Difference between revisions of "1999 AMC 8 Problems/Problem 23"

Latest revision as of 04:33, 12 March 2024

Problem

Square $ABCD$ has sides of length 3. Segments $CM$ and $CN$ divide the square's area into three equal parts. How long is segment $CM$ ?

$[asy] pair A,B,C,D,M,N; A = (0,0); B = (0,3); C = (3,3); D = (3,0); M = (0,1); N = (1,0); draw(A--B--C--D--cycle); draw(M--C--N); label("$A$",A,SW); label("$M$",M,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,SE); label("$N$",N,S); [/asy]$

$\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}$

Solution

Since the square has side length $3$ , the area of the entire square is $9$ .

The segments divide the square into 3 equal parts, so the area of each part is $9 \div 3 = 3$ .

Since $\triangle CBM$ has area $3$ and base $CB = 3$ , using the area formula for a triangle:

$A_{tri} = \frac{1}{2}bh$

$3 = \frac{1}{2}3h$

$h = 2$

Thus, height $BM = 2$ .

Since $\triangle CBM$ is a right triangle, $CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}$ .

Solution 2

Connect $AC$ , $S_\triangle AMC=S_\triangle ANC$ . To satisfied the three area is equal, we have $2S_\triangle AMC=S_\triangle BMC$ , $2S_\triangle ANC=S_\triangle DNC$ . Thus, $AM=AN=\frac{1}{2}BM=\frac{1}{2}AB=1$ . $BM=2,BC=3,MC=\boxed{\text{(C)}\ \sqrt{13}}$ .

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=AH5_Gol2GCM

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-Since the squares have side length <math>3</math>, the area of the entire square is <math>9</math>.
+==Problem==
+Square <math>ABCD</math> has sides of length 3. Segments <math>CM</math> and <math>CN</math> divide the square's area into three equal parts. How long is segment <math>CM</math>?
-The segments divide the square into 3 equal parts, so the area of each part is <math>3</math>.
+<asy>
+pair A,B,C,D,M,N;
+A = (0,0);
+B = (0,3);
+C = (3,3);
+D = (3,0);
+M = (0,1);
+N = (1,0);
+draw(A--B--C--D--cycle);
+draw(M--C--N);
+label("$A$",A,SW);
+label("$M$",M,W);
+label("$B$",B,NW);
+label("$C$",C,NE);
+label("$D$",D,SE);
+label("$N$",N,S);
+</asy>
-The base of a triangle is <math>3</math>, so the height must be <math>2</math>.
+<math>\text{(A)}\ \sqrt{10} \qquad \text{(B)}\ \sqrt{12} \qquad \text{(C)}\ \sqrt{13} \qquad \text{(D)}\ \sqrt{14} \qquad \text{(E)}\ \sqrt{15}</math>
-<math>3-2=1</math>.
+==Solution==
+Since the square has side length <math>3</math>, the area of the entire square is <math>9</math>.
-<math>CM=\sqrt {3^2+2^2</math>} = <math>\sqrt 13</math>
+The segments divide the square into 3 equal parts, so the area of each part is <math>9 \div 3 = 3</math>.
+Since <math>\triangle CBM</math> has area <math>3</math> and base <math>CB = 3</math>, using the area formula for a triangle:
+<math>A_{tri} = \frac{1}{2}bh</math>
+<math>3 = \frac{1}{2}3h</math>
+<math>h = 2</math>
+Thus, height <math>BM = 2</math>.
+Since <math>\triangle CBM</math> is a right triangle, <math>CM = \sqrt{BM^2 + BC^2} = \sqrt{2^2 + 3^2} = \boxed{\text{(C)}\ \sqrt{13}}</math>.
+==Solution 2==
+Connect <math>AC</math>, <math>S_\triangle AMC=S_\triangle ANC</math>. To satisfied the three area is equal, we have <math>2S_\triangle AMC=S_\triangle BMC</math>, <math>2S_\triangle ANC=S_\triangle DNC</math>. Thus, <math>AM=AN=\frac{1}{2}BM=\frac{1}{2}AB=1</math>. <math>BM=2,BC=3,MC=\boxed{\text{(C)}\ \sqrt{13}}</math>.
+== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
+https://www.youtube.com/watch?v=AH5_Gol2GCM
+==See Also==
+{{AMC8 box|year=1999|num-b=22|num-a=24}}
+{{MAA Notice}}

Page

Toolbox

Search