AMC10 2011 B

Answer (C): The given expression is equal to $12 / (9 − 9/12) = 4 / (3 − 3/4) = 16 − 9 / 12 = 7/12$.

Answer (E): The sum of her first 5 test scores is 385, yielding an average of 77. To raise her average to 80, her 6th test score must be the difference between 6 · 80 = 480 and 385, which is 95.

Answer (A): The smallest possible width for the rectangle is 2 − 0.5 = 1.5 inches. Similarly the smallest possible length is 2.5 inches. Hence the minimum area is (1.5)(2.5) = 3.75 square inches.

Answer (C): Bernardo has paid B − A dollars more than LeRoy. If LeRoy gives Bernardo half of that difference, (B−A)/2, then each will have paid the same amount.

Answer (E): Because 161 = 23 · 7, the only two digit factor of 161 is 23. The correct product must have been 32 · 7 = 224.

Answer (A): Let $x$ be Casper’s original number of candies. After the first day he was left with $x-\left(\frac{1}{3}x+2\right)=\frac{2}{3}x-2$ candies. On the second day he ate $\frac{1}{3}\left(\frac{2}{3}x-2\right)$ candies, gave away 4 candies, and was left with 8 candies. Therefore $$ \frac{2}{3}x-2-\left(\frac{1}{3}\left(\frac{2}{3}x-2\right)+4\right)=8. $$ Solving for $x$ results in $x=30$.

Answer (B): The degree measures of two of the angles have a sum of $\frac{6}{5} \cdot 90 = 108 $ and a positive difference of 30, so their measures are 69 and 39. The remaining angle has a degree measure of $ 180 - 108 = 72 $, which is the largest angle.

Answer (B): Because the beach was not crowded on June 10, at least one of the conditions was not met. That is, the weather might have been cooler than 80 °F and sunny, at least 80 °F and cloudy, or cooler than 80 °F and cloudy. The first possibility shows that (A) and (E) are invalid, the second shows that (C) is invalid, and the third shows that (D) is invalid. Only conclusion (B) is consistent with all three possibilities.

Answer (D): The area of $\triangle ABC$ is $\frac{1}{2}\cdot 3 \cdot 4=6$, so the area of $\triangle EBD$ is $\frac{1}{3}\cdot 6=2$. Note that $\triangle ABC$ and $\triangle EBD$ are right triangles with an angle in common, so they are similar. Therefore $BD$ and $DE$ are in the ratio $4$ to $3$. Let $BD=x$ and $DE=\frac{3}{4}x$. Then the area of $\triangle EBD$ can be expressed as $\frac{1}{2}\cdot x \cdot \frac{3}{4}x=\frac{3}{8}x^2$. Because $\triangle EBD$ has area $2$, solving yields $BD=\frac{4\sqrt{3}}{3}$.

Answer (B): The sum of the smallest ten elements is $1+10+100+\cdots+1{,}000{,}000{,}000=1{,}111{,}111{,}111.$ Hence the desired ratio is $\dfrac{10{,}000{,}000{,}000}{1{,}111{,}111{,}111}=\dfrac{9{,}999{,}999{,}999+1}{1{,}111{,}111{,}111}=9+\dfrac{1}{1{,}111{,}111{,}111}\approx 9.$ OR The sum of a finite geometric series of the form $a(1+r+r^2+\cdots+r^n)$ is $\dfrac{a}{1-r}(1-r^{n+1})$. The desired denominator $1+10+10^2+\cdots+10^9$ is a finite geometric series with $a=1$, $r=10$, and $n=9$. Therefore the ratio is $\dfrac{10^{10}}{1+10+10^2+\cdots+10^9}=\dfrac{10^{10}}{\dfrac{1}{1-10}(1-10^{10})}=\dfrac{10^{10}}{10^{10}-1}\cdot 9\approx \dfrac{10^{10}}{10^{10}}\cdot 9=9.$

Answer (D): If no more than 4 people have birthdays in any month, then at most 48 people would be accounted for. Therefore the statement is true for $n = 5$. The statement is false for $n \ge 6$ if, for example, 5 people have birthdays in each of the first 4 months of the year, and 4 people have birthdays in each of the last 8 months, for a total of $5 \cdot 4 + 4 \cdot 8 = 52$ people.

