Passing through point P makes PH perpendicular to X axis, and X axis is at point H. Let CN pass through point C perpendicular to point N of Y axis, and DQ pass through point D perpendicular to point CN and point Q.

When p moves on the CD, PC=t-20,

It is easy to prove that a triangle PMC is similar to a triangle DQC if PM is perpendicular to NC and intersects with point M.

t-20/ 10=MC/6

MC=4/5t- 16

Because PO=PQ, OH=QH 1/2 OQ.

Then oh = qh =1/2oq = Mn = cn-cm =14/5t+16 = 30-4/5t.

So 2 * (30-4/5t) = OQ =1+t.

t=59/( 13/5)=295/ 13

In fact, when P is calculated on AB and BC, it can be calculated as 5/3 on AB, but it doesn't meet the meaning of the question on BC, so I gave up.

The method is not necessarily simple. Please tell us if there is a simple method.