Two distinct circles $K_1$ and $K_2$ are drawn in the plane. They intersect at points $A$ and $B$, where $AB$ is the diameter of $K_1$. A point $P$ on $K_2$ and inside $K_1$ is also given. Using only a "T-square" (i.e. an instrument which can produce a straight line joining two points and the perpendicular to a line through a point on or off the line), find a construction for two points $C$ and $D$ on $K_1$ such that $CD$ is perpendicular to $AB$ and $\angle CPD$ is a right angle.
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Tags: AMC, USA(J)MO, USAMO, geometry unsolved, geometry
Faustus
31.01.2013 16:02
I'll work backwards(as they say that 's the best thing to do in constructions). I'll find a point $L$ which'll trivially give me $CD$. Suppose you already know $P$ and have drawn $CD$. Say $CD$ cuts $AB$ at $M$. Join and extend $PM$ to meet $K_2$ at the point $L$. In $K_1, DM.MC=AM.MB$. Again, in $K_2, AM.MB=PM.ML\implies PM=MS(%Error. "since" is a bad command. PM=DM=CM)$. To get from $L$ to $CD$: If we have $L$, we join $PL$ and it'll intersect $AB$ at $M$. From $M$, we draw a line perpendicular to $AB$, passing through $M$ to cut $K_1$ at $C$ and $D$. To get $L$: Join $AP$ and extend to meet $K_1$ at $Q$. Drop a perpendicular from $Q$ to $AB$ to meet $K_1$ at $R$. Join $AR$. Drop a perpendicular from $P$ to meet $AR$ at $T$. Drop a perpendicular from $T$ on $QR$ to meet $K_2$ at $L'$. It easy to show that $L=L'$. Peace. Faustus
Saucepan_man02
07.01.2025 09:35
Here we go: Construction: Extend line $AP$ to meet $K_1$ again at $M$. Draw perpendicular from $M$ onto $AB$ and let it intersect $K_1$ again at $N$. Let $\ell_1$ denote the perpendicular from $P$ onto $AB$ and let $P'=AN \cap \ell_1$ (Note that: $P, P'$ are reflections of each other with respect to $AB$). Draw a line $\ell_2$ perpendicular to $\ell_1$ at $P'$ and let $\ell_2$ intersect $K_2$ at $Z$ such that $A, Z$ are on the same side of $\ell_2$. Let $X = PZ \cap AB$. Draw a line $\ell_3$ perpendicular to $AB$ at $X$ meeting $K_1$ at $C, D$. We claim we found points $C, D$ with $\angle CPD = 90^\circ$. We will prove that the construction works: Note that: $P, P'$ are reflections of each other with respect to $AB$. Thus: $P, P', Z$ have same perpendicular distance from line $AB$. Note: $PX=XZ$ and $XC=XD$. Due to Power of Point: $$PX^2 = PX \cdot XZ = AX \cdot XB = CX \cdot XD = XC^2.$$Thus: $XP=XC=XD$ with $X, C, D$ are collinear. Thus: $P$ lies on the circle with diameter $CD$ or $\angle CPD = 90^\circ$.