Difference between revisions of "1960 IMO Problems/Problem 3"

Latest revision as of 07:51, 7 March 2025

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 $\tan {\angle PAR} = \tan (\angle RAD - \angle PAD) = \frac{\frac{PR}{h}}{1 + \frac{DP \cdot DR}{h^2}} = \frac{PR \cdot h}{h^2 + DP \cdot DR} = \frac{PR \cdot h}{AQ^2 - DQ^2 + DP \cdot DR} = \frac{PR \cdot h}{\frac{a^2}{4} - PQ^2} = \frac{\frac{a}{n} \cdot h}{\frac{a^2}{4} - \frac{a^2}{4n^2}} = \frac{4nh}{(n^2-1)a}.$

Solution 3

Let $\angle ACB = x$, and $\angle ABC = 90^\circ - x$. Let $M$ be the midpoint on the hypotenuse $BC$, and $Q$ and $P$ be points such that $PQ$ contains $BC$, with $Q$ closer to $C$ and $P$ closer to $B$. The midpoint will always be in the middle of line $QP$, unless $n$ is even or infinite, which it is not. Given such a triangle, we can express the altitude to the hypotenuse as:

$h = a \cos(x) \sin(x)$

Next, we shall denote line $AM$ as $f$, where $AM$ is the median to the hypotenuse. This means that line $AM = BM = CM$, and as $BM = \frac{a}{2}$, we have:

$f = \frac{a}{2}$

We know that $\angle MAB = 90^\circ - x$, and $\angle MAC = x$. This means that $\angle AMB = 2x$ and $\angle AMC = 180^\circ - 2x$. The length of $QP$ is $\frac{a}{n}$. Let $\angle QAM = k$ and $\angle PAM = z$, such that $\angle QAP$ (or $\alpha$) equals $k + z$. This means that $\angle AQM = 2x - k$, and $\angle APM = 180^\circ - 2x - z$.

As $M$ is in the middle of $QP$, we have $QM = PM = \frac{a}{2n}$. Applying the sine law on triangle $AQM$, we get:

$\frac{\sin(k)}{\frac{a}{2n}} = \frac{\sin(2x - k)}{\frac{a}{2}}$

Simplifying:

$\frac{2n \sin(k)}{a} = \frac{2 \sin(2x - k)}{a}$

$n \sin(k) = \sin(2x - k)$

Using the identity $\sin(2x - k) = \sin(2x) \cos(k) - \cos(2x) \sin(k)$, and since $\sin(2x) = 2 \sin(x) \cos(x)$, we substitute:

$\sin(2x) = \frac{2h}{a}$

Thus:

$n \sin(k) = \frac{2h}{a} \cos(k) - \cos(2x) \sin(k)$

Now, we know that:

$\cos(2x) = \frac{\sqrt{a^2 - 4h^2}}{a}$

Substituting this into the equation:

$n \sin(k) + \sin(k) \frac{\sqrt{a^2 - 4h^2}}{a} = \cos(k) \frac{2h}{a}$

Factoring out $\sin(k)$:

$\sin(k) \left( n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos(k) \frac{2h}{a}$

Thus:

$\tan(k) = \frac{2h}{an + \sqrt{a^2 - 4h^2}}$

By performing similar steps with $\tan(z)$, we can use the addition formula for $\tan(z+k)$ to find $\tan(\alpha)$, where $\alpha = z + k$.

