Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1997 AHSME Problems/Problem 15"
(Undo revision 214452 by Serengeti22 (talk))
(Tag: Undo)
 
(8 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Medians <math>BD</math> and <math>AE</math> of triangle <math>ABC</math> are perpendicular, <math>BD=8</math>, and <math>CE=12</math>.  The area of triangle <math>ABC</math> is  
+
Medians <math>BD</math> and <math>CE</math> of triangle <math>ABC</math> are perpendicular, <math>BD=8</math>, and <math>CE=12</math>.  The area of triangle <math>ABC</math> is  
  
 
<asy>
 
<asy>
Line 20: Line 20:
  
 
<math> \textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96 </math>
 
<math> \textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96 </math>
 +
 +
==Solution 1==
 +
 +
<asy>
 +
defaultpen(linewidth(.8pt));
 +
dotfactor=4;
 +
pair A = origin;
 +
pair B = (1.25,1);
 +
pair C = (2,0);
 +
pair D = midpoint(A--C);
 +
pair E = midpoint(A--B);
 +
pair F = midpoint(B--C);
 +
pair G = intersectionpoint(E--C,B--D);
 +
dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F);
 +
label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE);
 +
draw(A--B--C--cycle);
 +
draw(B--D);
 +
draw(E--C);
 +
draw(A--F);
 +
draw(rightanglemark(B,G,E,3));</asy>
 +
 +
One median divides a triangle into <math>2</math> equal areas, so all three medians will divide a triangle into <math>6</math> equal areas.
 +
 +
The median <math>CE</math> is divided into a <math>2:1</math> ratio at centroid <math>G</math>, so <math>GE = \frac{1}{3}\cdot CE = \frac{1}{3}\cdot 12 = 4</math>
 +
 +
Similarly, <math>BG = \frac{2}{3}\cdot 8 = \frac{16}{3}</math>
 +
 +
The area of the right triangle <math>\triangle BEG</math> is <math>\frac{1}{2}\cdot\frac{16}{3}\cdot 4</math>
 +
 +
The area of the whole figure is <math>6\cdot \frac{1}{2}\cdot\frac{16}{3}\cdot 4 = 64</math>, and the correct answer is <math>\boxed{D}</math>.
 +
 +
== Solution 2 ==
 +
Notice that if you were to draw in line ED, you would get an orthodiagonal quadrilateral with diagonals 8 and 12. The area is going to be equal to 48. Now we need to examine the triangle AED. If the area we are trying to find is denoted as A, we can tell that the area of AEC is A/2. The area of AED is going to be half of that since AD = DC so it would be A/4. This means that 48 is 3/4 of A, so naturally A is going to be 64. Giving <math>\boxed{D}</math>
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1997|num-b=10|num-a=12}}
+
{{AHSME box|year=1997|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 19:44, 15 February 2024

Problem

Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$, and $CE=12$. The area of triangle $ABC$ is

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]

$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$

Solution 1

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair F = midpoint(B--C); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(A--F); draw(rightanglemark(B,G,E,3));[/asy]

One median divides a triangle into $2$ equal areas, so all three medians will divide a triangle into $6$ equal areas.

The median $CE$ is divided into a $2:1$ ratio at centroid $G$, so $GE = \frac{1}{3}\cdot CE = \frac{1}{3}\cdot 12 = 4$

Similarly, $BG = \frac{2}{3}\cdot 8 = \frac{16}{3}$

The area of the right triangle $\triangle BEG$ is $\frac{1}{2}\cdot\frac{16}{3}\cdot 4$

The area of the whole figure is $6\cdot \frac{1}{2}\cdot\frac{16}{3}\cdot 4 = 64$, and the correct answer is $\boxed{D}$.

Solution 2

Notice that if you were to draw in line ED, you would get an orthodiagonal quadrilateral with diagonals 8 and 12. The area is going to be equal to 48. Now we need to examine the triangle AED. If the area we are trying to find is denoted as A, we can tell that the area of AEC is A/2. The area of AED is going to be half of that since AD = DC so it would be A/4. This means that 48 is 3/4 of A, so naturally A is going to be 64. Giving $\boxed{D}$

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AHSME_Problems/Problem_15&oldid=215559"