Let $x$, $y$, $z$, $t$ are real numbers. $x+y+z+t=6$ and $\sqrt{1-x^2}+\sqrt{4-y^2}+\sqrt{9-z^2}+\sqrt{16-t^2}=8$ Find each value of them as a number.

Question

It's so hard. How do you find it? It's just two equations but four variables. Can someone help me?

Answer 1

Consider following 4 vectors in $\mathbb{R}^2$, $$\vec{a} = (x,\sqrt{1-x^2}),\;\; \vec{b} = (y,\sqrt{4-y^2}),\;\; \vec{c} = (z,\sqrt{9-z^2})\;\;\text{ and }\;\; \vec{d} = (t,\sqrt{16-t^2}) $$ We have $|\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 3$ and $|\vec{d}| = 4$. The given conditions tell us $$\vec{a} + \vec{b} + \vec{c} + \vec{d} = (6,8)\\ \implies |\vec{a} + \vec{b} + \vec{c} + \vec{d}| = |(6,8)| = 10 = |\vec{a}| + |\vec{b}| + |\vec{c}| + |\vec{d}| $$ By triangle inequality, this is possible only when the 4 vectors are pointing in same direction. This implies $$x : y : z : t = |\vec{a}| : |\vec{b}| : |\vec{c}| : |\vec{d}| = 1 : 2 : 3 : 4 \\ \implies (x,y,z,t) = \left(\frac{3}{5}, \frac{6}{5}, \frac{9}{5}, \frac{12}{5}\right) $$

Answer 2

Hint: Think in a plane coordinates, $x,y,z,t$ can be placed consecutively on the $x$-axis, while $\sqrt{1-x^2},\dots$ can be placed consecutively on the $y$-axis.

Answer 3

By AM-GM $$8=\sqrt{(1-x)(1+x)}+\sqrt{(2-y)(2+y)}+\sqrt{(3-z)(3+z)}+\sqrt{(4-t)(4+t)}=$$ $$=2\sqrt{(1-x)\cdot\frac{1+x}{4}}+2\sqrt{(2-y)\cdot\frac{2+y}{4}}+2\sqrt{(3-z)\cdot\frac{3+z}{4}}+2\sqrt{(4-t)\cdot\frac{4+t}{4}}\leq$$ $$\leq1-x+\frac{1+x}{4}+2-y+\frac{2+y}{4}+3-z+\frac{3+z}{4}+4-t+\frac{4+t}{4}=8,$$ which gives $$1-x=\frac{1+x}{4},$$ $$2-y=\frac{2+y}{4},$$ $$3-z=\frac{3+z}{4}$$ and $$4-t=\frac{4+t}{4}$$ or $$(x,y,z,t)=\left(\frac{3}{5},\frac{6}{5},\frac{9}{5},\frac{12}{5}\right).$$