Kvant Math Problem 796
Consider a square $ABCD$ with a point $P$ inside it.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 10m09s
Source on kvant.digital
Problem
Point $P$ is located inside the square $ABCD$ such that $|AP|:|BP|:|CP|=1:2:3$. Find $\widehat{APB}$.
L. D. Kurlandchik
Exploration
Consider a square $ABCD$ with a point $P$ inside it. Label the vertices in order, and attempt to place $P$ such that $|AP| : |BP| : |CP| = 1:2:3$. Introduce coordinates: let $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$. Then $P=(x,y)$, with distances
$|AP|=\sqrt{x^2+y^2}, \quad |BP|=\sqrt{(x-1)^2+y^2}, \quad |CP|=\sqrt{(x-1)^2+(y-1)^2}.$
The ratio condition becomes $\sqrt{x^2+y^2} : \sqrt{(x-1)^2+y^2} : \sqrt{(x-1)^2+(y-1)^2} = 1:2:3$. Squaring to eliminate radicals, one obtains two equations. Attempting symmetry simplifications, assume $P$ lies along a line connecting opposite vertices or along a diagonal. The main challenge is solving the resulting nonlinear system. A likely key step is identifying the precise location of $P$ that satisfies all three distances in exact ratio. The desired angle $\widehat{APB}$ may reveal a simple value despite the complexity of coordinates.
Problem Understanding
We are asked to find the measure of angle $\widehat{APB}$ given a point $P$ inside a square with specified distance ratios to three vertices. This is a Type B problem, as the angle is determined by the distance ratio, not requiring classification or extremization. The core difficulty lies in translating the ratio conditions into a solvable geometric or algebraic system and extracting the angle. Intuition suggests that the ratios $1:2:3$ correspond to a specific placement along a line connecting $A$ and $C$, producing a simple angle $\widehat{APB}$, likely $\mathbf{60^\circ}$ or $\mathbf{120^\circ}$.
Proof Architecture
Lemma 1: The coordinates of $P$ satisfying $|AP| : |BP| : |CP| = 1:2:3$ lie on the line $y=x$. This follows from symmetry of the square and distance ratios.
Lemma 2: On the line $y=x$, the distance equations reduce to one quadratic for $x$, which has a unique solution inside the square.
Lemma 3: With $P$ determined, compute $\cos \widehat{APB}$ using the dot product formula for vectors $\overrightarrow{AP}$ and $\overrightarrow{BP}$. The hardest step is verifying that the ratio condition produces a solvable quadratic giving $x$ in $(0,1)$.
Solution
Place the square on the coordinate plane with $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$. Let $P=(x,y)$ inside the square. The distance ratio condition reads
$|AP| : |BP| : |CP| = \sqrt{x^2+y^2} : \sqrt{(x-1)^2+y^2} : \sqrt{(x-1)^2+(y-1)^2} = 1:2:3.$
Square the distances and introduce a common factor $k^2$ so that $|AP|^2=k^2$, $|BP|^2=4k^2$, $|CP|^2=9k^2$. Then
$x^2 + y^2 = k^2, \quad (x-1)^2 + y^2 = 4k^2, \quad (x-1)^2 + (y-1)^2 = 9k^2.$
Subtract the first equation from the second:
$(x-1)^2 + y^2 - (x^2+y^2) = 4k^2 - k^2 \implies x^2 - 2x + 1 + y^2 - x^2 - y^2 = 3k^2 \implies -2x+1 = 3k^2 \implies k^2 = \frac{1-2x}{3}.$
Subtract the second from the third:
$(x-1)^2 + (y-1)^2 - [(x-1)^2 + y^2] = 9k^2 - 4k^2 \implies y^2 - 2y + 1 - y^2 = 5k^2 \implies -2y + 1 = 5k^2 \implies k^2 = \frac{1-2y}{5}.$
Equate the two expressions for $k^2$:
$\frac{1-2x}{3} = \frac{1-2y}{5} \implies 5(1-2x)=3(1-2y) \implies 5-10x = 3 - 6y \implies 2 -10x = -6y \implies y = \frac{5x-1}{3}.$
To ensure $P$ is inside the square, $0< x <1$, $0< y <1$. Then $0< \frac{5x-1}{3}<1 \implies 0<5x-1<3 \implies 1/5 < x < 4/5$.
Now $k^2 = (1-2x)/3$. Also $y = (5x-1)/3$, so $|AP|^2 = x^2 + y^2 = x^2 + \left(\frac{5x-1}{3}\right)^2 = k^2 = \frac{1-2x}{3}$. Multiply both sides by 9:
$9x^2 + (5x-1)^2 = 3(1-2x) \implies 9x^2 + 25x^2 -10x +1 = 3-6x \implies 34x^2 -4x -2 = 0 \implies 17x^2 -2x -1 =0.$
Solve the quadratic: $x = \frac{2 \pm \sqrt{4 + 68}}{34} = \frac{2 \pm \sqrt{72}}{34} = \frac{2 \pm 6\sqrt{2}}{34} = \frac{1 \pm 3\sqrt{2}}{17}$. Only $0< x<1$ gives $x = \frac{1+3\sqrt{2}}{17}$. Then $y = \frac{5x-1}{3} = \frac{5(1+3\sqrt{2})/17 -1}{3} = \frac{(5+15\sqrt{2})/17 -1}{3} = \frac{(5+15\sqrt{2}-17)/17}{3} = \frac{-12+15\sqrt{2}}{51} = \frac{5\sqrt{2}-4}{17}$.
Thus $P=(\frac{1+3\sqrt{2}}{17}, \frac{5\sqrt{2}-4}{17})$. Compute vectors: $\overrightarrow{AP} = (x,y)$, $\overrightarrow{BP} = (x-1,y)$. Then
$\cos \widehat{APB} = \frac{\overrightarrow{AP}\cdot \overrightarrow{BP}}{|AP||BP|} = \frac{x(x-1)+y^2}{|AP||BP|}.$
Compute numerator: $x(x-1)+y^2 = x^2 - x + y^2 = (x^2+y^2)-x = k^2 - x = \frac{1-2x}{3}-x = \frac{1-2x-3x}{3} = \frac{1-5x}{3}$. Substituting $x = \frac{1+3\sqrt{2}}{17}$ gives $1-5x = 1 - 5(1+3\sqrt{2})/17 = 1 - (5+15\sqrt{2})/17 = (17-5-15\sqrt{2})/17 = (12-15\sqrt{2})/17$. Then numerator = $(12-15\sqrt{2})/51$. Denominators: $|AP| = k = \sqrt{(1-2x)/3}$, $|BP| = 2k = 2\sqrt{(1-2x)/3}$. Then $|AP||BP| = 2 (1-2x)/3 = 2(1-2x)/3 = 2(1-2(1+3\sqrt{2})/17)/3 = 2(1-2/17-6\sqrt{2}/17)/3 = 2(15-6\sqrt{2})/51 = (30-12\sqrt{2})/51$. Then $\cos \widehat{APB} = \frac{12-15\sqrt{2}}{30-12\sqrt{2}} = \frac{12-15\sqrt{2}}{30-12\sqrt{2}}$. Factor 3 from numerator and 6 from denominator