$ is equal to $320$
Therefore, the answer to the equation is $\boxed{330}$
| 330
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5,461
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4
| 6
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
|
Since $m+i\sqrt{n}$ is a common root and all the coefficients are real, $m-i\sqrt{n}$ must be a common root, too.
Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients).
First, however, the product of the two common roots is: \begin{align*} &&&(x-m-i\sqrt{n})(x-m+i\sqrt{n})\\ &=&&x^2-x(m+i\sqrt{n}+m-i\sqrt{n})+(m+i\sqrt{n})(m-i\sqrt{n})\\ &=&&x^2-2xm+(m^2-i^2n)\\ &=&&x^2-2xm+m^2+n \end{align*}
Now, let's equate the two forms of both the polynomials: \[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\] \[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\] Now we can match the zero coefficients. \[-2m+20=0\to m=10\text{ and}\] \[-42m+m^2+n=0\to-420+100+n=0\to n=320\text{.}\] Thus, $m+n=10+320=\boxed{330}$
| 330
|
5,462
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 1
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$ rd side is between $6$ and $14$ , exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the $3$ rd side. So the triangles' sides are between $6$ and $\sqrt{84}$ exclusive, and the larger bound is between $\sqrt{116}$ and $14$ , exclusive. The area of these triangles are from $0$ (straight line) to $2\sqrt{84}$ on the first "small bound" and the larger bound is between $0$ and $20$ $0 < s < 2\sqrt{84}$ is our first equation, and $0 < s < 20$ is our $2$ nd equation. Therefore, the area is between $\sqrt{336}$ and $\sqrt{400}$ , so our final answer is $\boxed{736}$
| 736
|
5,463
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 2
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:
For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:
Case (1): The longest side has length $\boldsymbol{10,}$ so $\boldsymbol{0<x<10.}$
By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$
By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\sqrt{84}.$
Taking the intersection produces $6<x<\sqrt{84}$ for this case.
At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.$ Together, we obtain $0<K<2\sqrt{84},$ or $K\in\left(0,2\sqrt{84}\right).$
Case (2): The longest side has length $\boldsymbol{x,}$ so $\boldsymbol{x\geq10.}$
By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$
By the Pythagorean Inequality Theorem, we have $4^2+10^2<x^2,$ from which $x>\sqrt{116}.$
Taking the intersection produces $\sqrt{116}<x<14$ for this case.
At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot10=20.$ Together, we obtain $0<K<20,$ or $K\in\left(0,20\right).$
Answer
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\left(0,2\sqrt{84}\right)$ or $\left(0,20\right).$ By the exclusive disjunction, the set of all such $s$ is \[[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),\] from which $a^2+b^2=\boxed{736}.$
| 736
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5,464
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 3
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
We have the diagram below.
[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("$A$",(0,0),SW); label("$B$",(1,2*sqrt(3)),N); label("$C$",(10,0),SE); label("$\theta$",(0,0),NE); label("$\alpha$",(1,2*sqrt(3)),SSE); label("$4$",(0,0)--(1,2*sqrt(3)),WNW); label("$10$",(0,0)--(10,0),S); [/asy]
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$ . This is because $s=20$ is attained at $\theta = 90^{\circ}$ , and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$ . This can be observed from \[s=\frac{1}{2}(4)(10)\sin\theta\] by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$
Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$ . This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$ , yileding $s=4\sqrt{21}$ . Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.
$\angle{ACB}$ cannot be obtuse since $AC>AB$
Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$ , the desired range is \[s\in [4\sqrt{21},20)\] giving \[a^2+b^2=\boxed{736}\Box\]
| 736
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5,465
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 4
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
Note: Archimedes15 Solution which I added an answer
here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most $1$ . Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \rightarrow x < 2\sqrt{21}$ . Using the same logic as the other case, the area is at most $4\sqrt{21}$ . Square and add $4\sqrt{21}$ and $20$ to get the right answer \[a^2+b^2= \boxed{736}\Box\]
| 736
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5,466
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 5
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red); dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill); Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label("$\angle C$ obtuse",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label("$\angle B$ obtuse",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that:
For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot("$D$", D, 1.5*dir(75), linewidth(0.8), UnFill); dot("$C_2$", C2, 1.5*N, linewidth(4.5)); dot("$C_1$", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas:
Finally, the set of all such $s$ is $[a,b)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$
| 736
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5,467
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 6
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
Let a triangle in $\tau(s)$ be $ABC$ , where $AB = 4$ and $BC = 10$ . We will proceed with two cases:
Case 1: $\angle ABC$ is obtuse. If $\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$ ; therefore, the area of the triangle will fall in the range of $(0, 20)$
Case 2: $\angle BAC$ is obtuse. Then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $\left(0, \sqrt{10^{2} - 4^{2}}\right)$ . Therefore, the area of the triangle will fall in the range of $\left(0, 2 \sqrt{84}\right)$
If $s < 2 \sqrt{84}$ , there will exist two types of triangles in $\tau(s)$ - one type with $\angle ABC$ obtuse; the other type with $\angle BAC$ obtuse. If $s \geq 2 \sqrt{84}$ , as we just found, $\angle BAC$ cannot be obtuse, so therefore, there is only one type of triangle - the one in which $\angle ABC$ is obtuse. Also, if $s > 20$ , no triangle exists with lengths $4$ and $10$ . Therefore, $s$ is in the range $\left[ 2 \sqrt{84}, 20\right)$ , so our answer is $\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}$
| 736
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5,468
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 7
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides $4$ and $10,$ when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$
The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10$ $AC\perp AB$ ).
The diagram shows triangles with equal heights. The yellow triangle $ABC'$ has the longest side $BC',$ the blue triangle $ABC$ has the longest side $AC.$ If $BC\perp AB,$ then $BC = \sqrt {AC^2 – AB^2} = 2 \sqrt{21}$ the area is equal to $4\sqrt{21}.$ In the interval, the blue triangle $ABC$ is acute-angled, the yellow triangle $ABC'$ is obtuse-angled. Their heights and areas are equal. The condition is met.
If the area is less than $4\sqrt{21},$ both triangles are obtuse, not equal, so the condition is not met.