Courtesy of Gordon Freeman

@@ Line 1: / Line 1: @@
-In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angel subtending, from <math>A</math>, that segment which contains the mdipoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse fo the triangle. Prove that:
+== Problem ==
+In a given right triangle <math>ABC</math>, the hypotenuse <math>BC</math>, of length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> an odd integer). Let <math>\alpha</math> be the acute angle subtending, from <math>A</math>, that segment which contains the midpoint of the hypotenuse. Let <math>h</math> be the length of the altitude to the hypotenuse of the triangle. Prove that:
 <center><math>
-\displaystyle\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
+\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
 </math>
 </center>
+== Solution ==
+Using coordinates, let <math>A=(0,0)</math>, <math>B=(b,0)</math>, and <math>C=(0,c)</math>. Also, let <math>PQ</math> be the segment that contains the midpoint of the hypotenuse with <math>P</math> closer to <math>B</math>.
+<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, <math>P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)</math>, and <math>Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n+1}{2}c\right)</math>.
+So, <math>\text{slope}</math><math>(PA)=\tan{\angle PAB}=\frac{c}{b}\cdot\frac{n-1}{n+1}</math>, and <math>\text{slope}</math><math>(QA)=\tan{\angle QAB}=\frac{c}{b}\cdot\frac{n+1}{n-1}</math>.
+Thus, <math>\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})}</math>
+<math>= \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}</math>.
+Since <math>[ABC]=\frac{1}{2}bc=\frac{1}{2}ah</math>, <math>bc=ah</math> and <math>\tan{\alpha}=\frac{4nh}{(n^2-1)a}</math> as desired.
+==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
+<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>
+==Solution 3==
+Let \(\angle ACB = x\), and \(\angle ABC = 90^\circ - x\). Let \(M\) be the midpoint on the hypotenuse \(BC\), and \(Q\) and \(P\) be points such that \(PQ\) contains \(BC\), with \(Q\) closer to \(C\) and \(P\) closer to \(B\). The midpoint will always be in the middle of line \(QP\), unless \(n\) is even or infinite, which it is not. Given such a triangle, we can express the altitude to the hypotenuse as:
+<cmath>
+h = a \cos(x) \sin(x)
+</cmath>
+Next, we shall denote line \(AM\) as \(f\), where \(AM\) is the median to the hypotenuse. This means that line \(AM = BM = CM\), and as \(BM = \frac{a}{2}\), we have:
+<cmath>
+f = \frac{a}{2}
+</cmath>
+We know that \(\angle MAB = 90^\circ - x\), and \(\angle MAC = x\). This means that \(\angle AMB = 2x\) and \(\angle AMC = 180^\circ - 2x\). The length of \(QP\) is \(\frac{a}{n}\). Let \(\angle QAM = k\) and \(\angle PAM = z\), such that \(\angle QAP\) (or \(\alpha\)) equals \(k + z\). This means that \(\angle AQM = 2x - k\), and \(\angle APM = 180^\circ - 2x - z\).
+As \(M\) is in the middle of \(QP\), we have \(QM = PM = \frac{a}{2n}\). Applying the sine law on triangle \(AQM\), we get:
+<cmath>
+\frac{\sin(k)}{\frac{a}{2n}} = \frac{\sin(2x - k)}{\frac{a}{2}}
+</cmath>
+Simplifying:
+<cmath>
+\frac{2n \sin(k)}{a} = \frac{2 \sin(2x - k)}{a}
+</cmath>
+<cmath>
+n \sin(k) = \sin(2x - k)
+</cmath>
+Using the identity \(\sin(2x - k) = \sin(2x) \cos(k) - \cos(2x) \sin(k)\), and since \(\sin(2x) = 2 \sin(x) \cos(x)\), we substitute:
+<cmath>
+\sin(2x) = \frac{2h}{a}
+</cmath>
+Thus:
+<cmath>
+n \sin(k) = \frac{2h}{a} \cos(k) - \cos(2x) \sin(k)
+</cmath>
+Now, we know that:
+<cmath>
+\cos(2x) = \frac{\sqrt{a^2 - 4h^2}}{a}
+</cmath>
+Substituting this into the equation:
+<cmath>
+n \sin(k) + \sin(k) \frac{\sqrt{a^2 - 4h^2}}{a} = \cos(k) \frac{2h}{a}
+</cmath>
+Factoring out \(\sin(k)\):
+<cmath>
+\sin(k) \left( n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos(k) \frac{2h}{a}
+</cmath>
+Thus:
+<cmath>
+\tan(k) = \frac{2h}{an + \sqrt{a^2 - 4h^2}}
+</cmath>
+By performing similar steps with \(\tan(z)\), we can use the addition formula for \(\tan(z+k)\) to find \(\tan(\alpha)\), where \(\alpha = z + k\).
+Courtesy of Gordon Freeman
+==See Also==
+{{IMO7 box|year=1960|num-b=2|num-a=4}}
+[[Category:Olympiad Geometry Problems]]

1960 IMO (Problems)
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6 • 7	Followed by Problem 4

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