Why can't a method be defined both for a struct and its pointer?

It can. Just define it on the struct and not the pointer. It will resolve both ways

Method Sets

The method set of the corresponding pointer type *T is the set of all methods with receiver *T or T (that is, it also contains the method set of T)

Try live: http://play.golang.org/p/PsNUerVyqp

package main

import (
    "fmt"
    "math"
    )

type Abser interface {
    Abs() float64
}

type Vertex struct {
    X, Y float64
}

func (v Vertex) Abs() float64 {
    return math.Sqrt(v.X*v.X + v.Y*v.Y)
}

func main() {
    v := Vertex{5, 10}
    v_ptr := &v
    fmt.Println(v.Abs())
    fmt.Println(v_ptr.Abs())
}

Update: As per comments I have created an extra example that actually makes use of the Abser interface to illustrate that both the value and the pointer satisfy the interface.

https://play.golang.org/p/Mls0d7_l4_t

While considering for example:

type T U

func (t *T) M() int { return 1 }

var t T

...we can now invoke M() on t by writing t.M() as the language permits to call a method with a pointer receiver even on its underlying (non pointer) typed instances, i.e. it becomes equivalent to (&t).M().

If it will be permitted to now additionaly define:

func (t T) M() int { return 2 }

...then there's no way to tell what is now t.M() supposed to return.