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: @@
+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>
-\section*{Proof for 1960 IMO Problem 3}
+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 \( \angle ACB = x \) and \( \angle ABC = 90^\circ - x \).
+<cmath>
-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 \( B \).
+f = \frac{a}{2}
-Since \( BC \) is divided into \( n \) equal parts, where \( n \) is odd, the length of each segment is:
+</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 \(\angle APM = 180^\circ - 2x - z\).
-\frac{a}{n}
-\]
-Because \( M \) is the midpoint of \( BC \), it must lie between two consecutive division points, meaning \( M \) is at the center of segment \( QP \).
+As \(M\) is in the middle of \(QP\), we have \(QM = PM = \frac{a}{2n}\). Applying the sine law on triangle \(AQM\), we get:
-### Step 1: Altitude to the Hypotenuse
+<cmath>
-From right triangle trigonometry, the altitude to the hypotenuse is given by:
+\frac{\sin(k)}{\frac{a}{2n}} = \frac{\sin(2x - k)}{\frac{a}{2}}
+</cmath>
-\[
+Simplifying:
-h = a \cos x \sin x
-\]
-### Step 2: Median to the Hypotenuse
+<cmath>
-Since \( AM \) is the median to the hypotenuse, we have:
+\frac{2n \sin(k)}{a} = \frac{2 \sin(2x - k)}{a}
+</cmath>
-\[
+<cmath>
-AM = BM = CM = \frac{a}{2}
+n \sin(k) = \sin(2x - k)
-\]
+</cmath>
-We also know that:
+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:
-\[
+<cmath>
-\angle MAB = 90^\circ - x, \quad \angle MAC = x
+\sin(2x) = \frac{2h}{a}
-\]
+</cmath>
-\[
-\angle AMB = 2x, \quad \angle AMC = 180^\circ - 2x
-\]
-### Step 3: Length of \( QP \) and Triangle Relations
+Thus:
-The length of \( QP \) is:
-\[
+<cmath>
-QP = \frac{a}{n}
+n \sin(k) = \frac{2h}{a} \cos(k) - \cos(2x) \sin(k)
-\]
+</cmath>
-Define:
+Now, we know that:
-\[
+<cmath>
-\angle QAM = k, \quad \angle PAM = z, \quad \angle QAP = \alpha = k + z
+\cos(2x) = \frac{\sqrt{a^2 - 4h^2}}{a}
-\]
+</cmath>
-From angle properties:
+Substituting this into the equation:
-\[
+<cmath>
-\angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z
+n \sin(k) + \sin(k) \frac{\sqrt{a^2 - 4h^2}}{a} = \cos(k) \frac{2h}{a}
-\]
+</cmath>
-Since \( M \) is the midpoint of \( QP \), we have:
+Factoring out \(\sin(k)\):
-\[
+<cmath>
-QM = PM = \frac{a}{2n}
+\sin(k) \left( n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos(k) \frac{2h}{a}
-\]
+</cmath>
-### Step 4: Applying the Law of Sines
+Thus:
-Applying the sine rule in \( \triangle AQM \):
-\[
+<cmath>
-\frac{\sin k}{a/2n} = \frac{\sin(2x - k)}{a/2}
+\tan(k) = \frac{2h}{an + \sqrt{a^2 - 4h^2}}
-\]
+</cmath>
-Multiplying both sides by \( \frac{2}{a} \):
+By performing similar steps with \(\tan(z)\), we can use the addition formula for \(\tan(z+k)\) to find \(\tan(\alpha)\), where \(\alpha = z + k\).
-\[
-\frac{2n \sin k}{a} = \frac{2 \sin(2x - k)}{a}
-\]
-which simplifies to:
-\[
-n \sin k = \sin(2x - k)
-\]
-Using the identity:
-\[
-\sin(2x - k) = \sin 2x \cos k - \sin k \cos 2x
-\]
-and the double-angle formula:
-\[
-\sin 2x = 2 \sin x \cos x = \frac{2h}{a}
-\]
-we substitute:
-\[
-\cos 2x = \frac{\sqrt{a^2 - 4h^2}}{a}
-\]
-into our equation:
-\[
-n \sin k + \sin k \frac{\sqrt{a^2 - 4h^2}}{a} = \cos k \frac{2h}{a}
-\]
-Factoring:
-\[
-\sin k \left( n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos k \frac{2h}{a}
-\]
-Dividing both sides by \( \cos k \):
-\[
-\tan k = \frac{2h}{an + \sqrt{a^2 - 4h^2}}
-\]
-### Step 5: Finding \( \tan \alpha \)
-Using a similar derivation for \( \tan z \), we apply the tangent sum identity:
-\[
-\tan (z + k) = \frac{\tan z + \tan k}{1 - \tan z \tan k}
-\]
-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