Difference between revisions of "2002 AMC 12B Problems/Problem 20"
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\qquad\mathrm{(E)}\ 32</math> | \qquad\mathrm{(E)}\ 32</math> | ||
− | == Solution == | + | == Solution 1 == |
[[Image:2002_12B_AMC-20.png]] | [[Image:2002_12B_AMC-20.png]] | ||
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Alternatively, we could note that since we found <math>x^2 + y^2 = 169</math>, segment <math>MN=13</math>. Right triangles <math>\triangle MON</math> and <math>\triangle XOY</math> are similar by Leg-Leg with a ratio of <math>\frac{1}{2}</math>, so <math>XY=2(MN)=\boxed{\mathrm{(B)}\ 26}</math> | Alternatively, we could note that since we found <math>x^2 + y^2 = 169</math>, segment <math>MN=13</math>. Right triangles <math>\triangle MON</math> and <math>\triangle XOY</math> are similar by Leg-Leg with a ratio of <math>\frac{1}{2}</math>, so <math>XY=2(MN)=\boxed{\mathrm{(B)}\ 26}</math> | ||
+ | |||
+ | |||
+ | == Solution 2 == | ||
+ | Let <math>XO=x</math> and <math>YO=y.</math> Then, <math>XY=\sqrt{x^2+y^2}.</math> | ||
+ | |||
+ | Since <math>XN=19</math> and <math>YM=22,</math> | ||
+ | <cmath>XN^2=19^2=x^2+(\dfrac{y}{2})^2)=\dfrac{x^2}{4}+y^2</cmath> | ||
+ | <cmath>YM^2=22^2=(\dfrac{x}{2})^2+y^2=\dfrac{x^2}{4}+y^2.</cmath> | ||
+ | |||
+ | Adding these up: | ||
+ | <cmath>19^2+22^2=\dfrac{4x^2+y^2}{4}+\dfrac{x^2+4y^2}{4}</cmath> | ||
+ | <cmath>845=\dfrac{5x^2+5y^2}{4}</cmath> | ||
+ | <cmath>3380=5x^2+5y^2</cmath> | ||
+ | <cmath>676=x^2+y^2.</cmath> | ||
+ | |||
+ | Then, we substitute: <math>XY=\sqrt{x^2+y^2}=\sqrt{676}=\boxed{26}.</math> | ||
+ | |||
+ | |||
+ | == Solution 3 (Solution 1 but shorter) == | ||
+ | Refer to the diagram in solution 1. <math>4x^2+y^2=361</math> and <math>4y^2+x^2=484</math>, so add them: <math>5x^2+5y^2=845</math> and divide by 5: <math>x^2+y^2=169</math> so <math>\dfrac{XY}{2}=\sqrt{169}=13</math> and so <math>XY=26</math>, or answer <math>B</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | Use the diagram in solution 1. Get <math>4x^2+y^2=361</math> and <math>4y^2+x^2=484</math>, and multiply the second equation by 4 to get <math>4x^2+16y^2=1936</math> and then subtract the first from the second. Get <math>15y^2=1575</math> and <math>y^2=105</math>. Repeat for the other variable to get <math>15x^2=960</math> and <math>x^2=64</math>. Now XY is equal to the square root of four times these quantities, so <math>(105+64) \cdot 4=676</math>, and <math>XY=26</math> | ||
+ | |||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/BIyhEjVp0iM?t=174 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=7wj6RupkO90 ~David | ||
== See also == | == See also == |
Latest revision as of 15:02, 17 August 2023
- The following problem is from both the 2002 AMC 12B #20 and 2002 AMC 10B #22, so both problems redirect to this page.
Contents
Problem
Let be a right-angled triangle with
. Let
and
be the midpoints of legs
and
, respectively. Given that
and
, find
.
Solution 1
Let ,
. By the Pythagorean Theorem on
respectively,
Summing these gives .
By the Pythagorean Theorem again, we have
Alternatively, we could note that since we found , segment
. Right triangles
and
are similar by Leg-Leg with a ratio of
, so
Solution 2
Let and
Then,
Since and
Adding these up:
Then, we substitute:
Solution 3 (Solution 1 but shorter)
Refer to the diagram in solution 1. and
, so add them:
and divide by 5:
so
and so
, or answer
.
Solution 4
Use the diagram in solution 1. Get and
, and multiply the second equation by 4 to get
and then subtract the first from the second. Get
and
. Repeat for the other variable to get
and
. Now XY is equal to the square root of four times these quantities, so
, and
Video Solution by OmegaLearn
https://youtu.be/BIyhEjVp0iM?t=174
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=7wj6RupkO90 ~David
See also
2002 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 |
2002 AMC 12B (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.