## Mission Math Tournament: Regular Round Part 1 Solutions

(With the exception of Regular Round #4, written by Jonathan Chen, all of these problems are written by me.)

1. What is the probability that a randomly selected letter in “RATIONAGE” is a vowel?

Solution: There are 5 vowels (A,I,O,A,E) out of 9 letters (R,A,T,I,O,N,A,G,E), and $\boxed{\frac{5}{9}}$ is already a simplified fraction.

2. A triangle with side lengths 5, 12, and 13 is inscribed in a circle. What is the circumference of this circle?

Solution: Since $5^2+12^2=13^2$, the triangle is a right triangle, with right angle opposite the side of length 13. By Thales’ Theorem, a right angle is subtended by a semicircle, so the edge of length 13 is a diameter of the circle the triangle is inscribed in. Since a circle’s circumference is π times its diameter, this circle’s circumference is $\boxed{13\pi}$.

3. Let a be the arithmetic mean of $24\sqrt{7}$ and $25\sqrt{7}$ and h be the harmonic mean of $24\sqrt{7}$ and $25\sqrt{7}$. Find the geometric mean of a and h.

Solution: The arithmetic mean of a and h is $\frac{a+h}{2}$. The harmonic mean of a and h is $\frac{2}{\frac{1}{a}+\frac{1}{h}}=\frac{2}{\frac{a+h}{ah}}=\frac{2ah}{a+h}$. The geometric mean of two positive numbers is the square root of their product. Thus, the geometric mean of the arithmetic and harmonic means of a and h is $\sqrt{\frac{a+h}{2}\left(\frac{2ah}{a+h}\right)}=\sqrt{ah}$. Using this result, the sought geometric mean is equal to $\sqrt{24\sqrt{7}*25\sqrt{7}}=\boxed{10\sqrt{42}}$.

4. There exists two numbers a and b such that $a+b=10$ and $\frac{1}{a}+\frac{1}{b}=\frac{1}{5}$. Find $a^3b+2a^2b^2+ab^3$.

Solution: $a^3b+2a^2b^2+ab^3=ab(a^2+2ab+b^2)=ab(a+b)^2$. We could thus compute the value of the original expression if we can find the values of a+b and ab, without needing to find what a and b are. We were already given the fact that $a+b=10$. Since $\frac{1}{a}+\frac{1}{b}=\frac{b}{ab}+\frac{a}{ab}=\frac{a+b}{ab}$, we now have $\frac{a+b}{ab}=\frac{1}{5}$, giving ab in terms of a+b, which we already know. Specifically, $ab=\frac{a+b}{\frac{1}{5}}=\frac{10}{\frac{1}{5}}=50$. Plugging the values of a+b and ab into the formula we found above, $ab(a+b)^2$, we find the sought value to equal $50(10)^2=\boxed{5000}$.

5. After removing face cards and jokers, how many of the remaining 40 standard poker cards have a front side with 180-degree rotational symmetry? Neglect imperfections in printing.

Solution: Note one important fact about suit pips: the diamond is 180-degree rotationally symmetric, whereas the spade, club, and heart are not! The 2, 3, 4, 5, 6, 7, 8, 9, and 10 of diamonds are 180-degree rotationally symmetric, whereas only the 2, 4, and 10 of spades, clubs, and hearts are, as in poker cards all pips along the middle of the card face in one direction. In total, there are $9+3+3+3=\boxed{18}$ of these cards.

6. Find all solutions $(x,y)$, with x and y both rational, to the equation $40x+ 72y=16x^2+79+24y^2$.

Solution: First, move all terms to one side to arrive at $16x^2-40x+24y^2-72y+79=0$. Moving the constant to the right side, we find that $16x^2-40x+24y^2-72y=-79$. Completing the square in both x and y, we arrive at $16x^2-40x+25+24y^2-72y+54=-79+25+54$, or $16x^2-40x+25+24y^2-72y+54=0$. Dividing the square trinomials of x and y by the respective variables’ coefficients gives $\frac{(x-5/4)^2}{16}+\frac{(y-3/2)^2}{24}=0$. Here, it can be identified that the conic relation is that of a point, but this equation can also be analyzed algebraically. Any real value squared is nonnegative, so both terms on the left hand side must equal 0 for this equation to be true. The only real solution, and also the only rational solution, is $\boxed{(5/4,3/2)}$.

