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 "2000 AIME II Problems/Problem 6"

m (whoops)
(→‎Solution)
 
(5 intermediate revisions by 5 users not shown)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
Let the shorter base have length <math>b</math> (so the longer has length <math>b+100</math>), and let the height be <math>h</math>. The length of the midline of the trapezoid is the average of its bases, which is <math>\frac{b+b+100}{2} = b+50</math>. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height <math>h</math>. Then,
+
Let the shorter base have length <math>b</math> (so the longer has length <math>b+100</math>), and let the height be <math>h</math>. The length of the midline of the trapezoid is the average of its bases, which is <math>\frac{b+b+100}{2} = b+50</math>. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height <math>h/2</math>. Then,
  
 
<center><asy>pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
 
<center><asy>pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
Line 14: Line 14:
 
</cmath>
 
</cmath>
  
−
We now construct the line which divides the rectangle into two regions of equal area. Suppose this line is a distance of <math>h_1</math> from the shorter base. By [[similar triangle]]s, we have <math>\frac{x - 75}{100} = \frac{h_1}{h}</math>. <font style="font-size:85%">Indeed, construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths <math>x_1, 75, x_2</math>. By similar triangles, we easily find that <math>\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}</math>, as desired.</font>
+
Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths <math>x_1, 75, x_2</math>. By similar triangles, we easily find that <math>\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}</math>.
  
 
<center><asy>pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
 
<center><asy>pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7);
Line 27: Line 27:
 
Substituting our expression for <math>\frac h{h_1}</math> from above, we find that
 
Substituting our expression for <math>\frac h{h_1}</math> from above, we find that
  
−
<cmath>x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 6125 + 12500 - 75x \Longrightarrow x^2 = 18125</cmath>
+
<cmath>x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125</cmath>
  
 
The answer is <math>\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}</math>.
 
The answer is <math>\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}</math>.
−
 
  
 
== See also ==
 
== See also ==
Line 36: Line 35:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:42, 20 December 2021

Problem

One base of a trapezoid is $100$ units longer than the other base. The segment that joins the midpoints of the legs divides the trapezoid into two regions whose areas are in the ratio $2: 3$. Let $x$ be the length of the segment joining the legs of the trapezoid that is parallel to the bases and that divides the trapezoid into two regions of equal area. Find the greatest integer that does not exceed $x^2/100$.

Solution

Let the shorter base have length $b$ (so the longer has length $b+100$), and let the height be $h$. The length of the midline of the trapezoid is the average of its bases, which is $\frac{b+b+100}{2} = b+50$. The two regions which the midline divides the trapezoid into are two smaller trapezoids, both with height $h/2$. Then,

[asy]pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7); pair A=(0,0),B=(175,0),C=(105,100),D=(30,100); D(A--B--C--D--cycle); D((A+D)/2 -- (B+C)/2, d); MP("b",(C+D)/2,N);MP("b+100",(A+B)/2); [/asy]

\[\frac{\frac 12 (h/2) (b + b+50)}{\frac 12 (h/2) (b + 50 + b + 100)} = \frac{2}{3} \Longrightarrow \frac{b + 75}{b + 25} = \frac 32 \Longrightarrow b = 75\]

Construct the perpendiculars from the vertices of the shorter base to the longer base. This splits the trapezoid into a rectangle and two triangles; it also splits the desired line segment into three partitions with lengths $x_1, 75, x_2$. By similar triangles, we easily find that $\frac{x - 75}{100} = \frac{x_1+x_2}{100} = \frac{h_1}{h}$.

[asy]pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7); pair A=(0,0),B=(175,0),C=(105,100),D=(30,100),E=D*(1.75-(18125)^.5/100),F=IP(B--C,E--(175,E.y)); D(A--B--C--D--cycle); MP("75",(C+D)/2,N);MP("175",(A+B)/2); D(C--(C.x,0),d);D(D--(D.x,0),d); D(E--F,d); D((-20,100)--(-20,0)); MP("h",(-20,50),(-1,0));MP("h_1",(C.x,(C.y+E.y)/2),(-1,0)); MP("x_1",((E.x+D.x)/2,E.y));MP("x_2",((F.x+C.x)/2,E.y)); [/asy]

The area of the region including the shorter base must be half of the area of the entire trapezoid, so

\[2 \cdot \frac 12 h_1 (75 + x) = \frac 12 h (75 + 175) \Longrightarrow x = 125 \cdot \frac{h}{h_1} - 75\]

Substituting our expression for $\frac h{h_1}$ from above, we find that

\[x = \frac{12500}{x-75} - 75 \Longrightarrow x^2 - 75x = 5625 + 12500 - 75x \Longrightarrow x^2 = 18125\]

The answer is $\left\lfloor\frac{x^2}{100}\right\rfloor = \boxed{181}$.

See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2000_AIME_II_Problems/Problem_6&oldid=168160"