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

Latest revision as of 08:57, 7 March 2025

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 $\alpha = z + k$.

@@ Line 1: / Line 1: @@
-== A proof of the 3rd question from the 1960 IMO ==
+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:
-\documentclass{article}
+<cmath>
-\usepackage{amsmath}
+h = a \cos(x) \sin(x)
-\begin{document}
+</cmath>
-Let \( \angle ACB = x \) and \( \angle ABC = 90^\circ - x \).
+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:
-Let \( M \) be the midpoint of the hypotenuse \( BC \), and let \( Q \) and \( P \) be points such that \( PQ \) contains \( BC \), with \( Q \) closer to \( C \) and \( P \) closer to \( B \).
-The midpoint \( M \) will always be in the middle of segment \( QP \), unless \( n \) is even or infinite, which it is not.
-Given such a triangle, we can express the altitude to the hypotenuse as:
+<cmath>
-\[
+f = \frac{a}{2}
-h = a \cos x \sin x
+</cmath>
-\]
-Now, let us denote the median \( AM \) by \( f \). Since \( AM \) is the median to the hypotenuse, we have:
+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 \(\angle APM = 180^\circ - 2x - z\).
-\[
-AM = BM = CM
-\]
-Since \( BM = \frac{a}{2} \), it follows that \( f = \frac{a}{2} \).
-Angles in the triangle give us:
+As \(M\) is in the middle of \(QP\), we have \(QM = PM = \frac{a}{2n}\). Applying the sine law on triangle \(AQM\), we get:
-\[
-\angle MAB = 90^\circ - x, \quad \angle MAC = x
-\]
-\[
-\angle AMB = 2x, \quad \angle AMC = 180^\circ - 2x
-\]
-The length of \( QP \) is given by:
+<cmath>
-\[
+\frac{\sin(k)}{\frac{a}{2n}} = \frac{\sin(2x - k)}{\frac{a}{2}}
-QP = \frac{a}{n}
+</cmath>
-\]
-Define \( \angle QAM = k \) and \( \angle PAM = z \), so that:
+Simplifying:
-\[
-\angle QAP = \alpha = k + z
-\]
-From angle properties, we get:
-\[
-\angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z
-\]
-Since \( M \) is the midpoint of \( QP \), we know:
+<cmath>
-\[
+\frac{2n \sin(k)}{a} = \frac{2 \sin(2x - k)}{a}
-QM = PM = \frac{a}{2n}
+</cmath>
-\]
-### Applying the Sine Rule in \( \triangle AQM \):
+<cmath>
-\[
+n \sin(k) = \sin(2x - k)
-\frac{\sin k}{a/2n} = \frac{\sin(2x - k)}{a/2}
+</cmath>
-\]
-Rearranging:
+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:
-\[
-\frac{2n \sin k}{a} = \frac{2 \sin(2x - k)}{a}
-\]
-\[
+<cmath>
-n \sin k = \sin(2x - k)
+\sin(2x) = \frac{2h}{a}
-\]
+</cmath>
-Using the sine subtraction identity:
+Thus:
-\[
-\sin(2x - k) = \sin 2x \cos k - \sin k \cos 2x
-\]
-Since \( \sin 2x = 2 \sin x \cos x \) and given that \( h = a \cos x \sin x \), we can substitute:
+<cmath>
-\[
+n \sin(k) = \frac{2h}{a} \cos(k) - \cos(2x) \sin(k)
-\sin x \cos x = \frac{2h}{a}
+</cmath>
-\]
-\[
-\cos 2x = \frac{\sqrt{a^2 - 4h^2}}{a}
-\]
-Substituting in our equation:
+Now, we know that:
-\[
-n \sin k + \sin k \frac{\sqrt{a^2 - 4h^2}}{a} = \cos k \frac{2h}{a}
-\]
-Factorizing:
+<cmath>
-\[
+\cos(2x) = \frac{\sqrt{a^2 - 4h^2}}{a}
-\sin k \left(n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos k \frac{2h}{a}
+</cmath>
-\]
-Dividing both sides by \( \cos k \):
+Substituting this into the equation:
-\[
-\tan k = \frac{2h}{an + \sqrt{a^2 - 4h^2}}
-\]
-Similarly, solving for \( \tan z \), we can apply the tangent sum identity:
+<cmath>
-\[
+n \sin(k) + \sin(k) \frac{\sqrt{a^2 - 4h^2}}{a} = \cos(k) \frac{2h}{a}
-\tan (z + k) = \frac{\tan z + \tan k}{1 - \tan z \tan k}
+</cmath>
-\]
-to find the value of \( \tan \alpha \), where \( \alpha = z + k \).
-\end{document}
+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 \(\alpha = z + k\).

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

Latest revision as of 08:57, 7 March 2025