Difference between revisions of "2024 AMC 10A Problems/Problem 22"
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[[File:2024 AMC 10 A Diagram Problem 22.png]] | [[File:2024 AMC 10 A Diagram Problem 22.png]] | ||
− | First, we should find the length of <math>AB</math>. In order to do this, as we see in the diagram, it can be split into 4 equal sections. Since diagram <math>K</math> shows us that it is made up of two <math>{30,60,90}</math> triangles, then the triangle outlined in red must be a <math>30,60,90</math> triangle, as <math>{30+30=60}</math>, and the two lines are perpendicular (it is proveable, but during competition, it is best to assume this is true, as the diagram is draw pretty well to scale). Also, since we know the length of the longest side is <math>{\sqrt3}</math>, then the side we are looking for, which is outlined in blue, must be <math>\frac{3}{2}</math> by the <math>{1,\sqrt3, 2}</math> relationship of <math>{30,60,90}</math> triangles. Therefore <math>AB</math>, which is the base of the triangle we are looking, for must be <math>6.</math> | + | First, we should find the length of <math>AB</math>. In order to do this, as we see in the diagram, it can be split into 4 equal sections. Since diagram <math>K</math> shows us that it is made up of two <math>{30,60,90}</math> triangles, then the triangle outlined in red must be a <math>30,60,90</math> triangle, as <math>{30+30=60}</math>, and the two lines are perpendicular (it is proveable, but during competition, it is best to assume this is true, as the diagram is draw pretty well to scale). Also, since we know the length of the longest side of the red triangle is <math>{\sqrt3}</math>, then the side we are looking for, which is outlined in blue, must be <math>\frac{3}{2}</math> by the <math>{1,\sqrt3, 2}</math> relationship of <math>{30,60,90}</math> triangles. Therefore <math>AB</math>, which is the base of the triangle we are looking, for must be <math>6.</math> |
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~[[User:Mathkiddus|mathkiddus]] | ~[[User:Mathkiddus|mathkiddus]] | ||
==Solution 7(coordinate bash)== | ==Solution 7(coordinate bash)== | ||
− | Assume we set A as (0, 0). The line AB can be split into 4 equal sections, easily noticed in the picture. Not that the first equal section, starting at A. When we follow the line AB, we see that it's a right angle, noticing its a 30, 60, 90 triangle. Because the hypothenuse of that triangle is <math>\sqrt{3}</math>. Using the 30, 60, 90 ratio, we see that one of those equal sections is 3/2. Making the coordinate of B, <math>(6, 0)</math>. | + | Assume we set A as <math>(0, 0)</math>. The line AB can be split into 4 equal sections, easily noticed in the picture. Not that the first equal section, starting at A. When we follow the line AB, we see that it's a right angle, noticing its a 30, 60, 90 triangle. Because the hypothenuse of that triangle is <math>\sqrt{3}</math>. Using the 30, 60, 90 ratio, we see that one of those equal sections is 3/2. Making the coordinate of B, <math>(6, 0)</math>. |
− | Now, we must find the coordinate of C. Note that when we drop C down to the perpindicular of AB, there is a <math>x</math> component to C, and a <math>y</math> component to C. To find the X component, we see that it's one equal section found before as 3/2, and another section equal to 1, given in the question, making it 5/2. The Y component can be seen as another two sections. The two green sections marked in solution 1. One is <math>\sqrt{3}/2</math> due to the same 30 60 90 triangle we found in the first part of the solution. And the other part is <math>\sqrt{3}</math>, given in the question. | + | Now, we must find the coordinate of C. Note that when we drop C down to the perpindicular of AB, there is a <math>x</math> component to C, and a <math>y</math> component to C. To find the X component, we see that it's one equal section found before as 3/2, and another section equal to 1, given in the question, making it <math>x</math> = 5/2. The Y component can be seen as another two sections. The two green sections marked in solution 1. One is <math>\sqrt{3}/2</math> due to the same 30 60 90 triangle we found in the first part of the solution. And the other part is <math>\sqrt{3}</math>, given in the question. Making <math>y</math> = <math>\frac{3}{2} \sqrt3</math>. |
+ | So we see that C = (<math>5/2</math>, <math>\frac{3}{2} \sqrt3</math>) | ||
Using shoelace theorem, we get to get <math>\boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.</math> | Using shoelace theorem, we get to get <math>\boxed{ \textbf{(B) } \frac{9}{2} \sqrt3}.</math> | ||
− | sorry for no diagrams : | + | sorry for no diagrams D: |
-marcus | -marcus | ||
+ | |||
+ | == Video Solution by Power Solve == | ||
+ | https://www.youtube.com/watch?v=bxC_UENbmNk | ||
== Video Solution by Pi Academy (Fast and Easy⚡️🚀) == | == Video Solution by Pi Academy (Fast and Easy⚡️🚀) == | ||
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~i_am_suk_at_math_2 | ~i_am_suk_at_math_2 | ||
− | + | ==Video Solution by SpreadTheMathLove== | |
+ | https://www.youtube.com/watch?v=6SQ74nt3ynw | ||
+ | ==Video Solution by Just Math⚡== | ||
+ | https://www.youtube.com/watch?v=OjdLNjiO5Qw | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}} | {{AMC10 box|year=2024|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:02, 17 November 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6(trig bash)
- 8 Solution 7(coordinate bash)
- 9 Video Solution by Power Solve
- 10 Video Solution by Pi Academy (Fast and Easy⚡️🚀)
- 11 Video Solution by Innovative Minds
- 12 Video Solution by SpreadTheMathLove
- 13 Video Solution by Just Math⚡
- 14 See also
Problem
Let be the kite formed by joining two right triangles with legs and along a common hypotenuse. Eight copies of are used to form the polygon shown below. What is the area of triangle ?
