2019 AMC 10 A

The first part can be rewritten as .The second part is.Adding these up gives .

The last three digits of for all are ,because there are at least three and three in its prime factorization. Because ,the answer is .

Solution

Let be the age of Ana and be the age of Bonita. Then,

and .

Substituting the second equation into the first gives us .

By using difference of squares and dividing, .Moreover, .

The answer is .

Solution (Guess and Check)

Simple guess and check works. Start with all the square numbers , , , , , , etc. (probably stop at around since at that point it wouldn't make sense). If Ana is , then Bonita is , so in the previous year, Ana's age was times greater than Bonita's. If Ana is , then Bonita is and Ana's age was times greater than Bonita's in the previous year, as required. The difference in the ages is .

By choosing the maximum number of balls while getting of each color, we could have chosen red balls, green balls, yellow balls, blue balls, white balls, and black balls, for a total of balls. Picking one more ball guarantees that we will get  balls of a color either red, green, or yellow. Thus the answer is .

We might at first think that the answer would be , because when . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence , , , , cancels out except . Thus, the answer is, intuitively, integers.

Though impractical, a proof of maximality can proceed as follows: Let the desired sequence of consecutive integers be , , , , where there are terms, and we want to maximize . Then the sum of the terms in this sequence is .Rearranging and factoring, this reduces to . Since must divide , and we know that is an attainable value of the sum, must be the maximum.

To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be if the middle two numbers are and , so the answer is .

This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is .

We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is .

The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is .

Let's first work out the slopeintercept form of all three lines: and implies so , while implies so . Also, implies . Thus the lines are , , and . Now we find the intersection points between each of the lines with , which are and . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base and height , whose area is .

Like in Solution , we determine the coordinates of the three vertices of the triangle. Now, using the Shoelace Theorem, we can directly find that the area is .

Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at and . Then apply Heron's Formula: the semiperimeter will be , so the area reduces nicely to a difference of squares, making it .

Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are , , and . We can now draw the bounding square with vertices , , and , and deduce that the triangle's area is .

Like in other solutions, we find that the three points of intersection are , , and . Using graph paper, we can see that this triangle has boundary lattice points and interior lattice points. By Pick's Theorem, the area is .

Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, , so the area is .

Like in other solutions, we find that the three points of intersection are , , and . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. .

Like in other solutions, we find the three points of intersection. Label these , , and . Then vectors and . The area of the triangle is half the magnitude of the cross product of these two vectors. .

Like in other solutions, we find that the three points of intersection are , , and . By the Pythagorean theorem, this is an isosceles triangle with base and equal length . The area of an isosceles triangle with base and equal length is . Plugging in and , .

Like in other solutions, we find the three points of intersection. Label these , , and . By the Pythagorean Theorem, and . By the Law of Cosines, Therefore, . By the extended Law of Sines ,Then the area is .

The area of a triangle formed by three lines, ,, is the absolute value of Plugging in the three lines, the area is the absolute value of

Statement is true. A rotation about the point half way between an upfacing square and a downfacing square will yield the same figure.Statement is also true. A translation to the left or right will place the image onto itself when the figures above and below the line realign (the figure goes on infinitely in both directions).Statement is false. A reflection across line will change the upfacing squares to downfacing squares and vice versa.

Finally, statement is also false because it will cause the diagonal lines extending from the squares to switch direction. Thus, only statements are true.

The sum of positive integers is , and we want this not to be a divisor of (the product of the first positive integers). Notice that if and only if were composite, all of its factors would be less than or equal to , so would be able to cancel with these factors in , and thus the sum would be a divisor. Hence in this case, must instead be prime. The greatest threedigit integer that is prime is , so we subtract to get .

As in Solution , we deduce that must be prime. If we can't immediately recall what the greatest threedigit prime is, we can instead use this result to eliminate answer choices as possible values of .Choices , , and don't work because is even, and choice does not work since is divisible by . Thus, the correct answer must be .

The number of tiles the bug visits is equal to plus the number of times it crosses a horizontal or vertical line. As it must cross horizontal lines and vertical lines, it must be that the bug visits a total of squares.

Note: The general formula for this is , because it is the number of vertical/horizontal lines crossed minus the number of corners crossed (to avoid double counting). In this particular problem, it was (since ), which is , but then you add because the first tile and the last tile are counted, which in the general formula are not counted.

We draw a diagram (optionally with grid paper and/or a ruler), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into sections, and just draw one of these feet by feet sections.