Answer (A): The only parts of the track that are longer walking on the outside edge rather than the inside edge are the two semicircular portions. If the radius of the inner semicircle is $r$, then the difference in the lengths of the two paths is $2\pi(r+6)-2\pi r=12\pi$. Let $x$ be her speed in meters per second. Then $36x=12\pi$, and $x=\frac{\pi}{3}$.

Answer (D): Consider all ordered pairs $(a,b)$ with each of the numbers $a$ and $b$ in the closed interval $[-20,10]$. These pairs fill a $30\times 30$ square in the coordinate plane, with an area of $900$ square units. Ordered pairs in the first and third quadrants have the desired property, namely $a\cdot b>0$. The areas of the portions of the $30\times 30$ square in the first and third quadrants are $10^2=100$ and $20^2=400$, respectively. Therefore the probability of a positive product is $\frac{100+400}{900}=\frac{5}{9}$.

Answer (C): Let $x$ and $y$ be the length and width of the parking lot, respectively. Then $xy = 168$ and $x^2 + y^2 = 25^2$. Note that $$(x+y)^2 = x^2 + y^2 + 2xy = 25^2 + 2\cdot 168 = 961.$$ Hence the perimeter is $2(x+y) = 2\cdot \sqrt{961} = 62$. Note that the dimensions of the parking lot are $7$ and $24$ meters.

Answer (E): If $x\neq 0$, then I is false: $x\odot(y+z)=\dfrac{x+(y+z)}{2}\neq\dfrac{x+y+x+z}{2}=\dfrac{x+y}{2}+\dfrac{x+z}{2}=(x\odot y)+(x\odot z).$ On the other hand, II and III are true for all values of $x$, $y$ and $z$: $x+(y\odot z)=x+\dfrac{y+z}{2}=\dfrac{2x+y+z}{2}=\dfrac{(x+y)+(x+z)}{2}=(x+y)\odot(x+z),$ and $x\odot(y\odot z)=\dfrac{x+\dfrac{y+z}{2}}{2}=\dfrac{\left(\dfrac{2x+y+z}{2}\right)}{2}=\dfrac{\dfrac{x+y}{2}+\dfrac{x+z}{2}}{2}=(x\odot y)\odot(x\odot z)$

Answer (A): Assume the octagon’s edge is 1. Then the corner triangles have hypotenuse 1 and thus legs $\frac{\sqrt{2}}{2}$ and area $\frac{1}{4}$ each; the four rectangles are $1$ by $\frac{\sqrt{2}}{2}$ and have area $\frac{\sqrt{2}}{2}$ each, and the center square has area $1$. The total area is $4\cdot\frac{1}{4}+4\cdot\frac{\sqrt{2}}{2}+1=2+2\sqrt{2}$. The probability that the dart hits the center square is $\frac{1}{2+2\sqrt{2}}=\frac{\sqrt{2}-1}{2}$.

Answer (C): Angle $EAB$ is $90^\circ$ because it subtends a diameter. Therefore angles $BEA$ and $ABE$ are $40^\circ$ and $50^\circ$, respectively. Angle $DEB$ is $50^\circ$ because $AB$ is parallel to $ED$. Also, $\angle DEB$ is supplementary to $\angle CDE$, so $\angle CDE = 130^\circ$. Because $EB$ and $DC$ are parallel chords, $ED = BC$ and $EBCD$ is an isosceles trapezoid. Thus $\angle BCD = \angle CDE = 130^\circ$. OR Let $O$ be the center of the circle. Establish, as in the first solution, that $\angle EAB = 90^\circ$, $\angle BEA = 40^\circ$, $\angle ABE = 50^\circ$, and $\angle DEB = 50^\circ$. Thus $AD$ is a diameter and $\angle AOE = 100^\circ$. By the Inscribed Angle Theorem $$ \angle BCD=\frac12(\angle BOA+\angle AOE+\angle EOD)=\frac12(80^\circ+100^\circ+80^\circ)=130^\circ. $$