Therefore, $s$ is in the range $\left[ 4 \sqrt{21}, 20\right)$ , so answer is $\left(4 \sqrt{21}\right)^{2} + 20^{2} = \boxed{736}$
| 736
|
5,469
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5
| 8
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$
|
If $4$ and $10$ are the shortest sides and $\angle C$ is the included angle, then the area is \[\frac{4\cdot10\cdot\sin\angle C}{2} = 20\sin\angle C.\] Because $0\leq\sin\angle C\leq1$ , the maximum value of $20\sin\angle C$ is $20$ , so $s\leq20$
If $4$ is a shortest side and $10$ is the longest side, the length of the other short side is $4\cos\angle C+2\sqrt{4\cos^2 \angle C+21}$ by law of cosines, and the area is $2\left(4\cos\angle C+2\sqrt{4\cos^2\angle C+21}\right)\sqrt{1-\cos\angle C}$ . Because $-1\le \cos\angle C\le 0$ , this is minimized if $\cos\angle C=0$ , where $s=4\sqrt{21}$
So, the answer is $20^2+\left(4\sqrt{21}\right)^2=\boxed{736}$
| 736
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5,470
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6
| 1
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For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
|
By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$ . Substituting into the equation and factoring, we get that $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ , so therefore $A \subseteq B$ or $B \subseteq A$ . WLOG $A\subseteq B$ , then for each element there are $3$ possibilities, either it is in both $A$ and $B$ , it is in $B$ but not $A$ , or it is in neither $A$ nor $B$ . This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could have also been the other way around. Now we need to subtract the overlaps where $A=B$ , and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\cdot 3^5 - 2^5 = \boxed{454}$
| 454
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5,471
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6
| 2
|
For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
|
We denote $\Omega = \left\{ 1 , 2 , 3 , 4 , 5 \right\}$ .
We denote $X = A \cap B$ $Y = A \backslash \left( A \cap B \right)$ $Z = B \backslash \left( A \cap B \right)$ $W = \Omega \backslash \left( A \cup B \right)$
Therefore, $X \cup Y \cup Z \cup W = \Omega$ and the intersection of any two out of sets $X$ $Y$ $Z$ $W$ is an empty set.
Therefore, $\left( X , Y , Z , W \right)$ is a partition of $\Omega$
Following from our definition of $X$ $Y$ $Z$ , we have $A \cup B = X \cup Y \cup Z$
Therefore, the equation
\[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
can be equivalently written as
\[\left( | X | + | Y | \right) \left( | X | + | Z | \right) = | X | \left( | X | + | Y | + | Z | \right) .\]
This equality can be simplified as
\[| Y | \cdot | Z | = 0 .\]
Therefore, we have the following three cases: (1) $|Y| = 0$ and $|Z| \neq 0$ , (2) $|Z| = 0$ and $|Y| \neq 0$ , (3) $|Y| = |Z| = 0$ .
Next, we analyze each of these cases, separately.
Case 1: $|Y| = 0$ and $|Z| \neq 0$
In this case, to count the number of solutions, we do the complementary counting.
First, we count the number of solutions that satisfy $|Y| = 0$
Hence, each number in $\Omega$ falls into exactly one out of these three sets: $X$ $Z$ $W$ .
Following from the rule of product, the number of solutions is $3^5$
Second, we count the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$
Hence, each number in $\Omega$ falls into exactly one out of these two sets: $X$ $W$ .
Following from the rule of product, the number of solutions is $2^5$
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy $|Y| = 0$ minus the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$ , i.e., $3^5 - 2^5$
Case 2: $|Z| = 0$ and $|Y| \neq 0$
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., $3^5 - 2^5$
Case 3: $|Y| = 0$ and $|Z| = 0$
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is $2^5$
By putting all cases together, following from the rule of sum, the total number of solutions is equal to
\begin{align*} \left( 3^5 - 2^5 \right) + \left( 3^5 - 2^5 \right) + 2^5 & = 2 \cdot 3^5 - 2^5 \\ & = \boxed{454}
| 454
|
5,472
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6
| 3
|
For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
|
The answer is \begin{align*} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\ &=2(2+1)^5-(1+1)^5 \\ &=2(243)-32 \\ &=\boxed{454} ~MRENTHUSIASM
| 454
|
5,473
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6
| 4
|
For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
|
The answer is \begin{align*} &\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ &=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\ &=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\ &=63+155+150+70+15+1 \\ &=\boxed{454} ~MRENTHUSIASM
| 454
|
5,474
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6
| 5
|
For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
|
Proceed with Solution 1 to get $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ . WLOG, assume $|A| = |A \cap B|$ . Thus, $A \subseteq B$
Since $A \subseteq B$ , if $|B| = n$ , there are $2^n$ possible sets $A$ , and there are also ${5 \choose n}$ ways of choosing such $B$
Therefore, the number of possible pairs of sets $(A, B)$ is
\[\sum_{k=0}^{5} 2^n {5 \choose n}\]
We can compute this manually since it's only from $k=0$ to $5$ , and computing gives us $243$ . We can double this result for $B \subseteq A$ , and we get $2(243) = 486$
However, we have double counted the cases where $A$ and $B$ are the same sets. There are $32$ possible such cases, so we subtract $32$ from $486$ to get $\boxed{454}$
| 454
|
5,475
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7
| 1
|
Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$
|
From the fourth equation we get $d=\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14$ . Hence $(abc)^2 - 14(abc)-120 = 0$ . Solving, we get $abc = -6$ or $abc = 20$ . From the first and second equation, we get $ab + bc + ca = ab-3c = -4 \Longrightarrow ab = 3c-4$ . If $abc=-6$ , substituting we get $c(3c-4)=-6$ . If you try solving this you see that this does not have real solutions in $c$ , so $abc$ must be $20$ . So $d=\frac{3}{2}$ . Since $c(3c-4)=20$ $c=-2$ or $c=\frac{10}{3}$ . If $c=\frac{10}{3}$ , then the system $a+b=-3$ and $ab = 6$ does not give you real solutions. So $c=-2$ . Since you already know $d=\frac{3}{2}$ and $c=-2$ , so you can solve for $a$ and $b$ pretty easily and see that $a^{2}+b^{2}+c^{2}+d^{2}=\frac{141}{4}$ . So the answer is $\boxed{145}$
| 145
|
5,476
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7
| 2
|
Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$
|
Note that $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$ . Hence, $ab = 3c - 4$
Rewriting $abc+bcd+cda+dab = 14$ , we get $ab(c+d) + cd(a+b) = 14$ .
Substitute $ab = 3c - 4$ and solving, we get \[3c^{2} - 4c - 4d - 14 = 0.\] We refer to this as Equation 1.
Note that $abcd = 30$ gives $(3c-4)cd = 30$ . So, $3c^{2}d - 4cd = 30$ , which implies $d(3c^{2} - 4c) = 30$ or \[3c^{2} - 4c = \frac{30}{d}.\] We refer to this as Equation 2.
Substituting Equation 2 into Equation 1 gives, $\frac{30}{d} - 4d - 14 = 0$
Solving this quadratic yields that $d \in \left\{-5, \frac{3}{2}\right\}$
Now we just try these two cases:
For $d = \frac{3}{2}$ substituting in Equation 1 gives a quadratic in $c$ which has roots $c \in \left\{\frac{10}{3}, -2\right\}$
Again trying cases, by letting $c = -2$ , we get $ab = 3c-4$ , Hence $ab = -10$ .
We know that $a + b = -3$ , Solving these we get $a = -5, b = 2$ or $a= 2, b = -5$ (doesn't matter due to symmetry in $a,b$ ).
