Difference between revisions of "1999 AIME Problems/Problem 8"
(solution) |
|||
(4 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math>\mathcal{T}</math> be the set of ordered triples <math>(x,y,z)</math> of nonnegative [[real number]]s that lie in the [[plane]] <math>x+y+z=1.</math> Let us say that <math>(x,y,z)</math> supports <math>(a,b,c)</math> when exactly two of the following are true: <math>x\ge a, y\ge b, z\ge c.</math> Let <math>\mathcal{S}</math> consist of those triples in <math>\mathcal{T}</math> that support <math>\left(\frac 12,\frac 13,\frac 16\right).</math> The area of <math>\mathcal{S}</math> divided by the area of <math>\mathcal{T}</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relatively prime positive integers | + | Let <math>\mathcal{T}</math> be the set of ordered triples <math>(x,y,z)</math> of nonnegative [[real number]]s that lie in the [[plane]] <math>x+y+z=1.</math> Let us say that <math>(x,y,z)</math> supports <math>(a,b,c)</math> when exactly two of the following are true: <math>x\ge a, y\ge b, z\ge c.</math> Let <math>\mathcal{S}</math> consist of those triples in <math>\mathcal{T}</math> that support <math>\left(\frac 12,\frac 13,\frac 16\right).</math> The area of <math>\mathcal{S}</math> divided by the area of <math>\mathcal{T}</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> |
== Solution == | == Solution == | ||
Line 7: | Line 7: | ||
[[Image:1999_AIME-8.png]] | [[Image:1999_AIME-8.png]] | ||
− | The region in <math>(x,y,z)</math> where <math>x \ge \frac{1}{2}, y \ge \frac{1}{3}</math> is that of a little triangle on the bottom of the above diagram, of <math>y \ge {1}{3}, z \ge \frac{1}{6}</math> is the triangle at the right, and <math>x \ge \frac 12, z \ge \frac 16</math> the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by <math>x+y+z=1,\ x,y,z \ge 0</math>. We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third. | + | The region in <math>(x,y,z)</math> where <math>x \ge \frac{1}{2}, y \ge \frac{1}{3}</math> is that of a little triangle on the bottom of the above diagram, of <math>y \ge \frac{1}{3}, z \ge \frac{1}{6}</math> is the triangle at the right, and <math>x \ge \frac 12, z \ge \frac 16</math> the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by <math>x+y+z=1,\ x,y,z \ge 0</math>. We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third. |
[[Image:1999_AIME-8a.png]] | [[Image:1999_AIME-8a.png]] | ||
− | The side length of the large equilateral triangle is <math>\sqrt{2}</math>, which we can find using 45-45-90 <math>\triangle</math> with the axes. Using the formula <math>A = \frac{s^2\sqrt{3}}{4}</math> for [[equilateral triangle]]s, the area of the large triangle is <math>\frac{(\sqrt{2})^2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}</math>. Since the lines of the smaller triangles are [[parallel]] to those of the large triangle, by corresponding angles we see that all of the triangles are [[similar triangles|similar]], so they are all | + | The side length of the large equilateral triangle is <math>\sqrt{2}</math>, which we can find using 45-45-90 <math>\triangle</math> with the axes. Using the formula <math>A = \frac{s^2\sqrt{3}}{4}</math> for [[equilateral triangle]]s, the area of the large triangle is <math>\frac{(\sqrt{2})^2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}</math>. Since the lines of the smaller triangles are [[parallel]] to those of the large triangle, by corresponding angles we see that all of the triangles are [[similar triangles|similar]], so they are all equilateral triangles. We can solve for their side lengths easily by subtraction, and we get <math>\frac{\sqrt{2}}{6}, \frac{\sqrt{2}}{3}, \frac{\sqrt{2}}{2}</math>. Calculating their areas, we get <math>\frac{\sqrt{3}}{8}, \frac{\sqrt{3}}{18}, \frac{\sqrt{3}}{72}</math>. The [[ratio]] <math>\frac{\mathcal{S}}{\mathcal{T}} = \frac{\frac{9\sqrt{3} + 4\sqrt{3} + \sqrt{3}}{72}}{\frac{\sqrt{3}}{2}} = \frac{14}{36} = \frac{7}{18}</math>, and the answer is <math>m + n = \boxed{025}</math>. |
− | To simplify the problem, we could used the fact that the area ratios are equal to the side ratios squared, and we get <math>\left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right^2 + \left(\frac{1}{6}\right)^2 = \frac{14}{36} = \frac{7}{18}</math>. | + | To simplify the problem, we could used the fact that the area ratios are equal to the side ratios squared, and we get <math>\left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{14}{36} = \frac{7}{18}</math>. |
== See also == | == See also == | ||
Line 19: | Line 19: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:40, 4 July 2013
Problem
Let be the set of ordered triples
of nonnegative real numbers that lie in the plane
Let us say that
supports
when exactly two of the following are true:
Let
consist of those triples in
that support
The area of
divided by the area of
is
where
and
are relatively prime positive integers. Find
Solution
This problem just requires a good diagram and strong 3D visualization.
The region in where
is that of a little triangle on the bottom of the above diagram, of
is the triangle at the right, and
the triangle on the left, where the triangles are coplanar with the large equilateral triangle formed by
. We can check that each of the three regions mentioned fall under exactly two of the inequalities and not the third.
The side length of the large equilateral triangle is , which we can find using 45-45-90
with the axes. Using the formula
for equilateral triangles, the area of the large triangle is
. Since the lines of the smaller triangles are parallel to those of the large triangle, by corresponding angles we see that all of the triangles are similar, so they are all equilateral triangles. We can solve for their side lengths easily by subtraction, and we get
. Calculating their areas, we get
. The ratio
, and the answer is
.
To simplify the problem, we could used the fact that the area ratios are equal to the side ratios squared, and we get .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.