7. Equilateral triangle ABC has side length $\sqrt{3}$. D is on $\overline{AC}$, and E and F are both on $\overline{BC}$, between B and C. Square DEFG has side length 1. Circle O is the smallest circle that passes through G and is tangent to $\overline{AB}$. This point of tangency is H. Find GH.

Solution: Draw altitude $\overline{AJ}$, and call its point of intersection with $\overline{BG}$ K. Label the intersection of $\overline{AB}$ and $\overline{BG}$ L. See the below diagram.

$\angle A\cong\angle B\cong\angle C$, because equilateral triangles are also equiangular, and $m\angle A=m\angle B=m\angle C=60^\circ$. Since opposite sides of a square are parallel, by theorems regarding parallel lines $m\angle A=m\angle{ADK}=m\angle{ALK}=60^\circ$. By vertical angles, $m\angle{ALK}=m\angle{GLH}=60^\circ$. The smallest circle through G and H is drawn when $\overline{GH}$ is the diameter. The diameter is perpendicular to a tangent at the point of tangency, so $m\angle{LHG}=90^\circ$, and $\triangle{GHL}$ is a 30º-60º-90º triangle, similar to $\triangle{AJB}$, $\triangle{AJC}$, $\triangle{AKD}$, and $\triangle{AKL}$.

Now that many 30º-60º-90º right triangles are established, repeatedly apply side-length ratios of 30º-60º-90º right triangles ($1:\sqrt{3}:2$) from here on. Since $AB=\sqrt{3}$, $AJ=\frac{3}{2}$. Since $JK=DE=1$, $AK=\frac{3}{2}-1=\frac{1}{2}$, and thus $DK=KL=\frac{\sqrt{3}}{6}$, and $DL=\frac{\sqrt{3}}{3}$. Since $DG=1$, $LG=1-\frac{\sqrt{3}}{3}$, and finally, $GH=\frac{\sqrt{3}}{2}(1-\frac{\sqrt{3}}{3})=\frac{\sqrt{3}}{2}-\frac{3}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\boxed{\frac{\sqrt{3}-1}{2}}$.

8. What is the 18th digit after the decimal point in the decimal representation of $(9-4\sqrt{5})^{9+4\sqrt{5}}$?

Solution: $9$ and $4\sqrt{5}$ are very close to each other in value. As the exponent in the given expression is significantly greater than 1, the expression itself has an even smaller value. Let’s try to see how small this value is.

Representing both $9$ and $4\sqrt{5}$ as the square root of a positive integer, we find the base to equal $\sqrt{81}-\sqrt{80}$. $\sqrt{81}-\sqrt{64}=1$, and the difference between square roots of consecutive integers decreases as the integers themselves increase, so $\sqrt{81}-\sqrt{80}<\frac{1}{17}<\frac{1}{16}$. The exponent, $9+4\sqrt{5}$, can be similarly written as $\sqrt{81}+\sqrt{80}>17=\sqrt{81}+\sqrt{64}$. Thus, $(9-4\sqrt{5})^{9+4\sqrt{5}}<\left(\frac{1}{16}\right)^{17}=\frac{1}{2^{68}}$.

Since $5^2<2^5$, $\frac{1}{2^{68}}=\frac{1}{2^{18}*2^{50}}<\frac{1}{2^{18}*5^{20}}=\frac{1}{10^{18}*5^2}<\frac{1}{10^{18}}$. Thus, $(9-4\sqrt{5})^{9+4\sqrt{5}}<\frac{1}{10^{18}}$, and the first nonzero digit of the expression has not even occurred by the 18th digit, and the 18th digit, like all digits between it and the decimal point (and a few after), is $\boxed{0}$.

9. Equilateral triangle XYZ has side lengths 60. Two line segments are drawn from X to $\overline{YZ}$, trisecting $\overline{YZ}$; two line segments are drawn from Y to $\overline{XZ}$, trisecting $\overline{XZ}$; and two line segments are drawn from Z to $\overline{XY}$, trisecting $\overline{XY}$. What is the perimeter of the smallest polygon described by these line segments that contains the center of the triangle?

Solution: We first draw a diagram, like the one below.

We see, after considering symmetry, that the smallest polygon described by these line segments that contains the center of the triangle is an equilateral, but not equiangular, hexagon. Thus, we can find the perimeter of this hexagon by finding the length of one of its sides and multiplying that length by six.

