1960 IMO Problems/Problem 3

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 ACB=x, and ABC=90x. 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=a2, we have:

\[f = \frac{a}{2}\]

We know that MAB=90x, and MAC=x. This means that AMB=2x and AMC=1802x. The length of QP is an. Let QAM=k and PAM=z, such that QAP (or α) equals k+z. This means that AQM=2xk, and APM=1802xz.

As M is in the middle of QP, we have QM=PM=a2n. 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(2xk)=sin(2x)cos(k)cos(2x)sin(k), and since sin(2x)=2sin(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(α), where α=z+k.

Courtesy of Gordon Freeman

Solution 4

Let $D$ and $E$ be points on $BC$ such that $\alpha=\angle DAE$, $D$ is closer to $B$ than to $C$, and $E$ is closer to $C$ than $B$. Furthermore, let $\alpha_1=\angle BAD$ and $\alpha_2=\angle CAE$, and let $\beta=\angle B$ and $\gamma=\angle C$. Applying Law of Sines on $\triangle BAD$ yields sinα1a(n1)2n=sin(180(α1+β))c2cna(n1)sinα1=sin(α1+β)2cna(n1)sinα1=sinα1cosβ+cosα1sinβ2cna(n1)sinα1=sinα1ca+cosα1ba2cnn1sinα1=csinα1+bcosα1cn+cn1sinα1=bcosα1tanα1=b(n1)c(n+1) Similarly, $\tan\alpha_2=\frac{c(n-1)}{b(n+1)}$. Next, we use the sum of tangents formula: tan(α1+α2)=b(n1)c(n+1)+c(n1)b(n+1)1b(n1)c(n+1)c(n1)b(n+1)=(bc+cb)n1n+11(n1)2(n+1)2=(bc+cb)(n21)(n+1)2(n1)2=(bc+cb)(n21)4n=(b2+c2)(n21)4nbc Since $\alpha+\alpha_1+\alpha_2=90^\circ$, we need to find $\tan\alpha=\tan\left(\frac{\pi}{2}-(\alpha_1+\alpha_2)\right)$. This is equal to $\cot(\alpha_1+\alpha_2)$, which is the reciprocal of $\tan(\alpha_1+\alpha_2)$, or $\frac{4nbc}{(b^2+c^2)(n^2-1)}$. Since $bc=ah$ due to triangle areas and $a^2=b^2+c^2$ due to the Pythagorean Theorem, we conclude that the expression is equal to $\frac{4nah}{a^2(n^2-1)}$, which simplifies to $\frac{4nh}{(n^2-1)a}$.

~eevee9406

See Also

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