Difference between revisions of "Talk:1960 IMO Problems/Problem 3"
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| − | + | \section*{Proof for 1960 IMO Problem 3} | |
| − | \ | ||
Let \( \angle ACB = x \) and \( \angle ABC = 90^\circ - x \). | Let \( \angle ACB = x \) and \( \angle ABC = 90^\circ - x \). | ||
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 \( B \). | 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 \( B \). | ||
| − | + | Since \( BC \) is divided into \( n \) equal parts, where \( n \) is odd, the length of each segment is: | |
| + | |||
| + | \[ | ||
| + | \frac{a}{n} | ||
| + | \] | ||
| + | |||
| + | Because \( M \) is the midpoint of \( BC \), it must lie between two consecutive division points, meaning \( M \) is at the center of segment \( QP \). | ||
| + | |||
| + | ### Step 1: Altitude to the Hypotenuse | ||
| + | From right triangle trigonometry, the altitude to the hypotenuse is given by: | ||
| − | |||
\[ | \[ | ||
h = a \cos x \sin x | h = a \cos x \sin x | ||
\] | \] | ||
| − | + | ### Step 2: Median to the Hypotenuse | |
| + | Since \( AM \) is the median to the hypotenuse, we have: | ||
| + | |||
\[ | \[ | ||
| − | AM = BM = CM | + | AM = BM = CM = \frac{a}{2} |
\] | \] | ||
| − | |||
| − | + | We also know that: | |
| + | |||
\[ | \[ | ||
\angle MAB = 90^\circ - x, \quad \angle MAC = x | \angle MAB = 90^\circ - x, \quad \angle MAC = x | ||
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\] | \] | ||
| − | The length of \( QP \) is | + | ### Step 3: Length of \( QP \) and Triangle Relations |
| + | The length of \( QP \) is: | ||
| + | |||
\[ | \[ | ||
QP = \frac{a}{n} | QP = \frac{a}{n} | ||
\] | \] | ||
| − | Define | + | Define: |
| + | |||
\[ | \[ | ||
| − | \angle QAP = \alpha = k + z | + | \angle QAM = k, \quad \angle PAM = z, \quad \angle QAP = \alpha = k + z |
\] | \] | ||
| − | From angle properties | + | |
| + | From angle properties: | ||
| + | |||
\[ | \[ | ||
\angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z | \angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z | ||
\] | \] | ||
| − | Since \( M \) is the midpoint of \( QP \), we | + | Since \( M \) is the midpoint of \( QP \), we have: |
| + | |||
\[ | \[ | ||
QM = PM = \frac{a}{2n} | QM = PM = \frac{a}{2n} | ||
\] | \] | ||
| − | ### Applying the | + | ### Step 4: Applying the Law of Sines |
| + | Applying the sine rule in \( \triangle AQM \): | ||
| + | |||
\[ | \[ | ||
\frac{\sin k}{a/2n} = \frac{\sin(2x - k)}{a/2} | \frac{\sin k}{a/2n} = \frac{\sin(2x - k)}{a/2} | ||
\] | \] | ||
| − | + | Multiplying both sides by \( \frac{2}{a} \): | |
| + | |||
\[ | \[ | ||
\frac{2n \sin k}{a} = \frac{2 \sin(2x - k)}{a} | \frac{2n \sin k}{a} = \frac{2 \sin(2x - k)}{a} | ||
\] | \] | ||
| + | |||
| + | which simplifies to: | ||
\[ | \[ | ||
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\] | \] | ||
| − | Using the | + | Using the identity: |
| + | |||
\[ | \[ | ||
\sin(2x - k) = \sin 2x \cos k - \sin k \cos 2x | \sin(2x - k) = \sin 2x \cos k - \sin k \cos 2x | ||
\] | \] | ||
| − | + | and the double-angle formula: | |
| + | |||
\[ | \[ | ||
| − | 2 \sin x \cos x = \frac{2h}{a} | + | \sin 2x = 2 \sin x \cos x = \frac{2h}{a} |
\] | \] | ||
| + | |||
| + | we substitute: | ||
| + | |||
\[ | \[ | ||
\cos 2x = \frac{\sqrt{a^2 - 4h^2}}{a} | \cos 2x = \frac{\sqrt{a^2 - 4h^2}}{a} | ||
\] | \] | ||
| − | + | into our equation: | |
| + | |||
\[ | \[ | ||
n \sin k + \sin k \frac{\sqrt{a^2 - 4h^2}}{a} = \cos k \frac{2h}{a} | n \sin k + \sin k \frac{\sqrt{a^2 - 4h^2}}{a} = \cos k \frac{2h}{a} | ||
\] | \] | ||
| − | + | Factoring: | |
| + | |||
\[ | \[ | ||
| − | \sin k \left(n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos k \frac{2h}{a} | + | \sin k \left( n + \frac{\sqrt{a^2 - 4h^2}}{a} \right) = \cos k \frac{2h}{a} |
\] | \] | ||
Dividing both sides by \( \cos k \): | Dividing both sides by \( \cos k \): | ||
| + | |||
\[ | \[ | ||
\tan k = \frac{2h}{an + \sqrt{a^2 - 4h^2}} | \tan k = \frac{2h}{an + \sqrt{a^2 - 4h^2}} | ||
\] | \] | ||
| − | + | ### Step 5: Finding \( \tan \alpha \) | |
| + | Using a similar derivation for \( \tan z \), we apply the tangent sum identity: | ||
| + | |||
\[ | \[ | ||
\tan (z + k) = \frac{\tan z + \tan k}{1 - \tan z \tan k} | \tan (z + k) = \frac{\tan z + \tan k}{1 - \tan z \tan k} | ||
\] | \] | ||
| − | |||
| − | \ | + | to find \( \tan \alpha \), where \( \alpha = z + k \). |
Revision as of 10:18, 5 March 2025
\section*{Proof for 1960 IMO Problem 3}
Let \( \angle ACB = x \) and \( \angle ABC = 90^\circ - x \). 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 \( B \). Since \( BC \) is divided into \( n \) equal parts, where \( n \) is odd, the length of each segment is:
\[ \frac{a}{n} \]
Because \( M \) is the midpoint of \( BC \), it must lie between two consecutive division points, meaning \( M \) is at the center of segment \( QP \).
- Step 1: Altitude to the Hypotenuse
From right triangle trigonometry, the altitude to the hypotenuse is given by:
\[ h = a \cos x \sin x \]
- Step 2: Median to the Hypotenuse
Since \( AM \) is the median to the hypotenuse, we have:
\[ AM = BM = CM = \frac{a}{2} \]
We also know that:
\[ \angle MAB = 90^\circ - x, \quad \angle MAC = x \] \[ \angle AMB = 2x, \quad \angle AMC = 180^\circ - 2x \]
- Step 3: Length of \( QP \) and Triangle Relations
The length of \( QP \) is:
\[ QP = \frac{a}{n} \]
Define:
\[ \angle QAM = k, \quad \angle PAM = z, \quad \angle QAP = \alpha = k + z \]
From angle properties:
\[ \angle AQM = 2x - k, \quad \angle APM = 180^\circ - 2x - z \]
Since \( M \) is the midpoint of \( QP \), we have:
\[ QM = PM = \frac{a}{2n} \]
- Step 4: Applying the Law of Sines
Applying the sine rule in \( \triangle AQM \):
\[ \frac{\sin k}{a/2n} = \frac{\sin(2x - k)}{a/2} \]
Multiplying both sides by \( \frac{2}{a} \):
\[ \frac{2n \sin k}{a} = \frac{2 \sin(2x - k)}{a} \]
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 \( \alpha = z + k \).