Solution 1
First, we should find the length of . In order to do this, as we see in the diagram, it can be split into 4 equal sections. Since diagram shows us that it is made up of two triangles, then the triangle outlined in red must be a triangle, as , and the two lines are perpendicular (it is proveable, but during competition, it is best to assume this is true, as the diagram is draw pretty well to scale). Also, since we know the length of the longest side of the red triangle is , then the side we are looking for, which is outlined in blue, must be by the relationship of triangles. Therefore , which is the base of the triangle we are looking, for must be
Now all we have to do is find the height. We can split the height into 2 sections, the green and the light green. The green section must be , as shows us. Also, the light green section must be equal to , as in the previous paragraph, the triangle outlined in red is . Then, the green section, which is the height, must be , which is just .
Then the area of the triangle must be , which is just
~Solution by HappySharks ~Minor Edits by mathkiddus
Solution 2
Let be quadrilateral . Drawing line splits the triangle into . Drawing the altitude from to point on line , we know is , is , and is .
Due to the many similarities present, we can find that is , and the height of is
is and the height of is .
Solving for the area of gives which is
~9897 (latex beginner here)
~i_am_suk_at_math(very minor latex edits)
Solution 3
Let's start by looking at kite . We can quickly deduce based off of the side lengths that the kite can be split into two triangles. Going back to the triangle , focus on side . There are kites, they are all either reflected over the line or a line perpendicular to , meaning the length of can be split up into 4 equal parts.
Pick out the bottom-left kite, and we can observe that the kite and the triangle formed by the intersection of the kite and share a degree angle. (this was deduced from the triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a angle. Because that is also a triangle with a hypotenuse of , so we find the length of AB to be , which is .
Then, we can drop an altitude from to . We know that will be equivalent to the sum of the longer side of the kite and the shorter side of the triangle formed by the intersection of the kite and . (Look at the line formed on the left of that drops down to if you are confused) We already have those values from the triangles, so we can just plug it into the triangle area formula, . We get
~YTH (Need help with Latex and formatting)
~WIP (Header)
~Tacos_are_yummy_1 ( & Formatting)
Solution 4
~mathboy282
Solution 5
Let the point of intersection of and the kite with as vertex be .
Let the left kite with as a vertex touch the kite with as vertex at point .
is a so and .
So, and , and the area is
~Mintylemon66
Solution 6(trig bash)
As stated in the solutions above we can easily find that is split into equal parts, so we have We can calculate by using similarity to find is a triangle, therefore we have and finally therefore Similarly we have is congruent to therefore Next we have is congruent to telling us Noticing is right, we apply Pythagorean theorem to to find Next we would like to calculate CJ, As said before so We know the inscribed angle between CG and GJ is and finally we know So we apply LoC on triangle CGJ in order to find CJ. Now since we have all side lengths of we can find Applying LoC again on we have, We can solve for the area using the sin area formula which is To find we use the well known fact So we find, Finally to wrap up we can find the area of using the sin area formula, Therefore our answer is
Solution 7(coordinate bash)
Assume we set A as . The line AB can be split into 4 equal sections, easily noticed in the picture. Not that the first equal section, starting at A. When we follow the line AB, we see that it's a right angle, noticing its a 30, 60, 90 triangle. Because the hypothenuse of that triangle is . Using the 30, 60, 90 ratio, we see that one of those equal sections is 3/2. Making the coordinate of B, .
Now, we must find the coordinate of C. Note that when we drop C down to the perpindicular of AB, there is a component to C, and a component to C. To find the X component, we see that it's one equal section found before as 3/2, and another section equal to 1, given in the question, making it = 5/2. The Y component can be seen as another two sections. The two green sections marked in solution 1. One is due to the same 30 60 90 triangle we found in the first part of the solution. And the other part is , given in the question. Making = . So we see that C = (, )
Using shoelace theorem, we get to get
sorry for no diagrams D:
-marcus
Video Solution by Power Solve
https://www.youtube.com/watch?v=bxC_UENbmNk
Video Solution by Pi Academy (Fast and Easy⚡️🚀)
https://youtu.be/hcpej1uRYu4?si=-DekbEWkAg_c6nsg
Video Solution by Innovative Minds
https://www.youtube.com/watch?v=bhC58BB3kJA
~i_am_suk_at_math_2
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=6SQ74nt3ynw
Video Solution by Just Math⚡
https://www.youtube.com/watch?v=OjdLNjiO5Qw
See also
2024 AMC 10A (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.