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 "1960 IMO Problems/Problem 3"
(→‎Solution 2)
 
Line 43: Line 43:
 
==Solution 2==
 
==Solution 2==
 
Let <math>P, Q, R</math> be points on side <math>BC</math> such that segment <math>PR</math> contains midpoint <math>Q</math>, with <math>P</math> closer to <math>C</math> and (without loss of generality) <math>AC \le AB</math>. Then if <math>AD</math> is an altitude, then <math>D</math> is between <math>P</math> and <math>C</math>. Combined with the obvious fact that <math>Q</math> is the midpoint of <math>PR</math> (for <math>n</math> is odd), we have
 
Let <math>P, Q, R</math> be points on side <math>BC</math> such that segment <math>PR</math> contains midpoint <math>Q</math>, with <math>P</math> closer to <math>C</math> and (without loss of generality) <math>AC \le AB</math>. Then if <math>AD</math> is an altitude, then <math>D</math> is between <math>P</math> and <math>C</math>. Combined with the obvious fact that <math>Q</math> is the midpoint of <math>PR</math> (for <math>n</math> is odd), we have
−
<cmath>\tan {\angle PAR} = \tan {\angle RAD - \angle PAD} = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4n}{a(n^2-1)}.</cmath>
+
<cmath>\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.</cmath>
  
 
==See Also==
 
==See Also==

Latest revision as of 23:14, 16 May 2015

Problem

In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that:

$\tan{\alpha}=\frac{4nh}{(n^2-1)a}.$

Solution

Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.

[asy] size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); dot(Q); label("A",A,SW); label("B",B,SE); label("C",C,NW); label("P",P,ENE); label("Q",Q,NNE); draw(A--B--C--cycle); draw(A--P); draw(A--Q); [/asy]

Then, $P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)$, and $Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)$.

So, $\text{slope}$$(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}$, and $\text{slope}$$(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}$.

Thus, $\tan{\alpha} = \tan{(\angle QAB - \angle PAB)} = \frac{(\frac{c}{b}\cdot\frac{n+1}{n-1})-(\frac{c}{b}\cdot\frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot\frac{n+1}{n-1})\cdot(\frac{c}{b}\cdot\frac{n-1}{n+1})}$ $= \frac{\frac{c}{b}\cdot\frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}} = \frac{4nbc}{(n^2-1)(b^2+c^2)}=\frac{4nbc}{(n^2-1)a^2}$.

Since $[ABC]=\frac{1}{2}bc=\frac{1}{2}ah$, $bc=ah$ and $\tan{\alpha}=\frac{4nh}{(n^2-1)a}$ as desired.

Solution 2

Let $P, Q, R$ be points on side $BC$ such that segment $PR$ contains midpoint $Q$, with $P$ closer to $C$ and (without loss of generality) $AC \le AB$. Then if $AD$ is an altitude, then $D$ is between $P$ and $C$. Combined with the obvious fact that $Q$ is the midpoint of $PR$ (for $n$ is odd), we have \[\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.\]

See Also

1960 IMO (Problems)
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Problem 4
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1960_IMO_Problems/Problem_3&oldid=70268"