Difference between revisions of "2017 AMC 10B Problems/Problem 22"
Ishankhare (talk | contribs) |
Andrew2019 (talk | contribs) (→Solution 6 (Coordinate bashing)) |
||
(33 intermediate revisions by 20 users not shown) | |||
Line 4: | Line 4: | ||
<math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math> | <math>\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | Notice that <math>ADE</math> and <math>ABC</math> are right triangles. Then <math>AE = \sqrt{7^2+5^2} = \sqrt{74}</math>. <math>\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}</math>, so <math>BC = \frac{20}{\sqrt{74}}</math>. We also find that <math>AC = \frac{28}{\sqrt{74}}</math> (You can also use power of point ~MATHWIZARD2010), and thus the area of <math>ABC</math> is <math>\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}</math>. | ||
− | + | <center><asy> | |
+ | size(10cm); | ||
+ | pair A, B, C, D, E, O; | ||
+ | A = (-2,0); | ||
+ | B = (2,0); | ||
+ | C = (2*cos(1.24),2*sin(1.24)); | ||
+ | D = (5,0); | ||
+ | E = (5,5); | ||
+ | O = (A+B)/2; | ||
+ | dot(A); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(D); | ||
+ | dot(E); | ||
+ | dot(O); | ||
+ | draw(Circle((A+B)/2,2)); | ||
+ | draw(A--D--E--C--A); | ||
+ | draw(C--B); | ||
+ | draw(rightanglemark(A,C,B,5)); | ||
+ | draw(rightanglemark(A,D,E,5)); | ||
+ | label("$A$",A,W); | ||
+ | label("$B$",B,SE); | ||
+ | label("$D$",D,SE); | ||
+ | label("$E$",E,NE); | ||
+ | label("$C$",C,N); | ||
+ | label("$2$",(O+B)/2,S); | ||
+ | label("$3$",(B+D)/2,S); | ||
+ | label("$5$",(D+E)/2,NE); | ||
+ | </asy></center> | ||
+ | |||
+ | ==Solution 2 (Similar Triangles) == | ||
+ | We note that <math>\triangle ACB \sim \triangle ADE</math> by <math>AA</math> similarity. Also, since the area of <math>\triangle ADE = \frac{7 \cdot 5}2 = \frac{35}2</math> and <math>AE = \sqrt{74}</math>, <math>\frac{[ABC]}{[ADE]} = \frac{[ABC]}{\frac{35}2} = \left(\frac{4}{\sqrt{74}}\right)^2</math>, so the area of <math>\triangle ABC = \boxed{\textbf{(D) } \frac{140}{37}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | As stated before, note that <math>\triangle ACB</math> is similar to <math>\triangle ADE</math>. By similarity, we note that <math>\frac{\overline{AC}}{\overline{BC}}</math> is equivalent to <math>\frac{7}{5}</math>. We set <math>\overline{AC}</math> to <math>7x</math> and <math>\overline{BC}</math> to <math>5x</math>. By the Pythagorean Theorem, <math>(7x)^2+(5x)^2 = 4^2</math>. Combining, <math>49x^2+25x^2=16</math>. We can add and divide to get <math>x^2=\frac{8}{37}</math>. We square root and rearrange to get <math>x=\frac{2\sqrt{74}}{37}</math>. We know that the legs of the triangle are <math>7x</math> and <math>5x</math>. Multiplying <math>x</math> by <math>7</math> and <math>5</math> eventually gives us <math>\frac {14\sqrt{74}}{37}</math> and <math>\frac {10\sqrt{74}}{37}</math>. We divide this by <math>2</math>, since <math>\frac{1}{2}bh</math> is the formula for a triangle. This gives us <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let's call the center of the circle that segment <math>AB</math> is the diameter of, <math>O</math>. Note that <math>\triangle ODE</math> is an isosceles right triangle. Solving for side <math>OE</math>, using the Pythagorean theorem, we find it to be <math>5\sqrt{2}</math>. Calling the point where segment <math>OE</math> intersects circle <math>O</math>, the point <math>I</math>, segment <math>IE</math> would be <math>5\sqrt{2}-2</math>. Also, noting that <math>\triangle ADE</math> is a right triangle, we solve for side <math>AE</math>, using the Pythagorean Theorem, and get <math>\sqrt{74}</math>. Using Power of Point on point <math>E</math>, we can solve for <math>CE</math>. We can subtract <math>CE</math> from <math>AE</math> to find <math>AC</math> and then solve for <math>CB</math> using Pythagorean theorem once more. | ||
+ | |||
+ | <math>(AE)(CE)</math> = (Diameter of circle <math>O</math> + <math>IE</math>)<math>(IE)</math> <math>\rightarrow</math> <math>{\sqrt{74}}(CE)</math> = <math>(5\sqrt{2}+2)(5\sqrt{2}-2)</math> <math>\Rightarrow</math> <math>CE</math> = <math>\frac{23\sqrt{74}}{37}</math> | ||
+ | |||
