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 "1996 AIME Problems/Problem 13"

(→‎Problem)
(→‎Solution 2)
 
(8 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
In triangle <math>ABC</math>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</math> bisects <math>\overline{BC}</math>, and <math>\angle ADB</math> is a right angle. The ratio
+
In [[triangle]] <math>ABC</math>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for which <math>\overline{AD}</math> [[bisect]]s <math>\overline{BC}</math>, and <math>\angle ADB</math> is a right angle. The ratio <math>\frac{[ADB]}{[ABC]}</math> can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
−
<cmath>\dfrac{Area(\triangle ADB)}{Area(\triangle ABC)}</cmath>
+
== Solution ==
 +
<center><asy>
 +
pointpen = black; pathpen = black + linewidth(0.7);
 +
pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E);
 +
D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B); D(MP("E",E));
 +
MP("\sqrt{30}",(A+B)/2,NW); MP("\sqrt{6}",(A+C)/2,SE); MP("\frac{\sqrt{15}}2",(E+C)/2); D(rightanglemark(B,D,A));
 +
</asy></center>
 +
Let <math>E</math> be the midpoint of <math>\overline{BC}</math>. Since <math>BE = EC</math>, then <math>\triangle ABE</math> and <math>\triangle AEC</math> share the same height and have equal bases, and thus have the same area. Similarly, <math>\triangle BDE</math> and <math>BAE</math> share the same height, and have bases in the ratio <math>DE : AE</math>, so <math>\frac{[BDE]}{[BAE]} = \frac{DE}{AE}</math> (see [[area ratios]]). Now,
 +
 
 +
<center><math>\dfrac{[ADB]}{[ABC]} = \frac{[ABE] + [BDE]}{2[ABE]} = \frac{1}{2} + \frac{DE}{2AE}.</math></center>
 +
 
 +
By [[Stewart's Theorem]], <math>AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}</math>, and by the [[Pythagorean Theorem]] on <math>\triangle ABD, \triangle EBD</math>,
 +
 
 +
<cmath>\begin{align*}
 +
BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\
 +
BD^2 + DE^2 &= \frac{15}{4} \\
 +
\end{align*}</cmath>
 +
 
 +
Subtracting the two equations yields <math>DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}</math>. Then <math>\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}</math>, and <math>m+n = \boxed{065}</math>.
 +
 
 +
 
 +
== Solution 2==
 +
Because the problem asks for a ratio, we can divide each side length by <math>\sqrt{3}</math> to make things simpler. We now have a triangle with sides <math>\sqrt{10}</math>, <math>\sqrt{5}</math>, and <math>\sqrt{2}</math>.
 +
 
 +
We use the same graph as above.
 +
 
 +
Draw perpendicular from <math>C</math> to <math>AE</math>. Denote this point as <math>F</math>. We know that <math>DE = EF = x</math> and <math>BD = CF = z</math> and also let <math>AE = y</math>.
 +
 
 +
Using Pythagorean theorem, we get three equations,
  
−
can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>
+
<center><math>(y+x)^2 + z^2 = 10</math>
 +
 
 +
<math>(y-x)^2 + z^2 = 2</math>
 +
 
 +
<math>x^2 + z^2 = \frac{5}{4}</math></center>
 +
 
 +
Adding the first and second, we obtain <math>x^2 + y^2 + z^2 = 6</math>, and then subtracting the third from this we find that <math>y = \frac{\sqrt{19}}{2}</math>. (Note, we could have used [[Stewart's Theorem]] to achieve this result).
 +
 
 +
Subtracting the first and second, we see that <math>xy = 2</math>, and then we find that <math>x = \frac{4}{\sqrt{19}}</math>  
 +
 
 +
Using base ratios, we then quickly find that the desired ratio is <math>\frac{27}{38}</math> so our answer is <math>\boxed{065}</math>
  
−
== Solution ==
 
−
{{solution}}
 
 
== See also ==
 
== See also ==
−
*[[1996 AIME Problems]]
+
{{AIME box|year=1996|num-b=12|num-a=14}}
  
−
{{AIME box|year=1996|num-b=12|num-a=14}}
+
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:18, 25 February 2016

Problem

In triangle $ABC$, $AB=\sqrt{30}$, $AC=\sqrt{6}$, and $BC=\sqrt{15}$. There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$, and $\angle ADB$ is a right angle. The ratio $\frac{[ADB]}{[ABC]}$ can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] pointpen = black; pathpen = black + linewidth(0.7); pair B=(0,0), C=(15^.5, 0), A=IP(CR(B,30^.5),CR(C,6^.5)), E=(B+C)/2, D=foot(B,A,E); D(MP("A",A)--MP("B",B,SW)--MP("C",C)--A--MP("D",D)--B); D(MP("E",E)); MP("\sqrt{30}",(A+B)/2,NW); MP("\sqrt{6}",(A+C)/2,SE); MP("\frac{\sqrt{15}}2",(E+C)/2); D(rightanglemark(B,D,A)); [/asy]

Let $E$ be the midpoint of $\overline{BC}$. Since $BE = EC$, then $\triangle ABE$ and $\triangle AEC$ share the same height and have equal bases, and thus have the same area. Similarly, $\triangle BDE$ and $BAE$ share the same height, and have bases in the ratio $DE : AE$, so $\frac{[BDE]}{[BAE]} = \frac{DE}{AE}$ (see area ratios). Now,

$\dfrac{[ADB]}{[ABC]} = \frac{[ABE] + [BDE]}{2[ABE]} = \frac{1}{2} + \frac{DE}{2AE}.$

By Stewart's Theorem, $AE = \frac{\sqrt{2(AB^2 + AC^2) - BC^2}}2 = \frac{\sqrt {57}}{2}$, and by the Pythagorean Theorem on $\triangle ABD, \triangle EBD$,

\begin{align*} BD^2 + \left(DE + \frac {\sqrt{57}}2\right)^2 &= 30 \\ BD^2 + DE^2 &= \frac{15}{4} \\ \end{align*}

Subtracting the two equations yields $DE\sqrt{57} + \frac{57}{4} = \frac{105}{4} \Longrightarrow DE = \frac{12}{\sqrt{57}}$. Then $\frac mn = \frac{1}{2} + \frac{DE}{2AE} = \frac{1}{2} + \frac{\frac{12}{\sqrt{57}}}{2 \cdot \frac{\sqrt{57}}{2}} = \frac{27}{38}$, and $m+n = \boxed{065}$.


Solution 2

Because the problem asks for a ratio, we can divide each side length by $\sqrt{3}$ to make things simpler. We now have a triangle with sides $\sqrt{10}$, $\sqrt{5}$, and $\sqrt{2}$.

We use the same graph as above.

Draw perpendicular from $C$ to $AE$. Denote this point as $F$. We know that $DE = EF = x$ and $BD = CF = z$ and also let $AE = y$.

Using Pythagorean theorem, we get three equations,

$(y+x)^2 + z^2 = 10$

$(y-x)^2 + z^2 = 2$

$x^2 + z^2 = \frac{5}{4}$

Adding the first and second, we obtain $x^2 + y^2 + z^2 = 6$, and then subtracting the third from this we find that $y = \frac{\sqrt{19}}{2}$. (Note, we could have used Stewart's Theorem to achieve this result).

Subtracting the first and second, we see that $xy = 2$, and then we find that $x = \frac{4}{\sqrt{19}}$

Using base ratios, we then quickly find that the desired ratio is $\frac{27}{38}$ so our answer is $\boxed{065}$

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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=1996_AIME_Problems/Problem_13&oldid=77032"