Difference between revisions of "2022 AIME I Problems/Problem 15"
Oxymoronic15 (talk | contribs) (→Solution 2 (pure algebraic trig, easy to follow)) |
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This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}.</math> | This is our answer in simplest form <math>\frac{m}{n}</math>, so <math>m + n = 1 + 32 = \boxed{033}.</math> | ||
-Oxymoronic15 | -Oxymoronic15 | ||
+ | |||
+ | ==solution 3== | ||
+ | Let <math>1-x=a;1-y=b;1-z=c</math>, rewrite those equations | ||
+ | |||
+ | <math>\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1</math>; | ||
+ | |||
+ | <math>\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}</math> | ||
+ | |||
+ | <math>\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}</math> | ||
+ | |||
+ | square both sides, get three equations: | ||
+ | |||
+ | <math>2ab-1=2\sqrt{(1-a^2)(1-b^2)}</math> | ||
+ | |||
+ | <math>2bc=2\sqrt{(1-b^2)(1-c^2)}</math> | ||
+ | |||
+ | <math>2ac+1=2\sqrt{(1-c^2)(1-a^2)}</math> | ||
+ | |||
+ | Getting that <math>a^2+b^2-ab=\frac{3}{4}</math> | ||
+ | |||
+ | <math>b^2+c^2=1</math> | ||
+ | |||
+ | <math>a^2+c^2+ac=\frac{3}{4}</math> | ||
+ | |||
+ | Subtract first and third equation, getting <math>(b+c)(b-c)=a(b+c)</math>, <math>a=b-c</math> | ||
+ | |||
+ | Put it in first equation, getting <math>b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}</math>, <math>bc=\frac{1}{4}</math> | ||
+ | |||
+ | Since <math>a^2=b^2+c^2-2bc=\frac{1}{2}</math>, the final answer is <math>\frac{1}{4}*\frac{1}{4}*\frac{1}{2}=\frac{1}{32}</math> the final answer is <math>\boxed{033}</math> |
Revision as of 21:58, 17 February 2022
Contents
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
solution 3
Let , rewrite those equations
;
square both sides, get three equations:
Getting that
Subtract first and third equation, getting ,
Put it in first equation, getting ,
Since , the final answer is the final answer is