While it appears that the line we drew comes very close to several points, we know that since and are relatively prime, it will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count from the diagram, we get squares, giving a total of .

Prime factorizing , we get . A perfect square must have even powers of its prime factors,so our possible choices for our exponents of a perfect square are , , , , for both and . This yields perfect squares.

Perfect cubes must have multiples of for each of their prime factors' exponents, so we have either , , , or for both and , which yields perfect cubes, for a total of . Subtracting the overcounted powers of (, , , and ), we get .

Observe that . Now divide into cases:

Case : The factor is . Then we can have , , , , , or .

Case : The factor is . This is the same as Case .

Case : The factor is some combination of and .

This would be easy if we could just have any combination, as that would simply give . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for .

is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.

is a "cube" because it would give a factor of this number that is a perfect cube. More generally,it is a multiple of .

is a "square"

is interesting, since it's both a "square" and a "cube". Don't count this as either because this would doublecount, so we will count this in another case.

is a "square"

is a "cube".

Now let's consider subcases:

Subcase : The squares are with each other.

Since we have square terms, and they would pair with other square terms, we get possibilities.

Subcase : The cubes are with each other.

Since we have cube terms, and they would pair with other cube terms, we get possibilities.

Subcase : A number pairs with .

Since any number can pair with (as it gives both a square and a cube), there would be possibilities. Remember however that there can be two different bases ( and ), and they would produce different results. Thus, there are in fact  possibilities.

Finally, summing the cases gives .

First of all, obviously has to be smaller than , since when calculating , we must take into account the , , and . So we can eliminate choices and . Since there are total entries, the median, , must be the one, at which point we note that is , so has to be the median (because is between and ). Now, the mean, , must be smaller than , since there are many fewer , , and . is less than ,because when calculating , we would include , , and . Thus the answer is .

As in Solution , we find that the median is . Then, looking at the modes , we realize that even if we were to have of each, their median would remain the same, being . As for the mean, we note that the mean of the first is simply the same as the median of them, which is . Hence, since we in fact have 's, 's, and 's, the mean has to be higher than . On the other hand, since there are fewer 's, 's, and 's than the rest of the numbers, the mean has to be lower than (the median). By comparing these values, the answer is .

We can solve this problem simply by carefully calculating each of the values, which turn out to be , , and . Thus the answer is

Drawing it out, we see and are right angles, as they are inscribed in a semicircle.Using the fact that it is an isosceles triangle, we find . We can find and by the triangle angle sum on and .

Then, we take triangle , and find

Alternatively, we could have used similar triangles. We start similarly to Solution .Drawing it out, we see and are right angles, as they are inscribed in a semicircle.Therefore, .

So, by AA Similarity, since and .Thus, we know , Finally, we deduce.

Through the property of angles formed by intersecting chords, we find that

Through the Outside Angles Theorem, we find that

Adding the two equations gives us

Since is the diameter, , and because is isosceles and , we have . Thus .

Notice that if , then and must be . Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that ≌.Thus , and so , which is .

Note: As in many elementary geometry problems, if you can't see how to solve it, you could simply draw an accurate diagram and measure the angle using a protractor as .

It is possible to obtain , , , , , and points of intersection, as demonstrated in the following figures:

It is clear that the maximum number of possible intersections is , since each pair of lines can intersect at most once. We now prove that it is impossible to obtain two intersections.

We proceed by contradiction. Assume a configuration of four lines exists such that there exist only two intersection points. Let these intersection points be and . Consider two cases:

Case : No line passes through both and

Then, since an intersection is obtained by an intersection between at least two lines, two lines pass through each of and . Then, since there can be no additional intersections, no line that passes through can intersect a line that passes through , and so each line that passes through must be parallel to every line that passes through . Then the two lines passing through are parallel to each other by transitivity of parallelism, so they coincide, contradiction.

Case : There is a line passing through and

Then there must be a line passing through , and a line passing through . These lines must be parallel. The fourth line must pass through either or . Without loss of generality, suppose passes through . Then since and cannot coincide, they cannot be parallel.

Then and cannot be parallel either, so they intersect, contradiction.

All possibilities have been exhausted, and thus we can conclude that two intersections is impossible.Our answer is given by the sum .

Using the recursive formula, we find , , and so on. It appears that , for all . Setting , we find , so the answer is .

To prove this formula, we use induction. We are given that and , which satisfy our formula. Now assume the formula holds true for all for some positive integer . By our assumption, and . Using the recursive formula,