So, this case yields solutions $(a,b,c,d) = \left(-5, 2 , -2, \frac{3}{2}\right)$
Similarly trying other three cases, we get no more solutions, Hence this is the solution for $(a,b,c,d)$
Finally, $a^{2} + b^{2} + c^{2} + d^{2} = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4} = \frac{m}{n}$
Therefore, $m + n = 141 + 4 = \boxed{145}$
| 145
|
5,477
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7
| 3
|
Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$
|
For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.
Rearranging $(2)$ and solving for $c,$ we have \begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \begin{align*} ab\left(\frac{ab+4}{3}\right)d&=30 \\ d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6) \end{align*} Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: \begin{align*} ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. \end{align*} Let $t=ab(ab+4),$ from which \[\frac{t}{3}-\frac{360}{t}=14.\] Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce \begin{align*} \left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\ \left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\ \underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\ ab&=6,-10. \end{align*} If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number.
If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\{a,b\}=\{-5,2\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\frac32,$ respectively. Together, we have \[a^2+b^2+c^2+d^2=\frac{141}{4},\] so the answer is $141+4=\boxed{145}.$
| 145
|
5,478
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7
| 4
|
Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$
|
Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: \[abc+d(ab+bc+ca)=14\] Then we plug in equation 2 to receive $abc-4d=14$ . By equation 4 we get $abc=\frac{30}{d}$ . Plugging in, we get $\frac{30}{d}-4d=14$ . Multiply by $d$ on both sides to get the quadratic equation $4d^2+14d-30=0$ . Solving using the quadratic equation, we receive $d=\frac{3}{2},d=-5$ . So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get \[ab+c(a+b)=-4\] After plugging in equation 1, we get $ab-3c=-4$ . Now we convert it into a quadratic to receive $3c^2-4c-abc=0$ . The value of $abc$ will depend on $d$ . So we obtain the discriminant $16+12abc$ .
Let d = -5.
Then $abc = \frac{30}{-5}$ , so $abc=-6$ , discriminant is $16-72$ , which makes this a dead end. Thus $d=\frac{3}{2}$ For $d=\frac{3}{2}$ , making $abc=20$ . This means the discriminant is just $256$ , so we obtain two values for $c$ as well. We get either $c=\frac{10}{3}$ or $c=-2$ . So, we must AGAIN test which one is correct.
We know $ab=3c-4$ , and $a+b=-3$ , so we use these values for testing.
Let $c=\frac{10}{3}$ .
Then $ab=6$ , so $a=\frac{6}{b}$ . We thus get $\frac{6}{b}+b=-3$ , which leads to the quadratic $b^2+3b+6$ . The discriminant for this is $9-24$ . That means this value of $c$ is wrong, so $c=-2$ . Thus we get polynomial $b^2+3b-10$ . The discriminant this time is $49$ , so we get two values for $b$ . Through simple inspection, you may see they are interchangeable, as if you take the value $b=2$ , you get $a=-5$ . If you take the value $b=-5$ , you get $a=2$ . So it doesn't matter. That means the sum of all their squares is \[\frac{9}{4}+4+4+25=\frac{141}{4}\] so the answer is $141+4=\boxed{145}.$ ~amcrunner
| 145
|
5,479
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7
| 5
|
Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$
|
Let the four equations from top to bottom be listed $(1)$ through $(4)$ respectively. Multiplying both sides of $(3)$ by $d$ and factoring some terms gives us $abcd + d^2(ab + ac + bc) = 14d$ . Substituting using equations $(4)$ and $(2)$ gives us $30 -4 d^2 = 14d$ , and solving gives us $d = -5$ or $d = \frac{3}{2}$ . Plugging this back into $(3)$ gives us $abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14$ , or using the other solution for $d$ gives us $abc - 6 = 14$ . Solving both of these equations gives us $abc = -6$ when $d = -5$ and $abc = 14$ when $d = \frac{3}{2}$
Multiplying both sides of $(2)$ by $c$ and factoring some terms gives us $abc + c^2 (a + b) = abc -3c^2 = -4c$ . Testing $abc = -6$ will give us an imaginary solution for $c$ , so therefore $abc = 14$ and $d = \frac{3}{2}$ . This gets us $14 - 3c^2 = -4c$ . Solving for $c$ gives us $c = \frac{3}{10}$ or $c = -2$ . With a bit of testing, we can see that the correct value of $c$ is $c=-2$ . Now we know $a+b = -3$ and $ab + bc + ca = ab + c(a+b) = ab + 6 = -4$ $ab = -10$ , and it is obvious that $a = -5$ and $b = 2$ or the other way around, and therefore, $a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4}$ , giving us the answer $141 + 4 = \boxed{145}$ ~hihitherethere
| 145
|
5,480
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_9
| 1
|
Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$
|
This solution refers to the Remarks section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 1} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 2} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.
We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\ && \\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex] 2 & 4,12,20,28 & 4 \\ && \\ [-2.25ex] 3 & 8,24 & 2 \\ && \\ [-2.25ex] 4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:
Together, the answer is $225+56+12+2=\boxed{295}.$
| 295
|
5,481
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10
| 1
|
Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
|
This solution refers to the Diagram section.
As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$ ) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\ell,$ so $\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\overline{O_1O_2}.$ We wish to find $T_3A.$ [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(O1--O2--O3--cycle^^O3--A,dashed); draw(T1--T2--T3--cycle^^T3--A,heavygreen); draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed); draw(L1--L2,L=Label("$\ell$",position=EndPoint,align=3*E),red); label("$\mathcal{P}$",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label("$\mathcal{Q}$",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); label("$\mathcal{R}$",O1,(0,-24,0)); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O_3$",O3,(0,-1.5,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); [/asy] Note that:
Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\mathcal{R}:$
We have the following diagram: [asy] size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O3, T1, T3, A, B, C, D; O1 = (0,-36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1)); C = (0,-36,-36*sqrt(1105)/23); D = (0,-36,-sqrt(1105)); draw(C--O1--O3--A^^D--O3--B,dashed); draw(T1--T3--A^^T3--B,heavygreen); draw(O1--T1^^O3--T3,mediumblue+dashed); draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label("$\ell$",position=EndPoint,align=3*E),red); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_3$",O3,(0,1,1),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$T_3$",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot("$A$",A,(0,0,-2),red+linewidth(4.5)); dot("$B$",B,(0,0,-2),red+linewidth(4.5)); dot("$C$",C,(0,0,-2),red+linewidth(4.5)); dot("$D$",D,(0,-2,0),linewidth(4.5)); [/asy] In cross-section $O_1O_3T_3T_1,$ since $\overline{O_1T_1}\parallel\overline{O_3T_3}$ as discussed, we obtain $\triangle O_1T_1B\sim\triangle O_3T_3B$ by AA, with the ratio of similitude $\frac{O_1T_1}{O_3T_3}=\frac{36}{13}.$ Therefore, we get $\frac{O_1B}{O_3B}=\frac{49+O_3B}{O_3B}=\frac{36}{13},$ or $O_3B=\frac{637}{23}.$
In cross-section $\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\triangle O_1DO_3,$ we have $O_1D=\sqrt{1105}.$ Moreover, since $\ell\perp\overline{O_1C}$ and $\overline{DO_3}\perp\overline{O_1C},$ we obtain $\ell\parallel\overline{DO_3}$ so that $\triangle O_1CB\sim\triangle O_1DO_3$ by AA, with the ratio of similitude $\frac{O_1B}{O_1O_3}=\frac{49+\frac{637}{23}}{49}.$ Therefore, we get $\frac{O_1C}{O_1D}=\frac{\sqrt{1105}+DC}{\sqrt{1105}}=\frac{49+\frac{637}{23}}{49},$ or $DC=\frac{13\sqrt{1105}}{23}.$
Finally, note that $\overline{O_3T_3}\perp\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\frac{13\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\triangle O_3T_3A$ gives $T_3A=\frac{312}{23},$ from which the answer is $312+23=\boxed{335}.$
| 335
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5,482
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10
| 2
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Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
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The centers of the three spheres form a $49$ $49$ $72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$ , which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints.
Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$ . Extend $\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$ , the center of the small sphere $C$ , we want to find $BD$
By Pythagoras, $AC=\sqrt{49^2-36^2}=\sqrt{1105}$ , and we know that $MA=36$ and $BC=13$ . We know that $\overline{MA}$ and $\overline{BC}$ must be parallel, using ratios we realize that $CD=\frac{13}{23}\sqrt{1105}$ . Apply the Pythagorean theorem to $\triangle BCD$ $BD=\frac{312}{23}$ , so $312 + 23 = \boxed{335}$
| 335
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5,483
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10
| 3
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Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
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This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((10,-3,-40)); triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; M = midpoint(T1--T2); draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(T2--T1--T3--A--cycle),cyan); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(T1--T2--T3--cycle^^M--A--T2,blue); dot("$O_1$",O1,(0,-1,1),linewidth(4.5)); dot("$O_2$",O2,(0,1,1),linewidth(4.5)); dot("$O$",O3,(0.5,-1,0),linewidth(4.5)); dot("$T_1$",T1,(0,-1,-1),blue+linewidth(4.5)); dot("$T_2$",T2,(0,1,-1),blue+linewidth(4.5)); dot("$T$",T3,(1,1,2),blue+linewidth(4.5)); dot("$M$",M,(0,0,5),blue+linewidth(4.5)); dot("$A$",A,(-0.5,-1.5,0),red+linewidth(4.5)); [/asy] The isosceles triangle of centers $O_1 O_2 O$ $O$ is the center of sphere of radii $13$ ) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$
Let $N$ be the midpoint $O_1 O_2$
The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13}$ and $T_1 T_2 = 72.$
Let $M$ be the midpoint $T_1 T_2.$
The height $TM$ is $\sqrt {12^2 \cdot 13 - 36^2} = 12 \sqrt {13-9} = 24.$
The tangents of the half-angle between the planes is $\frac {TO}{AT} = \frac {MN - TO}{TM},$ so $\frac {13}{AT} = \frac {36 - 13}{24},$ \[AT = \frac{24\cdot 13}{23} = \frac {312}{23} \implies 312 + 23 = \boxed{335}.\] vladimir.shelomovskii@gmail.com, vvsss
| 335
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5,484
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11
| 1
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A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$
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Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$
In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$ [asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label("$"+string(i+0.5)+"$",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label("$"+string(i+42.5)+"$",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label("$"+string(i+84.5)+"$",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy] From the second row of the table above, we perform casework on the possible maximum value of $S:$ \[\begin{array}{c||c|c|l} & & & \\ [-2.5ex] \textbf{Max Value} & \boldsymbol{S} & \textbf{Valid?} & \hspace{16.25mm}\textbf{Reasoning/Conclusion} \\ [0.5ex] \hline & & & \\ [-2ex] 49 & \{46,47,48,49\} & & \text{The student who gets } 46 \text{ will reply yes.} \\ 50 & \{47,48,49,50\} & \checkmark & \text{Another possibility is } S=\{89,90,91,92\}. \\ 51 & \{48,49,50,51\} & & \text{The student who gets } 51 \text{ will reply yes.} \\ 56 & \{53,54,55,56\} & & \text{The student who gets } 53 \text{ will reply yes.} \\ 57 & \{54,55,56,57\} & & \text{The student who gets } 57 \text{ will reply yes.} \\ 63 & \{60,61,62,63\} & & \text{The students who get } 60,61,62 \text{ will reply yes.} \\ 66 & \{63,64,65,66\} & & \text{The students who get } 64,65,66 \text{ will reply yes.} \\ 72 & \{69,70,71,72\} & & \text{The student who gets } 69 \text{ will reply yes.} \\ 73 & \{70,71,72,73\} & & \text{The student who gets } 73 \text{ will reply yes.} \\ 78 & \{75,76,77,78\} & & \text{The student who gets } 75 \text{ will reply yes.} \\ 79 & \{76,77,78,79\} & \checkmark & \text{Another possibility is } S=\{34,35,36,37\}. \\ 80 & \{77,78,79,80\} & & \text{The student who gets } 80 \text{ will reply yes.} \end{array}\] Finally, all possibilities for $S$ are $\{34,35,36,37\}, \{47,48,49,50\}, \{76,77,78,79\},$ and $\{89,90,91,92\},$ from which the answer is $37+50+79+92=\boxed{258}.$
| 258
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5,485
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11
| 2
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A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$
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We know right away that $42\not\in S$ and $84\not\in S$ as stated in Solution 1.
To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible sets of four consecutive integers containing 12 and 14; $\{11,12,13,14\}$ and $\{12,13,14,15\}$ . Here we get our first crucial idea; that if the multiples of 6 and 7 differ by two, there will be 2 possible sets of $S$ without any student input. Similarly, if they differ by 3, there will be only 1 possible set, and if they differ by 1, 3 possible sets.
Now we read the student input. Each student says they can’t figure out what $S$ is just based on the teacher’s information, which means each student has to have a number that would be in 2 or 3 of the possible sets (This is based off of the first line of student input). However, now that each student knows that all of them have numbers that fit into more than one possible set, this means that S cannot have two possible sets because otherwise, when shifting from one set to the other, one of the end numbers would not be in the shifted set, but we know each number has to fall in two or more possible sets. For example, take $\{11,12,13,14\}$ and $\{12,13,14,15\}$ . The numbers at the end, 11 and 15, only fall in one set, but each number must fall in at least two sets. This means that there must be three possible sets of S, in which case the actual S would be the middle S.