+ | <math>AC = AE - CE</math> <math>\rightarrow</math> <math>AC</math> = <math>{\sqrt74}</math> - <math>\frac{23\sqrt{74}}{37}</math> <math>\Rightarrow</math> <math>AC</math> = <math>\frac{14\sqrt{74}}{37}</math> | ||
+ | |||
+ | Now to solve for <math>CB</math>: | ||
+ | |||
+ | <math>AB^2</math> - <math>AC^2</math> = <math>CB^2</math> <math>\rightarrow</math> <math>4^2</math> + <math>\frac{14\sqrt{74}}{37}^2</math> = <math>CB^2</math> <math>\Rightarrow</math> <math>CB</math> = <math>\frac{10\sqrt{74}}{37}</math> | ||
+ | |||
+ | Note that <math>\triangle ABC</math> is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases <math>AC</math> and <math>BC</math>, we get the area of triangle <math>ABC</math> to be <math>\boxed{\textbf{(D) } \frac{140}{37}}</math>. | ||
+ | |||
+ | == Solution 5 (Coordinate Geo)== | ||
+ | |||
+ | Drawing the picture, we realize that the equation for the line from A to E is <math>y=\frac{5x}{7}</math>, and the equation for the circle is <math>(x-2)^2+y^2=4</math> | ||
+ | plugging in <math>\frac{5x}{7}</math> for y we get <math>x(74x-196)=0</math> so <math>x=\frac{98}{37}</math>, that means <math>y = \frac{98}{37} \cdot \frac{5}{7} = \frac{70}{37}</math> | ||
+ | |||
+ | the height is <math>\frac{70}{37}</math> and the base is <math>4</math>, so the area is <math>\boxed{\textbf{(D) } \frac{140}{37}}</math> | ||
+ | |||
+ | -harsha12345 | ||
+ | |||
+ | == Solution 6 (Coordinate bashing) == | ||
+ | We draw out the diagram, and let the center of the circle be the origin. <math>A</math> would then be <math>(0,-2)</math>, and <math>B</math> would be <math>(2,0)</math>. We find that the equation of the line <math>AE</math> is <math>\frac{5}{7}x + \frac{10}{7}</math>. The equation of the circle is <math>x^2+y^2=4</math>. We use substitution and bashing with the quadratic formula to get <math>x = \frac{24}{37}</math>. From this, we get <math>y = \frac{70}{37}</math> and get that the area is <math>\frac{70}{37}\cdot 4/2 = \boxed{\textbf{(D) } \frac{140}{37}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2017|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:16, 13 May 2024
Contents
Problem
The diameter of a circle of radius is extended to a point outside the circle so that . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the area of ?
Solution 1
Notice that and are right triangles. Then . , so . We also find that (You can also use power of point ~MATHWIZARD2010), and thus the area of is .
Solution 2 (Similar Triangles)
We note that by similarity. Also, since the area of and , , so the area of .
Solution 3
As stated before, note that is similar to . By similarity, we note that is equivalent to . We set to and to . By the Pythagorean Theorem, . Combining, . We can add and divide to get . We square root and rearrange to get . We know that the legs of the triangle are and . Multiplying by and eventually gives us and . We divide this by , since is the formula for a triangle. This gives us .
Solution 4
Let's call the center of the circle that segment is the diameter of, . Note that is an isosceles right triangle. Solving for side , using the Pythagorean theorem, we find it to be . Calling the point where segment intersects circle , the point , segment would be . Also, noting that is a right triangle, we solve for side , using the Pythagorean Theorem, and get . Using Power of Point on point , we can solve for . We can subtract from to find and then solve for using Pythagorean theorem once more.
= (Diameter of circle + ) = =
= - =
Now to solve for :
- = + = =
Note that is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases and , we get the area of triangle to be .
Solution 5 (Coordinate Geo)
Drawing the picture, we realize that the equation for the line from A to E is , and the equation for the circle is plugging in for y we get so , that means
the height is and the base is , so the area is
-harsha12345
Solution 6 (Coordinate bashing)
We draw out the diagram, and let the center of the circle be the origin. would then be , and would be . We find that the equation of the line is . The equation of the circle is . We use substitution and bashing with the quadratic formula to get . From this, we get and get that the area is .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.