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Midpoint in a weird configuration
Gimbrint   1
N 34 minutes ago by Beelzebub
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$. Point $L$ is chosen on arc $AC$, not containing $B$, so that, letting $BL$ intersect $AC$ at $S$, one has $AS<CS$. Points $D$ and $E$ lie on lines $AB$ and $BC$ respectively, such that $BELD$ is a parallelogram. Point $P$ is chosen on arc $BC$, not containing $A$, such that $\angle CBP=\angle BDE$. Line $AP$ intersects $EL$ at $X$, and line $CP$ intersects $DL$ at $Y$. Line $XY$ intersects $AB$, $BC$ and $BP$ at points $M$, $N$ and $T$ respectively.

Prove that $TN=TM$.
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Gimbrint
May 23, 2025
Beelzebub
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A touching question on perpendicular lines
Tintarn   2
N Apr 23, 2025 by pi_quadrat_sechstel
Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
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Tintarn
Mar 17, 2025
pi_quadrat_sechstel
Apr 23, 2025
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A touching question on perpendicular lines
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Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 3
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Tintarn
9045 posts
Tintarn
#1 h  Mar 17, 2025, 12:23 PM
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Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.
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Mathzeus1024
938 posts
Mathzeus1024
#2 h  Mar 22, 2025, 9:19 AM
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Let $k$ be the semicircle $y=\sqrt{R^2-x^2}$ such that we have the points: $A(-R,0); M(0,0); B(R,0); P(R\cos(t), R\sin(t))$ for $t \in (0,\pi)$. The centers of inscribed circles $k_{A}, k_{B}$ shall respectively lie on the angular bisectors $\overline{MC}, \overline{ME}$ through $M$:

$\overline{MC}: y = -\tan\left(\frac{\pi-t}{2}\right)x = -\cot\left(\frac{t}{2}\right)x$ (i);

$\overline{ME}: y = \tan\left(\frac{t}{2}\right)x$ (ii)

which in turn yield the points:

$C\left(-R\sin\left(\frac{t}{2}\right), R\cos\left(\frac{t}{2}\right)\right)$ (iii);

$E\left(R\cos\left(\frac{t}{2}\right), R\sin\left(\frac{t}{2}\right)\right)$ (iv).

If $k_{A}$ has radius $r<R$, contains (iii) above, and is modeled according to $\left[x + \tan\left(\frac{t}{2}\right)r\right]^2 + (y-r)^2=r^2$, then we obtain:

$x^2 +2rx\tan\left(\frac{t}{2}\right) + r^2\tan^{2}\left(\frac{t}{2}\right) + y^2-2ry+r^2=r^2$;

or $R^2 - 2Rr\sin\left(\frac{t}{2}\right)\tan\left(\frac{t}{2}\right)-2Rr\cos\left(\frac{t}{2}\right) + r^2\tan^{2}\left(\frac{t}{2}\right) =0$;

or $R^{2}\cos^{2}\left(\frac{t}{2}\right) - 2Rr\cos\left(\frac{t}{2}\right)\left[\sin^{2}\left(\frac{t}{2}\right) + \cos^{2}\left(\frac{t}{2}\right)\right] +r^{2}\sin^{2}\left(\frac{t}{2}\right)=0$;

or $R^{2}\cos^{2}\left(\frac{t}{2}\right) - 2Rr\cos\left(\frac{t}{2}\right) +r^{2}\sin^{2}\left(\frac{t}{2}\right)=0$;

or $r = \frac{2R\cos(t/2) \pm \sqrt{4R^{2}\cos^{2}(t/2)-4R^2\cos^{2}(t/2)\sin^{2}(t/2)}}{2\sin^{2}(t/2)} = R\cos(t/2)\left[\frac{1 \pm \cos(t/2)}{1-\cos^{2}(t/2)}\right]$;

or $r = R\cos(t/2)\left[\frac{1 \pm \cos(t/2)}{[1-\cos(t/2)][1+\cos(t/2)]}\right]$ (v).

Since $0 < r < R$, we only admit the negative root in the numerator of (v) to ultimate yield $r = \left[\frac{\cos(t/2)}{1+\cos(t/2)}\right]R$ (vi), which also yields the tangency point $D\left(-\left[\frac{\sin(t/2)}{1+\cos(t/2)}\right]R,0\right)$ (vii).

Next, we compute the slopes $m_{AE}, m_{CD}$:

$m_{AE} = \frac{R\sin(t/2)-0}{R\cos(t/2)+R} = \frac{\sin(t/2)}{\cos(t/2)+1}$ (viii);

$m_{CD} = \frac{R\cos(t/2)-0}{-R\sin(t/2)+ \left[\frac{\sin(t/2)}{1+\cos(t/2)}\right]R} = \frac{\cos(t/2)}{-\sin(t/2)+ \left[\frac{\sin(t/2)}{1+\cos(t/2)}\right]} = -\frac{\cos(t/2)+1}{\sin(t/2)}$ (ix).

and upon inspection $m_{AE} \cdot m_{CD} = -1 \Rightarrow \textcolor{red}{\overline{AE} \perp \overline{CD}}$.
This post has been edited 18 times. Last edited by Mathzeus1024, Mar 22, 2025, 10:23 AM
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pi_quadrat_sechstel
607 posts
pi_quadrat_sechstel
#3 h  Apr 23, 2025, 6:47 AM
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Tintarn wrote:
Let $k$ be a semicircle with diameter $AB$ and midpoint $M$. Let $P$ be a point on $k$ different from $A$ and $B$.

The circle $k_A$ touches $k$ in a point $C$, the segment $MA$ in a point $D$, and additionally the segment $MP$. The circle $k_B$ touches $k$ in a point $E$ and additionally the segments $MB$ and $MP$.

Show that the lines $AE$ and $CD$ are perpendicular.

Complete $k$ to the full circle and let $S$ be the midpoint of the arc $AB$ on the new half.

The homothety at $C$ sending $k_A$ to $k$ must send the "lowest" point $D$ to the "lowest" point $S$. So $CD=CS$.

We have congruences $AMC\sim PMC$ and $PME\sim BME$. Thus $\angle EMC=\frac{1}{2}\angle BMA=90^{\circ}$.

So a $90^{\circ}$-rotation at $M$ sends $A$ to $S$ and $E$ to $C$. We get $AE\perp CS=CD$.
This post has been edited 1 time. Last edited by pi_quadrat_sechstel, Apr 23, 2025, 6:48 AM
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