Take for example $\{33,34,35,36\}$ $\{34,35,36,37\}$ , and $\{35,36,37,38\}$ . 37 and 34 fall in two sets while 35 and 36 fall in all three sets, so the condition is met. Now, this means that the multiple of 6 and 7 must differ by 1.
Since 42 means the difference is 0, when you add/subtract 6 and 7, you will obtain the desired difference of 1, and similarly adding/subtracting 6 or 7 from 84 will also obtain the difference of 1. Thus there are four possible sets of $S$ $\{34,35,36,37\}$ $\{47,48,49,50\}$ $\{76,77,78,79\}$ and $\{89,90,91,92.\}$ . Therefore the sum of the greatest elements of the possible sets $S$ is $37+50+79+92=\boxed{258}$
| 258
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5,486
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11
| 3
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A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$
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In a solution that satisfies these constraints, the multiple of 6 must be adjacent to multiple of 7. The other two numbers must be on either side.
WLOG assume the set is $\{a,6j,7k,b\}$ . The student with numbers $a$ $6j$ , and $7k$ can think the set is $\{a-1, a,6j,7k\}$ or $\{a,6j,7k,b\}$ , and the students with number $6j$ $7k$ , and $b$ can think the set is $\{a,6j,7k,b\}$ or $\{6j,7k,b, b+1\}$ . Therefore, none of the students know the set for sure.
Playing around with the arrangement of the multiple of 6 and multiple of 7 shows that this is the only configuration viewed as ambiguous to all the students. (Therefore when they hear nobody else knows either, they can find out it is this configuration)
Considering $S$ as $\{a,6j,7k,b\}$ ,b is 2 mod 6 and 1 mod 7, so $b$ is 8 mod 42. (since it is all 2-digit, the values are either 50 or 92).
Similarly, considering $S$ as $\{a,7j,6k,b\}$ $b$ is 1 mod 6 and 2 mod 7, so $b$ is 37 mod 42. The values that satisfy that are 37 and 79.
The total sum of all these values is therefore $50+92+37+89=\boxed{258}$
| 258
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5,487
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https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11
| 4
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A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$
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Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$ . If one of the values in $S$ is $3$ or $4 \pmod{7}$ , observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \pmod{7}$ and $S$ contains a $0 \pmod{7}$ , then the values of $S$ must be $(b-3, b-2, b-1, b)$ . Similarly, if we are given a $b \equiv 4 \pmod{7}$ and we know that $0 \pmod{7}$ is in $S$ $S$ must be $(b, b+1, b+2, b+3).$ Hence, the only possibility for $a$ is $5, 6 \pmod{7}.$
In either case, we are guaranteed there is a $6, 0, 1 \pmod{7}$ value in $7$ . The difference comes down to if there is a $5 \pmod{7}$ value or a $1 \pmod{7}$ value. The person receiving such value will be able to determine all of $S$ but the $6, 0, 1 \pmod{7}$ people will not be able to differentiate the two cases ... yet.
Now consider which value among the consecutive integers is $c \equiv 3 \pmod{6}$ , if any. The person will know that $S$ is either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \pmod{6}$ value in $S$ . Neither the $6, 1 \pmod{7}$ person can be $3 \pmod{7}$ , else they can decipher what $S$ is right off the bat by considering which set has $0 \pmod{7}.$ This translates to the possible $5 \pmod{7}$ or $1 \pmod{7}$ person cannot be $0 \pmod{6}$ . We are given that $0 \pmod{7}$ and $0 \pmod{6}$ cannot be the same person. Hence we conclude one of the $6 \pmod{7}$ or $1 \pmod{7}$ must be the $0 \pmod{6}$ person.
Let the $6 \pmod{7}$ person be $0 \pmod{6}.$ Then--hypothetical-- $c \equiv 2 \pmod{7}$ person is $3 \pmod{6}.$ After the first round, the $6, 0, 1 \pmod{7}$ people realize that $2 \pmod{7}$ is not in $S$ else they would have deduced $S$ by noting $S$ was either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \pmod{6}$ and choosing the former based on where $0 \pmod{7}$ is. Hence they figure out $S$ by knowing $5 \pmod{7}.$ So $a \equiv 5 \pmod{7}$ and $a \equiv 5 \pmod{6}$ (from $6 \pmod{7}$ person being $0 \pmod{6}$ ).
Similarly, if $1 \pmod{7}$ person is $0 \pmod{6}$ we find that $a \equiv 6 \pmod{7}$ (so a $2 \pmod{7}$ is in $S$ ) and $a \equiv 4 \pmod{6}.$
By CRT, the possibilities are $a \equiv 5, 34 \pmod{42}.$ The sum of the greatest values of $S$ are the sum of $a + 3$ and so we get $(34 + 3) + (47 + 3) + (76 + 3) + (89 + 3) = \boxed{258}.$
| 258
|
5,488
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_13
| 1
|
Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$
|
Recall that $1000$ divides this expression if $8$ and $125$ both divide it. It should be fairly obvious that $n \geq 3$ ; so we may break up the initial condition into two sub-conditions.
(1) $5^n \equiv n \pmod{8}$ . Notice that the square of any odd integer is $1$ modulo $8$ (proof by plugging in $1^2,3^2,5^2,7^2$ into modulo $8$ ), so the LHS of this expression goes $5,1,5,1,\ldots$ , while the RHS goes $1,2,3,4,5,6,7,8,1,\ldots$ . The cycle length of the LHS is $2$ , RHS is $8$ , so the cycle length of the solution is $\operatorname{lcm}(2,8)=8$ . Indeed, the $n$ that solve this congruence are exactly those such that $n \equiv 5 \pmod{8}$
(2) $2^n \equiv n \pmod{125}$ . This is extremely computationally intensive if we try to calculate all $2^1,2^2,\ldots,2^{100} \pmod{125}$ , so we take a divide-and-conquer approach instead. In order for this expression to be true, $2^n \equiv n \pmod{5}$ is necessary; it shouldn't take too long for us to go through the $20$ possible LHS-RHS combinations, considering that $n$ must be odd. We only need to test $10$ values of $n$ and obtain $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$
With this in mind we consider $2^n \equiv n \pmod{25}$ . By the Generalized Fermat's Little Theorem, $2^{20} \equiv 1 \pmod{25}$ , but we already have $n$ modulo $20$ . Our calculation is greatly simplified. The LHS's cycle length is $20$ and the RHS's cycle length is $25$ , from which their least common multiple is $100$ . In this step we need to test all the numbers between $1$ to $100$ that $n \equiv 3 \pmod{20}$ or $n \equiv 17 \pmod{20}$ . In the case that $n \equiv 3 \pmod{20}$ , the RHS goes $3,23,43,63,83$ , and we need $2^n \equiv n \equiv 2^3 \pmod{25}$ ; clearly $n \equiv 83 \pmod{100}$ . In the case that $n \equiv 17 \pmod{20}$ , by a similar argument, $n \equiv 97 \pmod{100}$
In the final step, we need to calculate $2^{97}$ and $2^{83}$ modulo $125$
Note that $2^{97}\equiv2^{-3}$ ; because $8\cdot47=376\equiv1\pmod{125},$ we get $2^{97} \equiv 47\pmod{125}$
Note that $2^{83}$ is $2^{-17}=2^{-16}\cdot2^{-1}$ . We have \begin{align*} 2^{-1}&\equiv63, \\ 2^{-2}&\equiv63^2=3969\equiv-31, \\ 2^{-4}&\equiv(-31)^2=961\equiv-39, \\ 2^{-8}&\equiv1521\equiv21, \\ 2^{-16}&\equiv441, \\ 2^{-17}&\equiv63\cdot441\equiv7\cdot(-31)=-217\equiv33. \end{align*} This time, LHS cycle is $100$ , RHS cycle is $125$ , so we need to figure out $n$ modulo $500$ . It should be $n \equiv 283,297 \pmod{500}$
Put everything together. By the second subcondition, the only candidates less than $1000$ are $283,297,783,797$ . Apply the first subcondition, $n=\boxed{797}$ is the desired answer.
