Difference between revisions of "2000 AMC 10 Problems/Problem 7"
m (→Solution) |
|||
(3 intermediate revisions by 3 users not shown) | |||
Line 19: | Line 19: | ||
</asy> | </asy> | ||
− | <math>\ | + | <math>\textbf{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\textbf{(C)}\ 2+2\sqrt{2} \qquad\textbf{(D)}\ \frac{3+3\sqrt{5}}{2} \qquad\textbf{(E)}\ 2+\frac{5\sqrt{3}}{3}</math> |
==Solution== | ==Solution== | ||
Line 42: | Line 42: | ||
label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW); | label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW); | ||
label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N); | label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N); | ||
+ | label("$\sqrt{3}$",(1.7,0),S); | ||
+ | label("$2$",(3,1),W); | ||
</asy> | </asy> | ||
Line 55: | Line 57: | ||
Adding, we get <math>\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}</math>. | Adding, we get <math>\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | After computing <math>\overline{BP} = \frac{2\sqrt{3}}{3},</math> observe that triangle <math>\triangle BPD</math> is isosceles with <math>\angle DPB = \angle BPD.</math> Therefore, using <math>120 - 30 - 30</math> triangle properties, we see that the perimeter is just <math>(2+ \sqrt{3}) \cdot \frac{2\sqrt{3}}{3} = \boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}.</math> | ||
+ | |||
+ | ~Sliced_Bread | ||
+ | |||
+ | ==Video Solution by Daily Dose of Math== | ||
+ | |||
+ | https://youtu.be/wITXxUtZj3E?si=KuLEI9SeOFZTw05Y | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
==See Also== | ==See Also== |
Latest revision as of 00:40, 15 July 2024
Problem
In rectangle ,
,
is on
, and
and
trisect
. What is the perimeter of
?
Solution
.
Since is trisected,
.
Thus,
.
Adding, we get .
Solution 2
After computing observe that triangle
is isosceles with
Therefore, using
triangle properties, we see that the perimeter is just
~Sliced_Bread
Video Solution by Daily Dose of Math
https://youtu.be/wITXxUtZj3E?si=KuLEI9SeOFZTw05Y
~Thesmartgreekmathdude
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.