^{An outdated query, bringing again to life..}

Here is a solution:

which was principally discovered by trial-and-error technique together with a little bit of logic.

Explanation

I’ll typically seek advice from columns as A,B,C,.. (from left to proper), and rows as 1,2,3,.. (from high to backside), to specify corresponding cells.

The start line is clearly…

the bottom-left nook. Since 4 phrases overlap there, one phrase ought to comprise one other phrase inside it. There are three such phrases: TEEM, TEAM and TEAR these comprise TEE and TEA. The right phrase amongst these can’t lie horizontally or vertically since each A6 and B7 has the worth 1. Therefore it should lie alongside the diagonal. Thus A7 and B6 have to be T and E respectively.

As center cell might be both M or R, the phrase alongside the remaining cells of the diagonal ought to finish with these letters i.e. a type of three phrases aforementioned, or might be a 3-letter phrase if it doesn’t comprise the center cell; i.e. TEA or TEE. In each instances, E is the second letter (or center letter) thus F2 ought to be E.

Those are the letters we will deduce logically.

(This half was already talked about in Bob’s reply. I’m repeating it right here only for the completeness.)

From now on, there isn’t a apparent logical path.
So we’ve to take a threat. Guess!

Assume that the center cell is R. Then the phrase mendacity on the diagonal, ranging from A7 ought to be TEAR (C5 can’t be E, in any other case it’ll give just one phrase). Considering the principles given by OP, one phrase is counted solely as soon as, due to this fact the opposite phrase alongside the diagonal ought to be TEE (it doesn’t share the R).

Since we agreed to depend every phrase solely as soon as, we won’t use the identical phrase repeatedly deliberately. Thus the remaining cells ought to be stuffed with phrases these are a minimum of 4-letters lengthy. Keeping that in thoughts, to suit the density of purple cells on the diagonal, they need to be crammed as within the following diagram.

Then,

there are solely two attainable methods to do away with the center cell (proven in purple). But there is just one 4-letter phrase ending with R and it is already used.

Thus we’ve to surrender on this route. 🙁

Now think about the next grid.

B6 and D4 have glad the situation already, due to this fact I’ll name them locked cells. In order to proceed there ought to be phrases horizontally and vertically, ranging from A7. Take a take a look at D7. There ought to be two phrases by it. Since 3-letter phrase has been already used, there can’t be phrases alongside the dotted strains (horizontal/vertical). If there’s a phrase alongside the purple line, we won’t fill the cells beneath it. Hence the opposite phrase by D7 ought to lie alongside the gray arrow (route might be modified).

Now let’s attempt to fill the row7.

Starting with the utmost: not one of the 6-letter lengthy phrases (THEYRE, TACOkayED, TAUGHT) appears to suit, as a result of the letter which involves the orange cell will likely be solely used as soon as within the grid, violating the factors of that cell. (TAUGHT, written backward, additionally doesn’t match as a result of there isn’t a different phrase ending with U). And equally 5-letter lengthy phrases additionally don’t go well with. Thus it ought to be a 4-letter phrase. We must search for the phrases ending with identical letter. The attainable candidates are TACT, TAUT, TALE and TARE. We have to decide on just one from TAUT or TACT, trigger if we select each, there ought to be one other phrase beginning and ending with T, however we have no different phrase with that property. So I’ll go together with TAUT, TALE and TARE (This is non-obligatory).

Now think about C5.

Only one phrase is left to undergo it. We cannot select the horizontal or vertical paths (marked in dotted strains under) as a result of we’ve to order house for the longer phrases. Therefore that phrase ought to lie alongside the diagonal (line marked in black). Because of ‘A’, the one attainable candidates are TAIL and TEAR. But that is the one likelihood to satisfy the criterion of A3. Hence it ought to be TEAR.

Then,

…there have to be a phrase alongside the black line, since dotted strains aren’t attainable paths. And to fill B3, the phrase ought to be learn diagonally. Now, holding in thoughts that there are for much longer phrases these have not been used but, we will fill these cells.

Finally, simply by inspection, we will fill the remainder of grid

Voila!

Thanks for studying up up to now!

Addendum

I’ve to inform you one thing.

Did you discover that we may flip or change the place of some phrases to get different attainable solutions?

Apart from that, did you discover these two further phrases?

Ahhh…. this answer isn’t right!

Well, OP acknowledged that he counted one phrase solely as soon as, so in accordance with the given guidelines this answer is suitable.

But,

I imposed some constraints to get a novel answer.

1. Every letter within the grid ought to be used for phrases a minimum of as soon as.
2. If the identical phrase seems once more, it’s important to depend that additionally, as an example, right here it’s important to depend TEA in each TEAM and TEAR.
3. One phrase can seem greater than as soon as.

These constraints don’t make the puzzle tough, however make it simple.

Now let’s get again to the unique puzzle.

At the start, we will deduce three letters within the grid logically as we have achieved earlier. And there’s one other trace. Consider the phrases TEAM, TEEM and TEAR. These are the one phrases that accommodates one other phrase in it. So lots of these letters are counted twice, which means: they need to lie inside the caged areas under.

TEAM and TEEM ought to lie alongside the diagonal as they share the ending letter. T of TEAR is counted twice, due to this fact it have to be positioned in both of A2 or A4. But if it must be in A2, we’ve to fill the 2-letters beneath it with a 2-letter phrase, which is unimaginable. Hence we’ve the beginning cells accomplished as under. (There are two methods for the diagonal to be crammed)

There’s yet one more phrase to be positioned in A5-A7. The attainable options are TEA and TEE solely. (I’ll go together with one answer, hopefully it won’t have an effect on the following steps)

Now think about the attainable placement of the longest phrases.

If we enable the diagonal to be a 6-letter phrase, we will place solely one of many different two phrases.

If we place two of them within the 1st row and final row, there isn’t a house left for the opposite phrase.

If we repair one of many 6-letter phrases within the seventh row, there are three attainable methods to fill the the opposite two phrases.

In all three instances we will be unable to fill the the grid with remaining 4-letter phrases. (Note that letters can’t overlap on yellow cells).

Similarly, we will examine few different instances equivalent to under,

…solely to guarantee that the house isn’t sufficient for the remaining phrases.

Hence we will see,

sadly, there isn’t any answer for this puzzle. 🙁