| 797
|
5,489
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_13
| 2
|
Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$
|
We have that $2^n + 5^n \equiv n\pmod{1000}$ , or $2^n + 5^n \equiv n \pmod{8}$ and $2^n + 5^n \equiv n \pmod{125}$ by CRT. It is easy to check $n < 3$ don't work, so we have that $n \geq 3$ . Then, $2^n \equiv 0 \pmod{8}$ and $5^n \equiv 0 \pmod{125}$ , so we just have $5^n \equiv n \pmod{8}$ and $2^n \equiv n \pmod{125}$ . Let us consider both of these congruences separately.
First, we look at $5^n \equiv n \pmod{8}$ . By Euler's Totient Theorem (ETT), we have $5^4 \equiv 1 \pmod{8}$ , so $5^5 \equiv 5 \pmod{8}$ . On the RHS of the congruence, the possible values of $n$ are all nonnegative integers less than $8$ and on the RHS the only possible values are $5$ and $1$ . However, for $5^n$ to be $1 \pmod{8}$ we must have $n \equiv 0 \pmod{4}$ , a contradiction. So, the only possible values of $n$ are when $n \equiv 5 \pmod{8} \implies n = 8k+5$
Now we look at $2^n \equiv n \pmod{125}$ . Plugging in $n = 8k+5$ , we get $2^{8k+5} \equiv 8k+5 \pmod{125} \implies 2^{8k} \cdot 32 \equiv 8k+5 \pmod{125}$ . Note, for $2^n \equiv n\pmod{125}$ to be satisfied, we must have $2^n \equiv n \pmod{5}$ and $2^n \equiv n\pmod{25}$ . Since $2^{8k} \equiv 1\pmod{5}$ as $8k \equiv 0\pmod{4}$ , we have $2 \equiv -2k \pmod{5} \implies k = 5m-1$ . Then, $n = 8(5m-1) + 5 = 40m-3$ . Now, we get $2^{40m-3} \equiv 40m-3 \pmod{125}$ . Using the fact that $2^n \equiv n\pmod{25}$ , we get $2^{-3} \equiv 15m-3 \pmod{25}$ . The inverse of $2$ modulo $25$ is obviously $13$ , so $2^{-3} \equiv 13^3 \equiv 22 \pmod{25}$ , so $15m \equiv 0 \pmod{25} \implies m = 5s$ . Plugging in $m = 5s$ , we get $n = 200s - 3$
Now, we are finally ready to plug $n$ into the congruence modulo $125$ . Plugging in, we get $2^{200s-3} \equiv 200s - 3 \pmod{125}$ . By ETT, we get $2^{100} \equiv 1 \pmod{125}$ , so $2^{200s- 3} \equiv 2^{-3} \equiv 47 \pmod{125}$ . Then, $200s \equiv 50 \pmod{125} \implies s \equiv 4 \pmod{5} \implies s = 5y+4$ . Plugging this in, we get $n = 200(5y+4) - 3 = 1000y+797$ , implying the smallest value of $n$ is simply $\boxed{797}$
| 797
|
5,490
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14
| 1
|
Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
In this solution, all angle measures are in degrees.
Let $M$ be the midpoint of $\overline{BC}$ so that $\overline{OM}\perp\overline{BC}$ and $A,G,M$ are collinear. Let $\angle ABC=13k,\angle BCA=2k$ and $\angle XOY=17k.$
Note that:
Together, we conclude that $\triangle OAM \sim \triangle OXY$ by AA, so $\angle AOM = \angle XOY = 17k.$
Next, we express $\angle BAC$ in terms of $k.$ By angle addition, we have \begin{align*} \angle AOM &= \angle AOB + \angle BOM \\ &= 2\angle BCA + \frac12\angle BOC \hspace{10mm} &&\text{by Inscribed Angle Theorem and Perpendicular Bisector Property} \\ &= 2\angle BCA + \angle BAC. &&\text{by Inscribed Angle Theorem} \end{align*} Substituting back gives $17k=2(2k)+\angle BAC,$ from which $\angle BAC=13k.$
For the sum of the interior angles of $\triangle ABC,$ we get \begin{align*} \angle ABC + \angle BCA + \angle BAC &= 180 \\ 13k+2k+13k&=180 \\ 28k&=180 \\ k&=\frac{45}{7}. \end{align*} Finally, we obtain $\angle BAC=13k=\frac{585}{7},$ from which the answer is $585+7=\boxed{592}.$
| 592
|
5,491
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14
| 2
|
Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Let $M$ be the midpoint of $BC$ . Because $\angle{OAX}=\angle{OGX}=\angle{OGY}=\angle{OMY}=90^o$ $AXOG$ and $OMYG$ are cyclic, so $O$ is the center of the spiral similarity sending $AM$ to $XY$ , and $\angle{XOY}=\angle{AOM}$ . Because $\angle{AOM}=2\angle{BCA}+\angle{BAC}$ , it's easy to get $\frac{585}{7} \implies \boxed{592}$ from here.
