Difference between revisions of "2022 AIME I Problems/Problem 15"
Oxymoronic15 (talk | contribs) (→Solution 2 (pure algebraic trig, easy to follow)) |
Oxymoronic15 (talk | contribs) (→Solution 2 (pure algebraic trig, easy to follow)) |
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==Solution 2 (pure algebraic trig, easy to follow)== | ==Solution 2 (pure algebraic trig, easy to follow)== | ||
+ | (This eventually whittles down to the same concept as Solution 1) | ||
Note that in each equation in this system, it is possible to factor <math>\sqrt{x}</math>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math> from each term (on the left sides), since each of <math>x</math>, <math>y</math>, and <math>z</math> are positive real numbers. After factoring out accordingly from each terms one of <math>\sqrt{x}</math>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math>, the system should look like this: | Note that in each equation in this system, it is possible to factor <math>\sqrt{x}</math>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math> from each term (on the left sides), since each of <math>x</math>, <math>y</math>, and <math>z</math> are positive real numbers. After factoring out accordingly from each terms one of <math>\sqrt{x}</math>, <math>\sqrt{y}</math>, or <math>\sqrt{z}</math>, the system should look like this: |
Revision as of 20:53, 17 February 2022
Problem
Let and be positive real numbers satisfying the system of equations: Then can be written as where and are relatively prime positive integers. Find
Solution 1 (geometric interpretation)
First, we note that we can let a triangle exist with side lengths , , and opposite altitude . This shows that the third side, which is the nasty square-rooted sum, is going to have the length equal to the sum on the right - let this be for symmetry purposes. So, we note that if the angle opposite the side with length has a value of , then the altitude has length and thus so and the triangle side with length is equal to .
We can symmetrically apply this to the two other triangles, and since by law of sines, we have is the circumradius of that triangle. Hence. we calculate that with , and , the angles from the third side with respect to the circumcenter are , and . This means that by half angle arcs, we see that we have in some order, , , and (not necessarily this order, but here it does not matter due to symmetry), satisfying that , , and . Solving, we get , , and .
We notice that
- kevinmathz
Solution 2 (pure algebraic trig, easy to follow)
(This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor , , or from each term (on the left sides), since each of , , and are positive real numbers. After factoring out accordingly from each terms one of , , or , the system should look like this: This should give off tons of trigonometry vibes. To make the connection clear, , , and is a helpful substitution: From each equation can be factored out, and when every equation is divided by 2, we get: which simplifies to (using the Pythagorean identity ): which further simplifies to (using sine addition formula ): Without loss of generality, taking the inverse sine of each equation yields a simple system: giving solutions , , . Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: , , and . When plugging into the expression , noting that helps to simplify this expression into:
Now, all the cosines in here are fairly standard: , , and . With some final calculations: This is our answer in simplest form , so
-Oxymoronic15