Label the center of the triangle O. Label the midpoint of $\overline{YZ}$ A. Draw $\overline{XA}$, which is a perpendicular bisector, because a line segment from a vertex to the midpoint of the opposite side of an isosceles (including equilateral) triangle is an altitude and thus perpendicular to the side. By the properties of a 30º-60º-90º right triangle, $XA=30\sqrt{3}$. Label the points of trisection on $\overline{XZ}$ as B and C, with B closer to X. Since Drop a perpendicular from B to $\overline{YZ}$, labeling the point of contact W. Draw $\overline{WX}$. Label the point of trisection on $\overline{XY}$ closer to X as V. Drop a perpendicular from V to $\overline{AX}$, labeling the point of contact S. Label the point of intersection of $\overline{ZV}$ and $\overline{AX}$ M. Label the point of intersection of $\overline{ZV}$ and $\overline{BW}$ P. Label the point of intersection of $\overline{ZV}$ and $\overline{WX}$ N, which is a vertex of the inner hexagon and between M and P. Draw $\overline{CO}$. Draw $\overline{BS}$. See the diagram below.

$\angle AXZ$ is shared by $\triangle COX$ and $\triangle ZAX$. The centroid of any triangle is two-thirds of the way from a vertex to the midpoint of the opposite edge, so $OA=10\sqrt{3}$ and $\frac{XO}{OA}=2$. Since C is a point of trisection, $\frac{XC}{CZ}=2$. Thus, by SAS, $\triangle COX \sim \triangle ZAX$. Thus, since $\angle ZAX$ is a right angle, $\angle COX$ is a right angle. Additionally, since $AZ=30$, $OC=20$.

$\angle XZY$ is shared by $\triangle BZW$ and $\triangle XZA$. Both $\angle XAZ$ and $\angle BWZ$ are right angles, and are thus congruent. Thus, by AA, $\triangle BZW \sim \triangle XZA$. Since $XZ=60$, $BZ=40$, and $XA=30\sqrt{3}$. $BW=20\sqrt{3}$.

Since OPWA is a rectangle and $OA=10\sqrt{3}$, $PW=10\sqrt{3}$, and since $BW=20\sqrt{3}$, $BP=10\sqrt{3}$. Since B, S, and V are collinear because of symmetry, $SO=10\sqrt{3}$, and $AS=20\sqrt{3}$.

$\angle BYZ$ is shared by $\triangle MAY$ and $\triangle BWY$, and $\angle MAY$ and $\angle BWY$ are both right angles and thus congruent, so $\triangle MAY \sim \triangle BWY$. Hence, since $\frac{YA}{YW}=\frac{3}{4}$, $\frac{AM}{WB}=\frac{3}{4}$. Thus, since $BW=20\sqrt{3}$, $AM=15\sqrt{3}$, and since $AO=10\sqrt{3}$ and $AS=20\sqrt{3}$, $OM=MS=5\sqrt{3}$. By vertical angles, $\angle OMP \cong \angle SMV$, and $\angle MOP \cong \angle MSV$ because they’re both right angles. Thus, by ASA, $\triangle{MOP}\cong\triangle MSV$.

Since $AW=10$ and $AX=30\sqrt{3}$, by the Pythagorean Theorem, $WX=20\sqrt{7}$. By symmetry, $BY=20\sqrt{7}$, and by the similarity of triangles  MAY and BWY, $BM=5\sqrt{7}$. By symmetry, $VM=5\sqrt{7}$. By the congruence of triangles MOP and MSV, $MP=5\sqrt{7}$.

Since $XA=30\sqrt{3}$ and $AM=15\sqrt{3}$, $MX=15\sqrt{3}$. Since $XA||BW$, $\angle XMP \cong \angle WPM$, and $\angle MXW \cong \angle PWX$. Thus, by AA, $\triangle MXN \sim \triangle PWN$. Hence, because $\frac{MX}{PW}=\frac{15\sqrt{3}}{10\sqrt{3}}=\frac{3}{2}$, $\frac{MN}{NP}=\frac{3}{2}$. Thus, $\frac{MN}{MP}=\frac{3}{5}$. Since $MP=5\sqrt{7}$, this means $MN=\frac{3}{5}(5\sqrt{7})=3\sqrt{7}$. Now that we have found a side length of the equilateral hexagon, we sextoople it to find that this hexagon’s perimeter is $6(3\sqrt{7})=\boxed{18\sqrt{7}}$.

Problem 10 was dropped.