| 592
|
5,492
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14
| 3
|
Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Firstly, let $M$ be the midpoint of $BC$ . Then, $\angle OMB = 90^o$ . Now, note that since $\angle OGX = \angle XAO = 90^o$ , quadrilateral $AGOX$ is cyclic. Also, because $\angle OMY + \angle OGY = 180^o$ $OMYG$ is also cyclic. Now, we define some variables: let $\alpha$ be the constant such that $\angle ABC = 13\alpha, \angle ACB = 2\alpha,$ and $\angle XOY = 17\alpha$ . Also, let $\beta = \angle OMG = \angle OYG$ and $\theta = \angle OXG = \angle OAG$ (due to the fact that $AGOX$ and $OMYG$ are cyclic). Then, \[\angle XOY = 180 - \beta - \theta = 17\alpha \implies \beta + \theta = 180 - 17\alpha.\] Now, because $AX$ is tangent to the circumcircle at $A$ $\angle XAC = \angle CBA = 13\alpha$ , and $\angle CAO = \angle OAX - \angle CAX = 90 - 13\alpha$ . Finally, notice that $\angle AMB = \angle OMB - \angle OMG = 90 - \beta$ . Then, \[\angle BAM = 180 - \angle ABC - \angle AMB = 180 - 13\alpha - (90 - \beta) = 90 + \beta - 13\alpha.\] Thus, \[\angle BAC = \angle BAM + \angle MAO + \angle OAC = 90 + \beta - 13\alpha + \theta + 90 - 13\alpha = 180 - 26\alpha + (\beta + \theta),\] and \[180 = \angle BAC + 13\alpha + 2\alpha = 180 - 11\alpha + \beta + \theta \implies \beta + \theta = 11\alpha.\] However, from before, $\beta+\theta = 180 - 17 \alpha$ , so $11 \alpha = 180 - 17 \alpha \implies 180 = 28 \alpha \implies \alpha = \frac{180}{28}$ . To finish the problem, we simply compute \[\angle BAC = 180 - 15 \alpha = 180 \cdot \left(1 - \frac{15}{28}\right) = 180 \cdot \frac{13}{28} = \frac{585}{7},\] so our final answer is $585+7=\boxed{592}$
| 592
|
5,493
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14
| 4
|
Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$
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[asy] /* Made by MRENTHUSIASM */ size(375); pair A, B, C, O, G, X, Y; A = origin; B = (1,0); C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); O = circumcenter(A,B,C); G = centroid(A,B,C); Y = intersectionpoint(G--G+(100,0),B--C); X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); pair O1=circumcenter(O,G,A); real r1=length(O1-O); markscalefactor=3/160; filldraw(O--X--Y--cycle, rgb(255,255,0)); draw(rightanglemark(O,G,X),red); draw(A--O--B,fuchsia+0.4); draw(Arc(O1,r1,-40,50),royalblue+0.5); draw(circumcircle(O,G,Y), heavygreen+0.5); dot("$A$",A,1.5*dir(180+585/7),linewidth(4)); dot("$B$",B,1.5*dir(-585/7),linewidth(4)); dot("$C$",C,1.5N,linewidth(4)); dot("$O$",O,1.5N,linewidth(4)); dot("$G$",G,1.5S,linewidth(4)); dot("$Y$",Y,1.5E,linewidth(4)); dot("$X$",X,1.5W,linewidth(4)); draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); [/asy] $\angle OAX = \angle OGX = 90^\circ \implies$ quadrilateral $XAGO$ is cyclic $\implies$
$\angle GXO = \angle GAO,$ as they share the same intersept $\overset{\Large\frown} {GO}.$
$\angle OGY = \angle OMY = 90^\circ \implies$ quadrilateral $OGYM$ is cyclic $\implies$
$\angle GYO = \angle OMG,$ as they share the same intercept $\overset{\Large\frown} {GO}.$
In triangles $\triangle XOY$ and $\triangle AOM,$ two pairs of angles are equal, which means that the third angles $\angle XOY = \angle AOM$ are also equal.
$\angle ABC : \angle BCA : \angle AOM = 13 : 2 : 17,$ so $\angle AOM = \angle ABC + 2 \angle BCA.$
According to the Claim $\triangle ABC$ is isosceles, \[\angle ABC : \angle BCA : \angle BAC = 13 : 2 : 13.\] \[\angle BAC = \frac{13} {13 + 2 + 13} \cdot 180^\circ = \frac {585^\circ}{7} \implies 585 + 7 = \boxed{592}.\]
| 592
|
5,494
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14
| 5
|
Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$
|
Extend $XA$ and meet line $CB$ at $P$ . Extend $AG$ to meet $BC$ at $F$ . Since $AF$ is the median from $A$ to $BC$ $A,G,F$ are collinear. Furthermore, $OF$ is perpendicular to $BC$
Draw the circumcircle of $\triangle{XPY}$ , as $OA\bot XP, OG\bot XY, OF\bot PY$ $A,G,F$ are collinear, $O$ lies on $(XYP)$ as $AGF$ is the Simson line of $O$ with respect to $\triangle{XPY}$ . Thus, $\angle{P}=180-17x, \angle{PAB}=\angle{C}=2x, 180-15x=13x, x=\frac{45}{7}$ , the answer is $180-15\cdot \frac{45}{7}=\frac{585}{7}$ which is $\boxed{592}$
| 592
|
5,495
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15
| 1
|
Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$
|
Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$ . Note that this formula also returns the correct value when $n=(k+1)^2$ , but not when $n=k^2$ . Thus $f(n)=k^2+3k+2-n$ for $k^2<n \leq (k+1)^2$
If $2 \mid (k+1)^2-n$ $g(n)$ returns the same value as $f(n)$ . This is because the recursion once again stops at $(k+1)^2$ . We seek a case in which $f(n)<g(n)$ , so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$
Write $7f(n)=4g(n)$ , which simplifies to $3k^2+k-10=3n$ . Notice that we want the LHS expression to be divisible by 3; as a result, $k \equiv 1 \pmod{3}$ . We also want n to be strictly greater than $k^2$ , so $k-10>0, k>10$ . The LHS expression is always even (since $3k^2+k-10$ factors to $k(3k+1)-10$ , and one of $k$ and $3k+1$ will be even), so to ensure that $k$ and $n$ share the same parity, $k$ should be even. Then the least $k$ that satisfies these requirements is $k=16$ , giving $n=258$
Indeed - if we check our answer, it works. Therefore, the answer is $\boxed{258}$
| 258
|
5,496
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15
| 2
|
Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$
|
Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$ . If $k$ is odd, then $f(n)=g(n)$ . If $k$ is even, then \begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*} from which \begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*} Since $k$ is even, $m$ is even. Since $k\neq 0$ , the smallest $k$ is $2$ which produces the smallest $n$ \[k=2 \implies m=16 \implies n=16^2+2=\boxed{258}.\] ~Afo
| 258
|
5,497
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15
| 3
|
Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$
|
To begin, note that if $n$ is a perfect square, $f(n)=g(n)$ , so $f(n)/g(n)=1$ , so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$ , making the values of $f(n+k)$ and $g(n+k)$ integers. Using this notation, we see that $f(n)=k+f(n+k)$ , giving us a formula for the numerator of our ratio. However, since the function of $g(n)$ does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of $\sqrt{n+k}$ in $g(n)$ unless $k$ is an even number. However, this is impossible, since if $k$ was an even number, $f(n)=g(n)$ , giving a ratio of one. Thus, $k$ must be an odd number.