Answer (E): Sides AB and CD are parallel, so ∠CDM = ∠AMD. Because ∠AMD = ∠CMD, it follows that △CMD is isosceles and CD = CM = 6. Therefore △MCB is a 30–60–90° right triangle with ∠BMC = 30°. Finally, 2 · ∠AMD + 30°= ∠AMD + ∠CMD + 30°= 180°, so ∠AMD = 75°.

Answer (A): The right side of the equation is defined only when |x| ≥ 4. If x ≥ 4, the equation is equivalent to 5x + 8 = x² − 16, and the only solution with x ≥ 4 is x = 8. If x ≤ −4, the equation is equivalent to 8 − 5x = x² − 16, and the only solution with x ≤ −4 is x = −8. The product of the solutions is −8 · 8 = −64.

Answer (C): Let $E$ and $H$ be the midpoints of $\overline{AB}$ and $\overline{BC}$, respectively. The line drawn perpendicular to $AB$ through $E$ divides the rhombus into two regions: points that are closer to vertex $A$ than $B$, and points that are closer to vertex $B$ than $A$. Let $F$ be the intersection of this line with diagonal $AC$. Similarly, let point $G$ be the intersection of the diagonal $AC$ with the perpendicular to $BC$ drawn from the midpoint of $BC$. Then the desired region $R$ is the pentagon $BEFGH$. Note that $\triangle AFE$ is a $30-60-90^\circ$ triangle with $AE=1$. Hence the area of $\triangle AFE$ is $\frac12\cdot 1\cdot \frac{1}{\sqrt3}=\frac{\sqrt3}{6}$. Both $\triangle BFE$ and $\triangle BGH$ are congruent to $\triangle AFE$, so they have the same areas. Also $\angle FBG=120^\circ-\angle FBE-\angle GBH=60^\circ$, so $\triangle FBG$ is an equilateral triangle. In fact, the altitude from $B$ to $FG$ divides $\triangle FBG$ into two triangles, each congruent to $\triangle AFE$. Hence the area of $BEFGH$ is $4\cdot \frac{\sqrt3}{6}=\frac{2\sqrt3}{3}$.

Answer (B): The largest pairwise difference is 9, so $w-z=9$. Let $n$ be either $x$ or $y$. Because $n$ is between $w$ and $z$, $$ 9=w-z=(w-n)+(n-z). $$ Therefore the positive differences $w-n$ and $n-z$ must sum to 9. The given pairwise differences that sum to 9 are $3+6$ and $4+5$. The remaining pairwise difference must be $x-y=1$. The second largest pairwise difference is 6, so either $w-y=6$ or $x-z=6$. In the first case the set of four numbers may be expressed as $\{w,w-5,w-6,w-9\}$. Hence $4w-20=44$, so $w=16$. In the second case $w-x=3$, and the four numbers may be expressed as $\{w,w-3,w-4,w-9\}$. Therefore $4w-16=44$, so $w=15$. The sum of the possible values for $w$ is $16+15=31$. Note: The possible sets of four numbers are $\{16,11,10,7\}$ and $\{15,12,11,6\}$.

