Difference between revisions of "2018 AIME II Problems/Problem 12"
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==Solution 8 (Simple Geometry)== | ==Solution 8 (Simple Geometry)== | ||
− | [[File: | + | [[File:AIME-II-2018-12.png|400px|right]] |
<math>BP = PD</math> as in another solutions. | <math>BP = PD</math> as in another solutions. | ||
− | Let <math>D'</math> be the | + | Let <math>D'</math> be the reflection of <math>D</math> across <math>C</math>. |
− | Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars from <math>D,D' | + | Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars on <math>AC</math> from <math>D,D'</math>, and <math>B</math> respectively. |
− | <cmath>AB = CD = CD', BH = DE = D'E' \ | + | <cmath>\begin{align*} |
− | + | &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ | |
− | + | \Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\ | |
+ | \Rightarrow &\qquad \angle BAC = \angle ACD' \ | ||
+ | \Rightarrow &\qquad \triangle ABC = \triangle AD'C \\ | ||
+ | \Rightarrow &\qquad BC = AD'. | ||
+ | \end{align*}</cmath> | ||
The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>. | The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>. | ||
The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath> | The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath> |
Revision as of 22:01, 26 December 2022
Contents
[hide]Problem
Let be a convex quadrilateral with , , and . Assume that the diagonals of intersect at point , and that the sum of the areas of triangles and equals the sum of the areas of triangles and . Find the area of quadrilateral .
Diagram
Let and let . Let and let .
Solution 1
Let and let . Let and let . We easily get and .
We are given that , which we can now write as Either or . The former would imply that is a parallelogram, which it isn't; therefore we conclude and is the midpoint of . Let and . Then . On one hand, since , we have whereas, on the other hand, using cosine formula to get the length of , we get Eliminating in the above two equations and solving for we getwhich finally yields .
Solution 2
For reference, , so is the longest of the four sides of . Let be the length of the altitude from to , and let be the length of the altitude from to . Then, the triangle area equation becomes
What an important finding! Note that the opposite sides and have equal length, and note that diagonal bisects diagonal . This is very similar to what happens if were a parallelogram with , so let's extend to point , such that is a parallelogram. In other words, and Now, let's examine . Since , the triangle is isosceles, and . Note that in parallelogram , and are congruent, so and thus Define , so .
We use the Law of Cosines on and :
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot on gives and therefore . Seeing that , we conclude that is a 3-4-5 right triangle, so . Then, the area of is . Since , points and are equidistant from , so and hence -kgator
Just to be complete -- and can actually be equal. In this case, , but must be equal to . We get the same result. -Mathdummy.
Solution 3 (Another way to get the middle point)
So, let the area of triangles , , , . Suppose and , then it is easy to show that Also, because we will have So So So So As a result, Then, we have Combine the condition we can find out that so is the midpoint of
~Solution by (Frank FYC)
Solution 4 (With yet another way to get the middle point)
Denote by . Then . Using the formula for the area of a triangle, we get so Hence (note that makes no difference here). Now, assume that , , and . Using the cosine rule for and , it is clear that or Likewise, using the cosine rule for triangles and , It follows that Since , which simplifies to Plugging this back to equations , , and , it can be solved that . Then, the area of the quadrilateral is --Solution by MicGu
Solution 5
As in all other solutions, we can first find that either or , but it's an AIME problem, we can take , and assume the other choice will lead to the same result (which is true).
From , we have , and , therefore, By Law of Cosines, Square and , and add them, to get Solve, , -Mathdummy
Solution 6
Either or . Let . Applying Stewart's Theorem on and , dividing by and rearranging, Applying Stewart on and , Substituting equations 1 and 2 into 3 and rearranging, . By Law of Cosines on , so . Using to find unknown areas, .
-Solution by Gart
Solution 7
Now we prove P is the midpoint of . Denote the height from to as , height from to as .According to the problem, implies . Then according to basic congruent triangles we get Firstly, denote that . Applying Stewart theorem, getting that , denote Applying Stewart Theorem, getting solve for a, getting Now everything is clear, we can find using LOC, , the whole area is
~bluesoul
Solution 8 (Simple Geometry)
as in another solutions.
Let be the reflection of across . Let points and be the foot of perpendiculars on from , and respectively. The area of quadrilateral is equal to the area of triangle with sides . The semiperimeter is the area
vladimir.shelomovskii@gmail.com, vvsss
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.