Thus, since $k$ must be an odd number, regardless of whether $n$ is even or odd, to get an integral value in $g(n)$ , we must get to the next perfect square after $n+k$ . To make matters easier, let $z^2=n+k$ . Thus, in $g(n)$ , we want to achieve $(z+1)^2$
Expanding $(z+1)^2$ and substituting in the fact that $z=\sqrt{n+k}$ yields:
\[(z+1)^2=z^2+2z+1=n+k+2\sqrt{n+k}+1\]
Thus, we must add the quantity $k+2z+1$ to $n$ to achieve a integral value in the function $g(n)$ . Thus.
\[g(n)=(k+2z+1)+\sqrt{n+k+2\sqrt{n+k}+1}\]
However, note that the quantity within the square root is just $(z+1)^2$ , and so:
\[g(n)=k+3z+2\]
Thus, \[\frac{f(n)}{g(n)}=\frac{k+z}{k+3z+2}\]
Since we want this quantity to equal $\frac{4}{7}$ , we can set the above equation equal to this number and collect all the variables to one side to achieve
\[3k-5z=8\]
Substituting back in that $z=\sqrt{n+k}$ , and then separating variables and squaring yields that
\[9k^2-73k+64=25n\]
Now, if we treat $n$ as a constant, we can use the quadratic formula in respect to $k$ to get an equation for $k$ in terms of $n$ (without all the squares). Doing so yields
\[\frac{73\pm\sqrt{3025+900n}}{18}=k\]
Now, since $n$ and $k$ are integers, we want the quantity within the square root to be a perfect square. Note that $55^2=3025$ . Thus, assume that the quantity within the root is equal to the perfect square, $m^2$ . Thus, after using a difference of squares, we have \[(m-55)(m+55)=900n\] Since we want $n$ to be an integer, we know that the $LHS$ should be divisible by five, so, let's assume that we should have $m$ divisible by five. If so, the quantity $18k-73$ must be divisible by five, meaning that $k$ leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!).
Thus, we see that to achieve integers $n$ and $k$ that could potentially satisfy the problem statement, we must try the values of $k$ congruent to one modulo five. However, if we recall a statement made earlier in the solution, we see that we can skip all even values of $k$ produced by this modulo argument.
Also, note that $k=1,6$ won't work, as they are too small, and will give an erroneous value for $n$ . After trying $k=11,21,31$ , we see that $k=31$ will give a value of $m=485$ , which yields $n=\boxed{258}$ , which, if plugged in to for our equations of $f(n)$ and $g(n)$ , will yield the desired ratio, and we're done.
| 258
|
5,498
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15
| 4
|
Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$
|
First of all, if $n$ is a perfect square, $f(n)=g(n)=\sqrt{n}$ and their quotient is $1.$ So, for the rest of this solution, assume $n$ is not a perfect square.
Let $a^2$ be the smallest perfect square greater than $n$ and let $b^2$ be the smallest perfect square greater than $n$ with the same parity as $n,$ and note that either $b=a$ or $b=a+1.$ Notice that $(a-1)^2 < n < a^2.$
With a bit of inspection, it becomes clear that $f(n) = a+(a^2-n)$ and $g(n) = b+(b^2-n).$
If $a$ and $n$ have the same parity, we get $a=b$ so $f(n) = g(n)$ and their quotient is $1.$ So, for the rest of this solution, we let $a$ and $n$ have opposite parity. We have two cases to consider.
Case 1: $n$ is odd and $a$ is even
Here, we get $a=2k$ for some positive integer $k.$ Then, $b = 2k+1.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n) = 2k+2m+1$ and $g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.$
We set $\frac{2k+2m+1}{6k+2m+3}=\frac{4}{7},$ cross multiply, and rearrange to get $6m-10k=5.$ Since $k$ and $m$ are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions.
Case 2: $n$ is even and $a$ is odd
Here, we get $a=2k+1$ for some positive ineger $k.$ Then, $b=2k+2.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n)=2k+1+2m+1=2k+2m+2$ and $g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.$
We set $\frac{2k+2m+2}{6k+2m+6} = \frac{4}{7},$ cross multiply, and rearrange to get $5k=3m-5,$ or $k=\frac{3}{5}m-1.$ Since $k$ and $m$ are integers, $m$ must be a multiple of $5.$ Some possible solutions for $(k,m)$ with the least $k$ and $m$ are $(2,5), (5,10), (8,15),$ and $(11,20).$
We wish to minimize $k$ since $a=2k+1.$ One thing to keep in mind is the initial assumption $(a-1)^2 < n < a^2.$
The pair $(2,5)$ gives $a=2(2)+1=5$ and $n=5^2-(2(5)+1)=14.$ But $4^2<14<5^2$ is clearly false, so we discard this case.
The pair $(5,10)$ gives $a=2(5)+1=11$ and $n=11^2-(2(10)+1)=100,$ which is a perfect square and therefore can be discarded.
The pair $(8,15)$ gives $a=2(8)+1=17$ and $n=17^2-(2(15)+1)=258,$ which is between $16^2$ and $17^2$ so it is our smallest solution.
So, $\boxed{258}$ is the correct answer.
| 258
|
5,499
|
https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15
| 5
|
Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$
|
Say the answer is in the form n^2-x, then x must be odd or else f(x) = g(x). Say y = n^2-x. f(y) = x+n, g(y) = 3n+2+x. Because f(y)/g(y) = 4*(an integer)/7*(an integer), f(y) is 4*(an integer) so n must be odd or else f(y) would be odd. Solving for x in terms of n gives integer x = (5/3)n+8/3 which means n is 2 mod 3, because n is also odd, n is 5 mod 6. x must be less than 2n-1 or else the minimum square above y would be (n-1)^2. We set an inequality
(5/3)n+8/3<2n-1 => 5n+8<6n-3 => n>11.
Since n is 5 mod 6, n = 17 and x = 31 giving 17^2-31 = $\boxed{258}$
| 258
|
5,500
|
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_1
| 1
|
In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
[asy] size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, W); label("$E$", F, E); [/asy]
If we set $\angle{BAC}$ to $x$ , we can find all other angles through these two properties:
1. Angles in a triangle sum to $180^{\circ}$ .
2. The base angles of an isosceles triangle are congruent.
Now we angle chase. $\angle{ADE}=\angle{EAD}=x$ $\angle{AED} = 180-2x$ $\angle{BED}=\angle{EBD}=2x$ $\angle{EDB} = 180-4x$ $\angle{BDC} = \angle{BCD} = 3x$ $\angle{CBD} = 180-6x$ . Since $AB = AC$ as given by the problem, $\angle{ABC} = \angle{ACB}$ , so $180-4x=3x$ . Therefore, $x = 180/7^{\circ}$ , and our desired angle is \[180-4\left(\frac{180}{7}\right) = \frac{540}{7}\] for an answer of $\boxed{547}$
| 547
|
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