Difference between revisions of "2002 AMC 12B Problems/Problem 20"
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Erics son07 (talk | contribs) (→Solution 3 (Solution 1 but shorter)) |
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== Solution 3 (Solution 1 but shorter) == | == Solution 3 (Solution 1 but shorter) == | ||
− | Refer to the diagram in solution 1. <math>4x^2+y^2= | + | 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> | ||
== See also == | == See also == |
Revision as of 04:41, 3 September 2021
- 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
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.