Answer (A): Let $A$ be the apex of the pyramid, and let the base be the square $BCDE$. Then $AB=AD=1$ and $BD=\sqrt{2}$, so $\triangle BAD$ is an isosceles right triangle. Let the cube have edge length $x$. The intersection of the cube with the plane of $\triangle BAD$ is a rectangle with height $x$ and width $\sqrt{2}x$. It follows that $\sqrt{2}=BD=2x+\sqrt{2}x$, from which $x=\sqrt{2}-1$. Hence the cube has volume $$(\sqrt{2}-1)^3=(\sqrt{2})^3-3(\sqrt{2})^2+3\sqrt{2}-1=5\sqrt{2}-7.$$

Answer (D): In the expansion of $(2000+11)^{2011}$, all terms except $11^{2011}$ are divisible by $1000$, so the hundreds digit of $2011^{2011}$ is equal to that of $11^{2011}$. Furthermore, in the expansion of $(10+1)^{2011}$, all terms except $1^{2011}$, $\binom{2011}{1}(10)(1^{2010})$, and $\binom{2011}{2}(10)^2(1^{2009})$ are divisible by $1000$. Thus the hundreds digit of $11^{2011}$ is equal to that of $$ 1+\binom{2011}{1}(10)(1^{2010})+\binom{2011}{2}(10)^2(1^{2009}) $$ $$ =1+2011\cdot 10+2011\cdot 1005\cdot 100 $$ $$ =1+2011\cdot 100510. $$ Finally, the hundreds digit of this number is equal to that of $1+11\cdot 510=5611$, so the requested hundreds digit is $6$.

Answer (B): For $0<x\le 100$, the nearest lattice point directly above the line $y=\frac12x+2$ is $(x,\frac12x+3)$ if $x$ is even and $(x,\frac12x+\frac52)$ if $x$ is odd. The slope of the line that contains this point and $(0,2)$ is $\frac12+\frac1x$ if $x$ is even and $\frac12+\frac1{2x}$ if $x$ is odd. The minimum value of the slope is $\frac{51}{100}$ if $x$ is even and $\frac{50}{99}$ if $x$ is odd. Therefore the line $y=mx+2$ contains no lattice point with $0<x\le 100$ for $\frac12<m<\frac{50}{99}$.

Answer (D): Let $T_n=\triangle ABC$. Suppose $a=BC$, $b=AC$, and $c=AB$. Because $BD$ and $BE$ are both tangent to the incircle of $\triangle ABC$, it follows that $BD=BE$. Similarly, $AD=AF$ and $CE=CF$. Then $$ 2BE=BE+BD=BE+CE+BD+AD-(AF+CF)=a+c-b, $$ that is, $BE=\frac12(a+c-b)$. Similarly $AD=\frac12(b+c-a)$ and $CF=\frac12(a+b-c)$. In the given $\triangle ABC$, suppose that $AB=x+1$, $BC=x-1$, and $AC=x$. Using the formulas for $BE$, $AD$, and $CF$ derived before, it must be true that $$ BE=\frac12\big((x-1)+(x+1)-x\big)=\frac12x, $$ $$ AD=\frac12\big(x+(x+1)-(x-1)\big)=\frac12x+1,\ \text{and} $$ $$ CF=\frac12\big((x-1)+x-(x+1)\big)=\frac12x-1. $$ Hence both $(BC,CA,AB)$ and $(CF,BE,AD)$ are of the form $(y-1,y,y+1)$. This is independent of the values of $a$, $b$, and $c$, so it holds for all $T_n$. Furthermore, adding the formulas for $BE$, $AD$, and $CF$ shows that the perimeter of $T_{n+1}$ equals $\frac12(a+b+c)$, and consequently the perimeter of the last triangle $T_N$ in the sequence is $$ \frac{1}{2^{N-1}}(2011+2012+2013)=\frac{1509}{2^{N-3}}. $$ The last member $T_N$ of the sequence will fail to define a successor if for the first time the new lengths fail the Triangle Inequality, that is, if $$ -1+\frac{2012}{2^N}+\frac{2012}{2^N}\le 1+\frac{2012}{2^N}. $$ Equivalently, $2012\le 2^{N+1}$ which happens for the first time when $N=10$. Thus the required perimeter of $T_N$ is $\frac{1509}{2^7}=\frac{1509}{128}$.