Difference between revisions of "2018 AMC 10A Problems/Problem 10"
m (→Solution 4 (Geometric Interpretation)) |
|||
Line 37: | Line 37: | ||
===Solution 4 (Geometric Interpretation)=== | ===Solution 4 (Geometric Interpretation)=== | ||
− | Draw a right triangle <math>ABC</math> with a hypotenuse <math>AC</math> of length <math>7</math> and leg <math>AB</math> of length <math>x</math>. Draw <math>D</math> on <math>BC</math> such that <math>AD=5</math>. Note that <math>BC=\sqrt{49-x^2}</math> and <math>BD=\sqrt{25-x^2}</math>. Thus, from the given equation, <math>BC-BD=DC=3</math>. Using Law of Cosines on triangle <math>ADC</math>, we see that <math>\angle{ADC}=120^{\circ}</math> so <math>\angle{ADB}=60^{\circ}</math>. Since <math>ADB</math> is a <math>30-60-90</math> triangle, <math>\sqrt{25-x^2}=BD=\frac{5}{2}</math> and <math>\sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}</math>. Finally, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}</math> | + | Draw a right triangle <math>ABC</math> with a hypotenuse <math>AC</math> of length <math>7</math> and leg <math>AB</math> of length <math>x</math>. Draw <math>D</math> on <math>BC</math> such that <math>AD=5</math>. Note that <math>BC=\sqrt{49-x^2}</math> and <math>BD=\sqrt{25-x^2}</math>. Thus, from the given equation, <math>BC-BD=DC=3</math>. Using Law of Cosines on triangle <math>ADC</math>, we see that <math>\angle{ADC}=120^{\circ}</math> so <math>\angle{ADB}=60^{\circ}</math>. Since <math>ADB</math> is a <math>30-60-90</math> triangle, <math>\sqrt{25-x^2}=BD=\frac{5}{2}</math> and <math>\sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}</math>. Finally, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}</math>. |
+ | <asy> | ||
+ | var s = sqrt(3); | ||
+ | pair A = (-5*s/2, 0); | ||
+ | pair B = (0,0); | ||
+ | pair C = (0,5.5); | ||
+ | pair D = (0,2.5); | ||
+ | |||
+ | draw(A--B--C--A--D); | ||
+ | rightanglemark(A, B, D); | ||
+ | label("A", A, SW); | ||
+ | label("B", B, SE); | ||
+ | label("C", C, NE); | ||
+ | label("D", D, E); | ||
+ | label("7", (-5*s/4, 5.5/2), NW); | ||
+ | label("120$^\circ$", D, NW); | ||
+ | label("60$^\circ$", (0,2), SW); | ||
+ | label("$x$", 0.5*A, S); | ||
+ | draw(rightanglemark(A, B, C)); | ||
+ | |||
+ | draw(anglemark(A, D, B)); | ||
+ | markscalefactor = 0.04; | ||
+ | draw(anglemark(C, D, A)); | ||
+ | |||
+ | label("$\frac{5}{2}$", (0,1.25), E); | ||
+ | label("3", (0,4), E); | ||
+ | label("5", (-5*s/4, 5/4), N); | ||
+ | </asy> | ||
{{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}} | {{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}} |
Revision as of 10:44, 15 February 2018
Contents
Problem
Suppose that real number satisfies What is the value of ?
Solutions
Solution 1
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The terms cancel nicely.
Given that = 3,
Solution by PancakeMonster2004, explanations added by a1b2.
Solution 2
Let , and let . Then . Substituting, we get . Rearranging, we get . Squaring both sides and solving, we get and . Adding, we get that the answer is
Solution 3
Put the equations to one side. can be changed into .
We can square both sides, getting us
That simplifies out to Dividing both sides by 6 gets us .
Following that, we can square both sides again, resulting in the equation . Simplifying that, we get .
Substituting into the equation , we get . Immediately, we simplify into . The two numbers inside the square roots are simplified to be and , so you add them up:
Solution 4 (Geometric Interpretation)
Draw a right triangle with a hypotenuse of length and leg of length . Draw on such that . Note that and . Thus, from the given equation, . Using Law of Cosines on triangle , we see that so . Since is a triangle, and . Finally, .
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |