Difference between revisions of "2000 AMC 10 Problems/Problem 7"
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| − | <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== | ||
<asy> | <asy> | ||
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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 53: | Line 56: | ||
<math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>. | <math>BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}</math>. | ||
| − | Adding, we get <math>2+\frac{4\sqrt{3}}{3}</math> | + | Adding, we get <math>\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}</math>. |
| − | <math>\boxed{\ | + | == 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== | ||
| Line 61: | Line 73: | ||
{{AMC10 box|year=2000|num-b=6|num-a=8}} | {{AMC10 box|year=2000|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 23:40, 14 July 2024
Problem
In rectangle 
, 
, 
is on 
, and 
and 
trisect 
. What is the perimeter of 
?
![[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); [/asy]](http://latex.artofproblemsolving.com/1/0/a/10ad0d6c91539b40c085057a166894ac1cede67f.png)
Solution
![[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("$A$",(0,2),NW); label("$B$",(3.4,2),NE); label("$C$",(3.4,0),SE); label("$D$",(0,0),SW); label("$P$",(1.3,2),N); label("$1$",(0,1),W); label("$2$",(1.7,1),SE); label("$\frac{\sqrt{3}}{3}$",(0.65,2),N); label("$\frac{2\sqrt{3}}{3}$",(0.85,1),NW); label("$\frac{2\sqrt{3}}{3}$",(2.35,2),N); label("$\sqrt{3}$",(1.7,0),S); label("$2$",(3,1),W); [/asy]](http://latex.artofproblemsolving.com/7/f/2/7f2d07bc097212bf747f3f762ed1cc6b8fcb6214.png)
.
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. 
