Difference between revisions of "2018 AMC 12A Problems/Problem 20"
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== Solution 3 (Elementary) == | == Solution 3 (Elementary) == | ||
− | Like above, notice that <math>\triangle{EMI}</math> is isosceles and right, which means that <math>\dfrac{ME \cdot MI}{2} = 2</math>, so <math>MI^2=4</math> and <math>MI = 2</math>. Then construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math> as well as <math>\overline{MI}</math>. It's clear that <math>MG^2+GI^2 = MI^2</math> by Pythagorean, so knowing that <math>MG = \dfrac{AB}{2} = \dfrac{3}{2}</math> allows one to solve to get <math>GI = \dfrac{\sqrt{7}}{2}</math>. By just looking at the diagram, <math>CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}</math>. The answer is thus <math>3+7+2=\textbf{(D) }12</math>. | + | Like above, notice that <math>\triangle{EMI}</math> is isosceles and right, which means that <math>\dfrac{ME \cdot MI}{2} = 2</math>, so <math>MI^2=4</math> and <math>MI = 2</math>. Then construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math> as well as <math>\overline{MI}</math>. It's clear that <math>MG^2+GI^2 = MI^2</math> by Pythagorean, so knowing that <math>MG = \dfrac{AB}{2} = \dfrac{3}{2}</math> allows one to solve to get <math>GI = \dfrac{\sqrt{7}}{2}</math>. By just looking at the diagram, <math>CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}</math>. The answer is thus <math>3+7+2=\boxed{\textbf{(D) }12}</math>. |
== Solution 4 (Coordinate Geometry) == | == Solution 4 (Coordinate Geometry) == |
Revision as of 23:40, 17 August 2021
Contents
Problem
Triangle is an isosceles right triangle with . Let be the midpoint of hypotenuse . Points and lie on sides and , respectively, so that and is a cyclic quadrilateral. Given that triangle has area , the length can be written as , where , , and are positive integers and is not divisible by the square of any prime. What is the value of ?
Diagram
Solution 1
Observe that is isosceles right ( is the midpoint of diameter arc since ), so . With , we can use Law of Cosines to determine that . The same calculations hold for also, and since , we deduce that is the smaller root, giving the answer of .
Solution 2 (Ptolemy)
We first claim that is isosceles and right.
Proof: Construct and . Since bisects , one can deduce that . Then by AAS it is clear that and therefore is isosceles. Since quadrilateral is cyclic, one can deduce that . Q.E.D.
Since the area of is 2, we can find that ,
Since is the mid-point of , it is clear that .
Now let and . By Ptolemy's Theorem, in cyclic quadrilateral , we have . By Pythagorean Theorem, we have . One can solve the simultaneous system and find . Then by deducting the length of from 3 we get , giving the answer of . (Surefire2019)
Solution 3 (Elementary)
Like above, notice that is isosceles and right, which means that , so and . Then construct and as well as . It's clear that by Pythagorean, so knowing that allows one to solve to get . By just looking at the diagram, . The answer is thus .
Solution 4 (Coordinate Geometry)
Let lie on , on , on , and on . Since is cyclic, (which is opposite of another right angle) must be a right angle; therefore, . Compute the dot product to arrive at the relation . We can set up another equation involving the area of using the Shoelace Theorem. This is . Multiplying, substituting for , and simplifying, we get . Thus, . But , meaning , and the final answer is .
Solution 5 (Quick)
From cyclic we get and , so is an isosceles right triangle.
From we get .
Notice , because , , and .
Let , so .
By Pythagoras on we have , and solve this to get for a final answer of .
Solution 6 (Bash)
Let , . Because opposite angles in a cyclic quadrilateral are supplementary, we have . By the law of cosines, we have , and . Notice that , where is the origin of the circle mentioned in the problem. Thus . By the Pythagorean Theorem, we have . By the Pythagorean Theorem, we have . Thus we have . We know that . We take the smaller solution because we have , and we want , not , thus . Thus our final answer is .
-vsamc
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=4465
~ pi_is_3.14
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.